Олимпиадад бэлтгэгчдэд тусламж (Тэнцэтгэл биш 1)
TRANSCRIPT
Inequalities
Methods and
Olympiad Problems
Contents
1 Part I 3
1.1 Squares are positive . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Some special inequalities and identities for two numbers . . . . 10
1.3 Some special inequalities and identities for more numbers . . . 13
1.4 The mathematical induction . . . . . . . . . . . . . . . . . . . . 17
1.5 The AM-GM inequality . . . . . . . . . . . . . . . . . . . . . . 25
1.6 The quadratic trinomial . . . . . . . . . . . . . . . . . . . . . . 29
1.7 Cauchy-Schwartz Inequality . . . . . . . . . . . . . . . . . . . . 41
1.8 Young Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 48
1.9 Advanced techniques with Cauchy-Buniakowski-Schwarz andHolder Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 50
1.10 The principle of extremality and monotonicity . . . . . . . . . . 84
1.11 Breaking the inequality . . . . . . . . . . . . . . . . . . . . . . 86
1.12 Separating the squares . . . . . . . . . . . . . . . . . . . . . . . 92
1.13 The Dual Principle . . . . . . . . . . . . . . . . . . . . . . . . . 101
1.14 Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
1.15 Homogenization and dehomogenization . . . . . . . . . . . . . . 109
1.16 Unimonotonic sequences . . . . . . . . . . . . . . . . . . . . . . 111
1.17 Working backwards . . . . . . . . . . . . . . . . . . . . . . . . . 128
1.18 Mixing variables . . . . . . . . . . . . . . . . . . . . . . . . . . 129
1.19 Limits in inequalities . . . . . . . . . . . . . . . . . . . . . . . . 133
1.20 Derivatives in Inequalities . . . . . . . . . . . . . . . . . . . . . 135
1.21 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
1.22 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 141
1.23 The shrinking principle and Karamata’s Inequality . . . . . . . 144
1.24 Schur’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 148
1.25 The generalized Means . . . . . . . . . . . . . . . . . . . . . . . 149
1.26 Inequalities between the symmetric sums . . . . . . . . . . . . . 150
1.27 The pqr technique . . . . . . . . . . . . . . . . . . . . . . . . . 151
1.28 The tangent line technique and its extensions . . . . . . . . . . 159
1.29 Using identities to prove inequalities . . . . . . . . . . . . . . . 178
2 Problems 185
3 Solutions 225
1
4 Anexa 1 4894.1 Metoda compararii . . . . . . . . . . . . . . . . . . . . . . . . . 4894.2 Metoda majorarii si minorarii . . . . . . . . . . . . . . . . . . . 4994.3 Metoda coeficientilor nedeterminati . . . . . . . . . . . . . . . . 5044.4 Metoda normalizarii . . . . . . . . . . . . . . . . . . . . . . . . 5104.5 Metoda omogenizarii . . . . . . . . . . . . . . . . . . . . . . . . 515
2
Chapter 1
Part I
1.1 Squares are positive
Perhaps the first inequality one learns is ”The square of any real number isnonnegative”. It’s a very simple one, but despite its simplicity it is one ofthe most important inequalities, used frequently by beginners and advancedproblem solvers alike. It lies at the base of virtually every inequality: forexample, the inequality a+b
2 ≥√ab can be rewritten as 1
2(√a−√b)2 ≥ 0. But
writing an expression as a square, or as a sum of squares, is usually far fromobvious. It requires a certain level of intuition and creativity, but perhaps moreimportantly, experience. We thus start with some introductory problems.
Example 1. If a, b, c are arbitrary real numbers, then a2+b2+c2 ≥ ab+bc+ca.
Solution: Let us first see what we have here: squares of numbers versus theirpairwise products. Next, we ask ourselves: ”Have I ever seen things like thesebefore?”. Well, yes. Namely, we’ve just seen that (x − y)2 = x2 − 2xy +y2, where x, y are two (randomly chosen) real numbers. Motivated by thisidentity, we multiply by two our inequality, and see that it is equivalent with2(a2 + b2 + c2) ≥ 2(ab+ bc+ ca). We now transfer everything to the left side.In this case,
2a2 + b2 + c2 − 2(ab+ bc+ ca) =
= (a2 − 2ab+ b2) + (b2 − 2bc+ c2) + (c2 − 2ca+ a2)
=[(a− b)2 + (b− c)2 + (c− a)2
],
which is clearly nonnegative, and thus our inequality is proven.
Example 2. If a, b are real numbers such that a+ b = 2, then a4 + b4 ≥ 2.
Solution: Since 2(a4 + b4)− (a2 + b2)2 = a4 − 2a2b2 + b4 = (a2 − b2)2 ≥ 0, we
get that a4 + b4 ≥ (a2 + b2)2
2. Similarly, 2(a2 + b2)− (a+ b)2 = a2−2ab+ b2 =
(a−b)2 ≥ 0, and therefore a2 +b2 ≥ (a+ b)2
2= 2.Thus a4 +b4 ≥ (a2 + b2)2
2≥
22
2 = 2.
3
Example 3. Let a, b, c > 0. Prove that a3 + b3 + c3 + ab2 + bc2 + ca2 ≥
2(a2b+ b2c+ c2a).Solution: Note that a3 +ab2−2a2b = a(a2−2ab+b2)2 = a(a−b)2 ≥ 0. In thiscase, we analogously establish that b3+bc2−2b2c ≥ 0 and c3+ca2−2c2a ≥ 0.Summing up all these three inequalities yields our result.
Example 4. Let a, b, x, y be arbitrary real numbers. Show that (a2 + b2)(x2 +y2) ≥ (ax+ by)2.
Solution:Transferring the terms to the left side, the inequality rewrites as(a2 + b2)(x2 + y2)− (ax+ by)2 ≥ 0. Denoting the quantity from the left handside by P , we get
P = (a2 + b2)(x2 + y2)− (ax+ by)2
= a2x2 + a2y2 + b2x2 + b2y2 − (a2x2 + b2y2 + 2abxy)
= a2y2 + b2x2 − 2abxy = (ay − bx)2 ≥ 0.
Remarks. Notice that this inequality is a particular case of the following:
(a21 + a22 + . . .+ a2n)(b21 + b22 + . . .+ b2n) ≥ (a1b1 + a2b2 + . . .+ anbn)2,
where a1, a2, . . . , an and b1, b2, . . . , bn are arbitrary real numbers. Thislast one is known in literature as the Cauchy-Schwartz inequality, to which wehave dedicated the next chapter. Anyway, before this, let us see some other(. . .)2 ≥ 0 problems.
Exercises
1. Show that a2 + b2 + c2 + 3 ≥ 2(a+ b+ c).
Solution: Moving everything to the left side, we get
a2 + b2 + c2 + 3− 2(a+ b+ c) = (a2 − 2a+ 1) + (b2 − 2b+ 1) + (c2 − 2c+ 1)
= (a− 1)2 + (b− 1)2 + (c− 1)2 ≥ 0,
which proves our inequality.
2. If x, y are real numbers, show that x2 + y2 ≥ (x−y)22
Solution: This is equivalent to x2+y2+2xy2 ≥ 0 i.e. (x+y)2
2 ≥ 0.
3. Show that x4 + y4 + z2 + 1 ≥ 2x(xy2 − x+ z + 1).
Solution: Moving everything to the left side, and denoting by P thequantity from the left hand side, we get
P = x4 + y4 + z2 + 1− 2x(xy2 − x+ z + 1)
= (x4 − 2x2y2 + y4) + (x2 − 2xz + z2) + (x2 − 2x+ 1)
= (x2 − y2)2 + (x− z)2 + (x− 1)2 ≥ 0.
4
4. Show that for any real x, we have (x− 1)(x− 3)(x− 4)(x− 6) + 10 > 0.
Solution: Factoring our expression, we obtain that
(x− 1)(x− 3)(x− 4)(x− 6) + 10 = (x2 − 7x+ 6)(x2 − 7x+ 12) + 10
= (x2 − 7x+ 6)2 + 6(x2 − 7x+ 6) + 10
= (x2 − 7x+ 9)2 + 1 > 0.
5. If a and b are strictly (nonzero) positive real numbers, prove thata√b
+
b√a≥√a+√b.
Solution: Moving everything to the left side, it follows that
a√b
+b√a−√a−√b =
(a√b−√b
)+
(b√a−√a
)=
a− b√b
+b− a√a
= (a− b)(
1√b− 1√
a
)
=
(√a−√b)2 (√
a+√b)
√ab
≥ 0.
6. Show that for any positive real number x, we have x5− x2− 3x+ 5 > 0.
Solution: Note that the following chain of equalities holds:
x5 − x2 − 3x+ 5 = x2(x3 − 1)− 3(x− 1) + 2
= (x− 1)[x2(x2 + x+ 1)− 3
]+ 2
= (x− 1)[(x4 − 1) + (x3 − 1) + (x2 − 1)
]+ 2
= (x− 1)2(x3 + 2x2 + 3x+ 3) + 2.
Since this last term is nonnegative, we deduce our inequality.
7. If x, y are two real numbers, prove that x4 + y4 ≥ x3y + y3x.
Solution: Since x2 + xy + y2 = 14(x− y)2 + 3
4(x+ y)2 ≥ 0, we have that
x4 + y4 − x3y − y3x = (x4 − x3y) + (y4 − y3x) = x3(x− y) + y3(y − x)
= (x− y)(x3 − y3) = (x− y)2(x2 + xy + y2) ≥ 0.
8. Let a, b, c,m, n, p be real numbers (a, b, c 6= 0), such that ap−2bn+cm =0 and ac = b2. Show that n2 ≥ mp.
Solution: For any two real numbers x and y, we have that (x+y)2−4xy =x2 − 2xy + y2 = (x − y)2 ≥ 0, or equivalently, (x + y)2 ≥ 4xy. In thiscase,
4b2n2 = (ap+ cm)2 ≥ 4ap · cm = 4acpm = 4b2mp,
and thus n2 ≥ mp.
5
9. Show that a2 + b2 + c2 + d2 + e2 ≥ a(b+ c+ d+ e).
Solution: Moving everything to the left side, it follows that
A = a2 + b2 + c2 + d2 + e2 − a(b+ c+ d+ e)
=
(a2
4− ab+ b2
)+
(a2
4− ac+ c2
)+
(a2
4− ad+ d2
)+
(a2
4− ae+ e2
)=
(a2− b)2
+(a
2− c)2
+(a
2− d)2
+(a
2− e)2≥ 0.
10. If a, b, c are nonnegative real numbers, show that
a3
a2 + ab+ b2+
b3
b2 + bc+ c2+
c3
c2 + ca+ a2≥ a+ b+ c
3.
Solution: Sincea3
a2 + ab+ b2− 2a− b
3=
(a− b)2(a+ b)
3(a2 + ab+ b2)≥ 0, we get
a3
a2 + ab+ b2+
b3
b2 + bc+ c2+
c3
c2 + ca+ a2≥ 2a− b
3+
2b− c3
+2c− a
3
=a+ b+ c
3.
11. Let a, b, c be three arbitrary real numbers. Prove that√a2 + ab+ b2 +
√b2 + bc+ c2 +
√c2 + ca+ a2 ≥
√3(a+ b+ c).
Solution: Since a2 + ab + b2 = 14(a − b)2 + 3
4(a + b)2 ≥ 34(a + b)2, we
deduce that√a2 + ab+ b2+
√b2 + bc+ c2+
√c2 + ca+ a2 ≥
√3
2(|a+ b|+ |b+ c|+ |c+ a|) .
Now, for any real number x we have |x| ≥ x, and therefore
√3
2(|a+ b|+ |b+ c|+ |c+ a|) ≥
√3
2(a+ b+ b+ c+ c+ a)
=√
3(a+ b+ c),
which yields our result.
12. Prove that 19x2 + 54y2 + 16z2 + 36xy − 24yz − 16zx ≥ 0.
Solution: The inequality follows from the chain of equalities writtenbelow:
P = 19x2 + 54y2 + 16z2 + 36xy − 24yz − 16zx
= (9x2 + 36xy + 36y2) + (18y2 − 24yz + 8z2) + (8x2 − 16xz + 8z2) + 2x2
= 9(x+ 2y)2 + 2(3y − 2z)2 + 8(x− z)2 + 2x2 ≥ 0.
6
13. Let x, y be real numbers such that xy ≥ 1. Show that1
x2 + 1+
1
y2 + 1≥
2
xy + 1.
Solution: Note that
1
x2 + 1+
1
y2 + 1− 2
xy + 1=
(1
x2 + 1− 1
xy + 1
)+
(1
y2 + 1− 1
xy + 1
)=
x(y − x)
(xy + 1)(x2 + 1)+
y(x− y)
(xy + 1)(y2 + 1)
=x− yxy + 1
(y
y2 + 1− x
x2 + 1
)=
(x− y)2(xy − 1)
(xy + 1)(x2 + 1)(y2 + 1),
which is clearly nonnegative. This proves the inequality in question.
14. If a, b, c are positive real numbers, prove thata
b+ c+
b
c+ a+
c
a+ b≥ 3
2 .
Solution: We have
P =a
b+ c+
b
c+ a+
c
a+ b− 3
2
=
(a
b+ c− 1
2
)+
(b
c+ a− 1
2
)+
(c
a+ b− 1
2
)=
(a− b) + (a− c)2(b+ c)
+(b− c) + (b− a)
2(c+ a)+
(c− a) + (c− b)2(a+ b)
=a− b
2
(1
b+ c− 1
c+ a
)+b− c
2
(1
c+ a− 1
a+ b
)+c− a
2
(1
a+ b− 1
b+ c
)=
(a− b)2
2(b+ c)(a+ c)+
(b− c)2
2(a+ b)(a+ c)+
(c− a)2
2(a+ b)(b+ c),
which is clearly nonnegative.
15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Provethat √
a+(b− c)2
4+
√b+
(c− a)2
4+
√c+
(a− b)24
≤ 2.
Solution: Squaring both sides, the inequality to prove rewrites as
∑cyc
(a+
(b− c)2
4
)+ 2∑cyc
√(b+
(c− a)2
4
)(c+
(a− b)24
)≤ 4.
Now, notice that (b− c)2(1 + a− b− c) = 2a(b− c)2 ≥ 0, which gives usthat
2
√(b+
(c− a)2
4
)(c+
(a− b)24
)≤ b+ c+
(a− b)(a− c)2
.
7
Thus it suffices to show that∑cyc
(a+
(b− c)2
4
)+∑cyc
(b+ c+
(a− b)(a− c)2
)≤ 4,
which since∑cyc
(a − b)(a − c) = a2 + b2 + c2 − ab − bc − ca, reduces to
3(a+ b+ c) + a2 + b2 + c2 − ab− bc− ca ≤ 4, or equivalently, a2 + b2 +c2 − ab− bc− ca ≤ 1. This is obviously true, because
1 = (a+ b+ c)2 = (a2 + b2 + c2 − ab− bc− ca) + 3(ab+ bc+ ca)
≥ a2 + b2 + c2 − ab− bc− ca.
16. Let x, y be two strictly (nonzero) positive real numbers such that x2 +y3 ≥ x3 + y4. Prove that x3 + y3 ≤ x2 + y2 ≤ x+ y ≤ 2.
Solution: Since y2 − y3 − (y3 − y4) = y2(y − 1)2 ≥ 0, we have that
x2 − x3 + y2 − y3 ≥ x2 − x3 + y3 − y4 = (x2 + y3)− (x3 + y4) ≥ 0,
and therefore x3 + y3 ≤ x2 + y2. Now, since x − x2 − (x2 − x3) =x(x − 1)2 ≥ 0, and y − y2 − (y3 − y4) = y(y − 1)2(y + 1) ≥ 0, weget x2 + y2 ≤ x + y, and thus the problem reduces to proving thatx + y ≤ 2. Now, note that x − (1 − x2 + x3) = −(x − 1)2(x + 1) ≤ 0,y − (1− y3 + y4) = −(y − 1)2(y2 + y + 1) ≤ 0, and therefore, it followsthat
x+ y ≤ 1− x2 + x3 + 1− y3 + y4 = 2 + (x3 + y4)− (x2 + y3) ≤ 2.
17. Let a, b, c > 0. Show that ab
(b
c− 1
)+ bc
( ca− 1)
+ ca(ab− 1)≥ 0.
Solution: Moving everything to the left side, it follows that
P = ab
(b
c− 1
)+ bc
( ca− 1)
+ ca(ab− 1)
=ab2
c+bc2
a+ca2
b− ab− bc− ca
=
(ab2
c+ ca− 2ab
)+
(bc2
a+ ab− 2bc
)+
(ca2
b+ bc− 2ca
)=
a(b− c)2
c+b(c− a)2
a+c(a− b)2
b≥ 0.
18. Consider three positive real numbers a, b, c such that a > c and b > c.Prove that √
c(a− c) +√c(b− c) ≤
√ab.
Solution: According to Example 4, we get[√c(a− c) +
√c(b− c)
]2=
(√c ·√a− c+
√b− c ·
√c)2
≤ (c+ b− c) (a− c+ c) = ab.
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19. Let a, b, c be nonnegative real numbers, from which at least two arenonzero. Prove that√
a(b+ c)
b2 + c2+
√b(c+ a)
c2 + a2+
√c(a+ b)
a2 + b2≥ 2.
Solution: Notice that for any two nonnegative real numbers x, y, we havex+ y− 2
√xy =
(√x−√y
)2 ≥ 0, and thus x+ y ≥ 2√xy. Now, assume
that a ≥ b ≥ c; then√a(b+ c)
b2 + c2≥
√a(b+ c)
b2 + bc=
√a
b,√
b(c+ a)
c2 + a2≥
√b(c+ a)
ca+ a2=
√b
a,
and therefore√a(b+ c)
b2 + c2+
√b(c+ a)
c2 + a2+
√c(a+ b)
a2 + b2≥√a
b+
√b
a≥ 2
√√a
b·√b
a= 2.
20. Let a, b, c be nonnegative real numbers, from which at least two arenonzero. Prove that√
a(b+ c)
a2 + bc+
√b(c+ a)
b2 + ca+
√c(a+ b)
c2 + ab≥ 2.
Solution: Assume that a ≥ b ≥ c. Since (x+y)2 = x2+y2+2xy ≥ x2+y2,for all x, y ≥ 0, we have√
a(b+ c)
a2 + bc+
√c(a+ b)
c2 + ab≥√a(b+ c)
a2 + bc+c(a+ b)
c2 + ab,
and we will now show thata(b+ c)
a2 + bc+c(a+ b)
c2 + ab≥ b2 + ca
b(c+ a), or equiv-
alently,c(a+ b)
c2 + ab≥ b2 + ca
b(c+ a)− a(b+ c)
a2 + bc. This rewrites as
c(a+ b)
c2 + ab≥
c(a+ b)(a− b)2
b(c+ a)(a2 + bc), or b(c+ a)(a2 + bc) ≥ (a− b)2(ab+ c2) which is obvi-
ously true since b(c+a) = ab+ bc ≥ ab+ c2, and a2 + bc ≥ a2 ≥ (a− b)2.Hence √
a(b+ c)
a2 + bc+
√c(a+ b)
c2 + ab≥
√b2 + ca
b(c+ a),
and therefore the problem reduces to
√b(c+ a)
b2 + ca+
√b2 + ca
b(c+ a)≥ 2. In
the proof of Exercise 19, we have showed that x + y ≥ 2√xy for any
nonnegative real numbers x, y. Therefore√b(c+ a)
b2 + ca+
√b2 + ca
b(c+ a)≥ 2
√√√√√b(c+ a)
b2 + ca·
√b2 + ca
b(c+ a)= 2,
which completes our proof.
9
1.2 Some special inequalities and identities for twonumbers
We have already met the arithmetic mean of two numbers:
AM =a+ b
2
and the geometric mean of two numbers:
GM =√ab
We can also consider the harmonic mean
HM =2
1
a+
1
b
=2ab
a+ b
and the quadratic mean of two numbers
QM =
√a2 + b2
2
For positive a, b, these means satisfy the following chain of inequalities:
QM ≥ AM ≥ GM ≥ HM.
Proof : From the transitivity of ≥ it suffices to prove
QM ≥ AM, AM ≥ GM, GM ≥ HM.
Let’s take them step by step:
a) QM ≥ AM . This is equivalent to
QM2 ≥ AM2 ora2 + b2
2≥(a+ b
2
)2
=a2 + 2ab+ b2
4.
Multiplying by 4 we must prove 2(a2+b2) ≥ a2+b2+2ab or a2+b2 ≥ 2abwhich is just the inequality AM ≥ GM for a2, b2.
b) AM ≥ GM was proven by us in the previous topic.
c) GM ≥ HM . This is equivalent to1
GM≤ 1
HM, or to
1√ab≤
1
a+
1
b2
which is just AM ≥ GM for1
a,1
b.
[Remark: This chain of inequalities can be generalized in the following form:Let a1, a2, . . . , an be positive real numbers, and for every k ∈ R consider thegeneralized means
mk =k
√ak1 + ak2 + . . .+ akn
n
10
if k 6= 0 andm0 = n
√a1a2 . . . an
Then mk is increasing as a function in k: that is
mα ≥ mβ
if α ≥ β
Our inequality chain QM ≥ AM ≥ GM ≥ HM is just a special case of thisinequality for n− 2 and k = 3, 2, 1, 0.The proof of this inequality requires more advanced methods, and we willreturn to it in the further sections.]
We now move to identities. There is one basic identity for two numbers: theNewton Binomial Theorem:
(a+ b)n =
n∑i=0
(n
i
)aibn−i
where(ni
)=
n!
i!(n− i)!(recall that n! = 1× 2× . . .× n).
Particularly(a+ b)2 = a2 + 2ab+ b2
(a+ b)3 = a3 + 3a2b+ 3ab2 + b3
(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4
(a− b)2 = a2 − 2ab+ b2
(a− b)3 = a3 − 3a2b+ 3ab2 − b3(a− b)4 = a4 − 4a3b+ 6a2b2 − 4ab3 + b4
etc.Another well-known identity is
an − bn = (a− b)(an−1 + an−2b+ . . .+ bn−2a+ bn−1),
and for n odd,
an + bn = (a+ b)(an−1 − an−2b+ . . .− abn−2 + bn−1).
Note: the above identities are often used for b = 1. This is because ho-mogeneous inequalities in two numbers a, b can be reduced to inequalities injust one variable x by letting x = a
b (but more on that in the section abouthomogeneity).Finally, a very simple but sometimes crucial inequality is the triangle inequal-ity
|x|+ |y| ≥ |x+ y|.
Exercises
1. Let a > b > 0. Show that 2(a2 − b2)2 ≤ a2(a− b)2 + a2(a− b)2
Proof: This is equivalent to 4(a−b)2(a+b)24a2
≤ (a−b)2+(a−b)22 which is just the
inequality QM ≥ HM squared for the numbers a+ b, a− b. Of course,
11
this inequality can also be shown by opening the brackets and usingthe AM-GM inequality or turning it into a sum of squares. That’s notsurprising since all the inequalities AM ≥ GM,GM ≥ HM,QM ≥ AMwere proven using the same method, which is essentially that the squareof the difference of two numbers is non-negative.
2. If m,n are natural numbers and x > 0 thenxmn − 1
m≥ xn − 1
x.
Solution: We can extract the common factor xn − 1 to get
xmn − 1
m− xn − 1
x=
1
nx(xn − 1)
[x(xn(m−1) + . . .+ 1)−m
]=
1
nx(x− 1)(xn−1 + . . .+ 1)
[x(xn(m−1) + . . .+ 1)−m
].
Now as the right parenthesis vanishes for x = 1, extracting x− 1 out ofit produces
1
nx(xn − 1)(xn(m−1) − 1 + xn(m−2) − 1 + . . .+ 1− 1) =
=1
nx(x− 1)2(xn−1 + . . .+ 1)
[(xn(m−1) + . . .+ 1) + (xn(m−2) + . . .+ 1) + . . .
],
and this is clearly positive.
3. For a1, a2, a3, . . . , an ∈ R show that
min1≤i<k≤n
(ak − ai)2 ≤12
n(n2 − 1)
(n∑k=1
a2k
).
Solution: Suppose min{|ak − ai|} = a and a1 < a2 < . . . < an. Then,we clearly have |al − ak| = |al − al−1|+ . . .+ |ak+1 − ak| ≥ |l − k|a.
Next we use the inequality x2 + y2 ≥ (x− y)2
2- proven in the exercises
of the previous section. So we have a2i +a2n−i+1 ≥(n−2i+1)2
2 a2. Summingover all i = 1, 2, . . . , n and dividing by 2 we get the desired result.
4. Let a, b, c be real numbers such that |ax2 + bx+ c| ≤ 1 whenever |x| ≤ 1.Show that |cx2 + bx+ a| ≤ 2 whenever |x| ≤ 1.
Solution: By setting x = 0 we get |c| ≤ 1 and by setting x = ±1 we get|a+ b+ c|, |a− b+ c| ≤ 1. Then as |x| ≤ 1, |x2 − 1| ≤ 1 and then
|cx2 + bx+ a| = |c(x2 − 1) + c+ bx+ a| ≤ |c| · |x2 − 1|+ |c+ bx+ a|.
Since c+ bx+ a is between c+ b+ a and c− b+ a we get
|c+ bx+ a| ≤ max{|c+ b+ a|, |c− b+ a|} ≤ 1
and hence |cx2 + bx+ a| ≤ 1 + 1 = 2 as desired.
12
1.3 Some special inequalities and identities for morenumbers
Naturally there are more, but we shall announce only some of them, mostlyfor three numbers.Let’s take the inequality a2 + b2 ≥ 2ab. We can also deduce b2 + c2 ≥ 2bc, a2 +c2 ≥ 2ac. Summing all this inequalities we deduce 2(a2 + b2 + c2) ≥ 2(ab +bc+ ca) so
a2 + b2 + c2 ≥ ab+ bc+ ca
a very nice inequality in three numbers.Since (a+ b+ c)2 = a2 + b2 + c2 + 2(ab+ bc+ ca) we deduce
3(a2 + b2 + c2) ≥ (a+ b+ c)2 ≥ 3(ab+ bc+ ca).
Now for positive a, b, c, we multiply a2 + b2 + c2 − ab− bc− ca by a+ b+ c toget a3 + b3 + c3 − 3abc = (a+ b+ c)(a2 + b2 + c2 − ab− ca− bc) ≥ 0 so
a3 + b3 + c3 ≥ 3abc−AM-GM inequality for three numbers.
If we let a3 = x, b3 = y, c3 = z we deduce another form for it:
x+ y + z
3≥ 3√xyz or (x+ y + z)3 ≥ 27xyz
Particularly,
a2b ≤ 2a3 + b3
3, b2c ≤ 2b3 + c3
3, c2a ≤ 2c3 + a3
3.
Summinga2b+ b2c+ c2a ≤ a3 + b3 + c3
and analogouslyab2 + bc2 + ca2 ≤ a3 + b3 + c3
.From the other side, by applying the AM-GM inequality we can see
a2b+ b2c+ c2a ≥ 3abc, ab2 + bc2 + ca2 ≥ 3abc.
The expression a3 + b3 + c3 − (a+ b+ c)3 vanishes when two of a, b, c sum tozero, so it’s divisible by (a+ b)(b+ c)(c+ a).We deduce then (by direct computation) the identity
a3 + b3 + c3 − (a+ b+ c)3 = −3(a+ b)(b+ c)(c+ a)
.We wish to compare a2b+ b2c+ c2a with ab2 + bc2 + ca2, and have the identity
a2b+ b2c+ c2a− ab2 − bc2 − ca2 = (a− b)(b− c)(a− c)
which casts a light on the relationship between the two expressions accordingto the relative order of a, b, c.An analogous identity is a3b+ b3c+ c3a− ab3 − bc3 − ca3 = (a− b)(b− c)(a−c)(a+ b+ c).
13
We note that (a+ b)(b+ c)(c+ a) = a2b+ b2a+ b2c+ c2b+ a2c+ c2a+ 2abcand
(a+b+c)(ab+bc+ca) = a2b+b2a+b2c+c2b+3abc = (a+b)(b+c)(c+a)+abc.
Since abc ≤ (a+ b)(b+ c)(c+ a)
8(by multiplying the AM-GM inequality for
the three pairs (a, b), (b, c), (c, a)), we deduce
(a+ b)(b+ c)(c+ a) ≥ 8
9(a+ b+ c)(ab+ bc+ ca).
Consider the product (b+ c−a)(a+ c− b)(a+ b− c). At most one of b+ c−a,a+ c− b, a+ b− c is negative because the sum of any two of the is positive.So, if this product is positive, then they are all positive. we can deduce√
(b+ c− a)(a+ c− b) ≤ (b+ c− a) + (a+ c− b)2
= c.
Multiplying this by the two analogous inequalities we get
(b+ c− a)(a+ c− b)(a+ b− c) ≤ abc.
This is true also when the RHS is negative. Now opening the brackets werewrite the inequality as
a3 + b3 + c3 + 3abc ≥ a2b+ b2a+ b2c+ c2b+ a2c+ c2a
which is the famous Schur’s inequality.
Examples
1. Factorize a4b+ b4c+ c4a− ab4 − bc4 − ca4
Answer : (a− b)(b− c)(a− c)(a2 + b2 + c2 + ab+ bc+ ca).
2. Factorize 2(a5 + b5 + c5)− 5abc(a2 + b2 + c2)
Answer : (a+ b+ c)[2(a2 + b2 + c2)− 5(ab+ bc+ ca)
].
3. Show that a2 + b2 + c2 − ab− bc− ca ≥ 3
4(a− b)2
Solution: We have a2+b2+c2−ab−bc−ca =1
2
[(a− b)2 + (b− c)2 + (c− a)2
]and (b− c)2 + (c− a)2 ≥ 1
2(b− c+ c− a)2 =
(a− b)2
2, and from here the
conclusion follows.
4. Show that if a, b, c are real numbers any two of which do not sum tozero, then
a5 + b5 + c5 − (a+ b+ c)5
a3 + b3 + c3 − (a+ b+ c)3≥ 10
9(a+ b+ c)2.
Solution: We already know that
a3 + b3 + c3 − (a+ b+ c)3 = −3(a+ b)(b+ c)(c+ a),
14
and we establish an analogous identity for degree five:
a5+b5+c5−(a+b+c)5 = 5(a+b)(b+c)(c+a)(a2+b2+c2+ab+bc+ca)
(prove it!). Hence the ratio is
5
3(a2 + b2 + c2 + ab+ bc+ ca) =
5
6(a2 + b2 + c2) +
5
6(a+ b+ c)2
≥ 5
18(a+ b+ c)2 +
5
6(a+ b+ c)2
=10
9(a+ b+ c)2,
as desired.
5. If a is the smallest of the positive reals a, b, c then show that
a3 + b3 + c3 − 3abc ≥ 2
(b+ c
2− a)3
.
Solution: This factorizes as
2(a+ b+ c)[(a− b)2 + (b− c)2 + (c− a)2
]≥ [(b− a) + (c− a)]3 .
If we denote x = b− a, y = c− a then a+ b+ c > x+ y and
(a− b)2 + (b− c)2 + (c− a)2 ≥ x2 + y2
and so the inequality follows from 2(x+ y)(x2 + y2) ≥ (x+ y)3 which bycancelling common terms turns to x3 + y3 ≥ x2y + xy2 which is true.
6. Let x, y, z be positive numbers such that x2 + y2 + z2 = 1. Determine
the minimum value ofxy
z+xz
y+yz
x.
Solution: We havexy
z+xz
y+yz
x=x2y2 + y2z2 + z2x2
xyz.
Now we use the famous inequality (a+b+c)2 ≥ 3(ab+bc+ca) for a = uv,b = vw, c = uw, deducing that (uv + uw + vw)2 ≥ 3(u + v + w)uvw.Applying this for u = x2, v = y2, w = z2, we deduce that
(x2y2 + z2x2 + y2z2)2 ≥ 3x2y2z2(x2 + y2 + z2).
Taking square root and keeping in mind that x2 + y2 + z2 = 1 we have
x2y2 + y2z2 + z2x2 ≥√
3xyz.
The value√
3 can be achieved, when x = y = z =1√3
.
7. Let 0 < x < y < z < t such that x+t = y+z. Show that for any naturalnumber n ≥ 2, xn + tn > yn + zn.
Solution: Let a = y − x = t− z. We rewrite the conclusion as
(x+ a)n − xn < (z + a)n − zn,
15
which by Newton Binomial Theorem is equivalent to
nxn−1a+n(n− 1)
2xn−2a+ . . .+an < nzn−1a+
n(n− 1)
2zn−2a+ . . .+an
which is clearly true from z > x.
8. Let a, b, c be length of the sides of triangle; and let t ≥ 1 be real number.Prove that:
(t+ 1)
(a2
c+b2
a+c2
b
)≥ b2
c+c2
a+a2
b+ t(a+ b+ c).
Solution: It suffices to prove it for t = 1 since
a2
c+b2
a+c2
b≥ a+ b+ c.
This latter inequality can be proven in many ways. A simple way is tonote that a2−c(2a−b) = (a−c)2 ≥ 0 hence c(2a−c) ≤ a2 so a2
c ≥ 2a−cand similarly b2
a ≤ 2b−a, c2b ≥ 2c−b and we sum these three inequalitiesto obtain the desired result.
So it suffices to show 2(a2
c + b2
a + c2
b ) ≥ c2
b + b2
c + c2
a + a + b + c. Wemultiply this latter inequality by abc to turn it into the equivalent form2(a3b+b3c+c3a) ≥ ab3+bc3+ca3+abc(a+b+c). This can be rewrittenas ab3 + bc3 + ca3− a3b− b3c− c3a ≤ a3b+ b3c+ c3a− abc(a+ b+ c) i.e.
(a− b)(b− c)(c− a)(a+ b+ c) ≤ a3b+ b3c+ c3a− abc(a+ b+ c).
The right-hand side is non-negative since we’ve just shown that a2
c + b2
a +c2
b ≥ a+b+c, thus is suffices to consider the case (a−b)(b−c)(c−a) > 0.
In this case we can suppose a < b < c (check!) and so the RHS is
b(b2 − a2)(c− a) + a(c− b)(c2 − b2)
By the AM-GM inequality this is greater than or equal to
2(c− b)√ab(c− b)(b− a)(c+ b)(b+ a)
and so by squaring both sides and canceling reducing the common factorsit remains to prove that
4ab(c+ b)(b+ a) ≥ (c− b)(b− a)(a+ b+ c)2
Indeed, a ≥ c− b, b ≥ b− a, 4(c+ b)(b+ a)− (a+ b+ c)2 = 4c(b+ a)−(a + b + c)2 + 4b(b + a) = 4b(b + a) − (a + b − c)2 ≥ 4b(b + a) − a2 > 0and by multiplying we get the result.
16
1.4 The mathematical induction
The Principle of Mathematical Induction is formulated as follows:
Suppose we have to prove an affirmation A(n) depending on n ∈ N.
If we can prove it for some k (the basis), and then we can prove that A(n+ 1)(the induction step) follows from (the induction assumption) A(n) then A(n)is true for any n.
Another variation is to show that A(n+1) follows from A(k), A(k+1), . . . A(n)- this is called ”strong induction” as opposed to the ”weak induction” methodabove. Of course, strong induction become weak induction one we replace thepropositions A(n) with the equivalent set of propositions B(n) that state thatA(k), A(k + 1), . . . , A(n) are all true.
Consider for example Bernoulli’s Inequality
If α ≥ 0 then (1 + α)n ≥ 1 + nα.
For n = 1 this is actually an identity (the basis).
Now if (1 + α)n ≥ 1 + nα (the induction assumption) then
(1+α)n+1 = (1+α)n(1+α) ≥ (1+nα)(1+α) = 1+(n+1)α+α2 ≥ 1+(n+1)α
(The induction step). Hence the affirmation is true for all n.
We may have noted that the inequality a3 + b3 + c3 ≥ 3abc resembles theAM-GM inequality, generalized for three numbers. Indeed, if we put a3 =
x, b3 = y, c3 = z we transform it tox+ y + z
3≥ 3√xyz.
A natural question appears: is this inequality true for any number of variables,
i.e., is it true that for positive x1, x2, . . . , xn we havex1 + x2 + . . .+ xn
n≥
n√x1x2 . . . xn?
We know it’s true for n = 4, and we know how to deduce it from the two-variables case:
a+ b+ c+ d ≥ 2√ab+ 2
√cd ≥ 4
√√ab√cd = 4
4√abcd.
We can prove the inequality for n = 2m the very same way using induction:
x1 + . . .+ x2m = (x1 + . . .+ x2m−1) + (x2m−1+1 + . . .+ x2m) ≥
≥ 2m−1 2m−1√x1x2 . . . x2m−1 + 2m−1 2m−1√x2m−1+1 . . . x2m .
Applying the AM-GM Inequality for two numbers, we deduce this quantity isat greater or equal than 2m 2m
√x1x2 . . . x2m . So, we can prove the inequality for
any power of 2. To pass to the general case, pick up such a m with 2m > n andconsider the system b1, b2, . . . , b2m with bi = ai if i ≤ n and bi = n
√a1a2 . . . an
otherwise. The geometric mean of the system is still n√a1a2 . . . an and so we
deduce
b1 + b2 + . . .+ b2m ≥ 2m n√a1a2 . . . an
so
a1 + a2 + . . .+ an + (2m − n) n√a1a2 . . . an ≥ 2m n
√a1a2 . . . an
17
soa1 + a2 + . . .+ an
n≥ n√a1a2 . . . an.
This inequality is called the general Arithmetic Mean-Geometric Mean (AM-GM) Inequality.
The method we used is called Cauchy Induction, which is based on provingthe inequality for an infinite class of numbers (in our case powers of two) anthen showing that if m < n and for n the proposition is true, then for m italso must be true. In a simpler way, if for n the assertion holds, then it holdsfor n− 1.
The mathematical induction is a tool that can be used in inequalities withmany variables. Sometimes the induction step is quite clear, but it can not beproven, In this case it’s good to sharpen the induction assumption.
Look at the following problem:
1
2· 3
4· . . . · 2n− 1
2n<
1√3n.
The induction seems obvious: we prove the basis and then we show that fromn−1 to n, the LHS increases less than RHS. This is wrong, however: we need to
prove2n− 1
2n<
√n− 1√n
. Squaring this is equivalent to (2n−1)2n < 4n2(n−1)
or 4n3 − 4n2 + n < 4n3 − 4n2, which is false. In this case, sharpening the
assumption saves us: we show that actually1
2· 3
4. . . · 2n− 1
2n≤ 1√
3n+ 1.
The basis still holds, and the inductive step can now be performed, as it is
equivalent to2n− 1
2n≤√
3n− 2√3n+ 1
or (2n − 1)2(3n + 1) ≤ (2n)2(3n − 2) or
12n3 − 12n2 − n− 1 ≤ 12n3 − 8n2, true.
Exercises
1. Show that 2(√n+ 1− 1) ≤ 1 +
1√2
+ . . .+1√n< 2√n.
Solution: The basis is easy to check. The induction step is to prove
2(√n+ 1−
√n) <
1√n< 2(√n−√n− 1).
These inequalities transform to 2[√
n(n+ 1)− n]< 1 < 2
[n−
√n(n− 1)
]or√n(n+ 1) < n+
1
2and
√n(n− 1) < n− 1
2which follow by squaring.
2. Show that xn =1
n+ 1+
1
n+ 2+ . . .+
1
2n<
7
10.
Solution: As xn+1−xn =1
2n+ 1+
1
2n+ 2− 1
n+ 1=
1
(2n+ 1)(2n+ 2)we
cannot apply induction in the most trivial way. So we need to strengthen
18
somehow the condition. We might replace xn by xn +k
n. For yn to be
decreasing we need to check
yn − yn+1 =1
(2n+ 1)(2n+ 2)− k
n(n+ 1)=
(1− 4k)n− 2k
2n(n+ 1)(2n+ 1)< 0,
which holds for k =1
4. For this new sequence, the induction step is
already proven and the basis holds from n = 4.
3. Assume that x1, x2, . . . , xn, y1, y2, . . . , yn are positive numbers with
y1 ≤ y2 ≤ . . . ≤ yn andx1y1≥ x2y2≥ . . . ≥ xn
yn.
Then show that
x1y1
+ . . .+xnyn≥ n · x1 + x2 + . . .+ xn
y1 + y2 + . . .+ yn.
Solution: It’s natural to denote zi =xiyi
.
For n = 2 the inequality isx1y1
+x2y2≥ 2 · x1 + x2
y1 + y2or
x1y2(y1 + y2) + x2y1(y1 + y2) ≥ 2(x1 + x2)y1y2
or x1y22 +x2y
21 ≥ (x1 +x2)(y1y2) or (y2− y1)
(x1y1− x2y2
)y1y2 ≥ 0. Note
that for n = 2 this is a necessary and sufficient condition that y2 − y1and
x1y1− x2y2
have the same sign.
Now let’s perform the induction step. By the induction assumption
x1y1
+ . . .+xnyn≥ x1y1
+ (n− 1) · x2 + . . .+ xny2 + . . .+ yn
.
So we are left to prove that
x1y1
+ (n− 1) · x2 + . . .+ xny2 + . . .+ yn
≥ n · x1 + x2 + . . .+ xny1 + y2 + . . .+ yn
,
which is in turn equivalent to
x1(y2+. . .+yn)(y1+y2+. . .+yn)+(n−1)(x2+. . .+xn)(y1+y2+. . .+yn)y1 ≥
≥ n(x1 + . . .+ xn)y1(y2 + . . .+ yn).
If we denote x =x2 + . . .+ xn
n− 1, y =
y2 + . . .+ ynn− 1
then y1 ≤ y and
x
y≤ x1y1
and the inequality to prove rewrites as
x1(n− 1)y [y1 + (n− 1)y] + (n− 1)(n− 1)x [y1 + (n− 1)y] y1 ≥
≥ n(x1 + (n− 1)x)(n− 1)yy1,
19
or x1y [y1 + (n− 1)y]+(n−1)xy1 [y1 + (n− 1)y] ≥ n [x1 + (n− 1)x] yy1.
By cancelling the common terms we get
−(n− 1)x1y1y + (n− 1)y2x1 + (n− 1)y21x− (n− 1)xyy1 ≥ 0
or (n− 1)(y − y1)(x1y1− x
y
)yy1 ≥ 0, true.
4. Show that1
22+
1
32+ . . .+
1
n2< 1.
Solution: The sequence is increasing so we must sharpen the condition.
1
22+ . . . +
1
n2< 1 − 1
nis good, because
1
(n+ 1)2<
1
n− 1
n+ 1. (Note
that this inequality can give us a direct proof without induction).
5. Show that if x1, x2, . . . , xn ∈(0, 12)
then
x1x2 . . . xn(x1 + x2 + . . .+ xn)n
≤ (1− x1) . . . (1− xn)
((1− x1) + . . .+ (1− xn))n.
Solution: This is an application for Cauchy induction. Firstly, we provethe inequality by induction for n = 2k. If n = 2 we must prove
x1x2(x1 + x2)2
≤ (1− x1)(1− x2)(2− x1 − x2)2
,
or x1x2[4− 4(x1 + x2) + (x1 + x2)
2]≤ (x1+x2)
2 [1− (x1 + x2) + x1x2]which turns to 4(x1 − x2)2(1− x1 − x2) ≥ 0 which is true.
To pass from n to 2n we see that
x1x2 . . . xnxn+1 . . . x2n(x1 + . . .+ xn + xn+1 + . . .+ x2n)2n
=
=x1x2 . . . xn
(x1 + x2 + . . .+ xn)nxn+1 . . . x2n
(xn+1 + . . .+ x2n)n·
· (x1 + . . .+ xn)n(xn+1 + . . .+ x2n)n
(x1 + . . .+ xn + xn+1 + . . .+ x2n)2n
=x1x2 . . . xn
(x1 + x2 + . . .+ xn)nxn+1 . . . x2n
(xn+1 + . . .+ x2n)n·
·
x1 + . . .+ xn
n
xn+1 + . . .+ x2nn(
x1 + . . .+ xnn
+xn+1 + . . .+ x2n
n
)2
n
≤ (1− x1) . . . (1− xn)
[(1− x1) + . . .+ (1− xn)]n(1− xn+1) . . . (1− x2n)
[(1− xn+1) + . . .+ (1− x2n)]n·
·
(
1− x1 + . . .+ xnn
)(1− xn+1 + . . .+ x2n
n
)[(
1− x1 + . . .+ xnn
)+
(1− xn+1 + . . .+ x2n
2n
)]2n
=(1− x1) . . . (1− x2n)
[(1− x1) + . . .+ (1− x2n)]2n.
20
Now the backward step: assume that our assumption holds for n and
let’s prove it for n − 1. To do this, take xn =x1 + . . .+ xn−1
n− 1(in such
exercises it’s convenient to set the ”free” variable equal to some mean ofthe other variables).Then
x1x2 . . . xn−1x1 + . . .+ xn−1
n− 1(x1 + x2 + . . .+ xn−1 +
x1 + . . .+ xn−1n− 1
)n ≤
≤(1− x1)(1− x2) . . . (1− xm−1)(1−
x1 + . . .+ xn−1n− 1
)[(1− x1) + . . .+ (1− xn−1) +
(1− x1 + . . .+ xn−1
n− 1
)]n ,which after reducing the common terms in the numerators and denomi-nators becomes exactly what we wanted.
6. Let ai ≥ 1 and |ai − ai+1| < 1 for 1 ≤ i ≤ i− 1. Show that
a1a2
+a2a3
+ . . .+an−1an
+ana1
< 2n− 1.
Solution: The idea is to use induction on n. To do this, we must deletea variable and show that the LHS decreases by at most 2. However weneed to satisfy the inequality |ai − ai+1| < 1 which can be fulfilled onlyif we remove an−1. Now if we remove an−1 from the sequence, the newsequence a1, a2, . . . , an−2, an consists of n− 1 terms. Moreover the RHShas changed by
an−2an− an−1
an− an−2an−1
=an−2an−1 − a2n−1 − an−2
an−1an
= −2 +an−1an − (an−1 − an−2)(an−1 − an)
anan−1> −2.
so the induction step is fulfilled. For n = 1 the condition clearly holds(with equality).
7. Show that for any n ≥ 4, x1, x2, . . . , xn ∈ R+ we have
(x1 + x2 + . . .+ xn)2 ≥ 4(x1x2 + x2x3 + . . .+ xn−1xn + xnx1).
Solution: For n = 4 this is (x1 + x3 − x2 − x4)2 ≥ 0.
Now let n ≥ 5 and denote
f(x1, x2, . . . , xn) = (x1 +x2 + . . .+xn)2−4(x1x2 + . . .+xn−1xn+xnx1).
Then f(x1 +x2, x3, x4, . . . , xn) = f(x1, x2, . . . , xn) +x3x1 +x2xn−x1x2.Now if x3 ≥ x2 (which we may assume as the inequality is cyclic) we get
f(x1 + x2, x3, x4, . . . , xn) ≥ f(x1, x2, . . . , xn)
and so we are done by induction.
21
8. Let a1, a2, . . . , an ∈ (0, 1) with a1a2 . . . an = An. Show that
1
1 + a1+ . . .+
1
1 + an≤ n
1 +A.
Solution: If n = 2 the inequality is equivalent to1
1 + x2+
1
1 + y2≤
2
1 + xyor by clearing denominators to (2+x2+y2)(1+xy)−2(1+x2)(1+
y2) ≤ 0 or (x− y)2(xy − 1) ≤ 0, which is true. Now set b =anan+1
A.
Then a1a2 . . . an−1b = An so by induction assumption we have
1
1 + a1+ . . .+
1
1 + an−1+
1
1 + b≤ n
1 +A
and we are left to prove that
1
1 + a1+. . .+
1
1 + an−1+
1
1 + b+
1
1 +A≥ 1
1 + a1+. . .+
1
1 + an−1+
1
1 + an+
1
1 + an+1,
or1
1 + an+
1
1 + an+1≤ 1
1 +A+
A
A+ anan+1.
By clearing denominators this reduce to (an−A)(an+1−A)(1−anan+1) ≤0. As an, an+1 < 1 this inequality will be fulfilled iff one of an, an+1 isgreater than A ad the other is smaller (or equal). But we can clearlyassume this as the inequality is cyclic and we can’t have all ai greaterthan A or all ai smaller than A.
9. Let S(n) = 11+22+. . .+nn. Show that1
nn>
1
S(n+ 1)+
1
S(n+ 2)+. . ..
Solution: The quickest idea that comes to our mind is to use ”induction”,
i.e. prove that1
S(n)<
1
nn− 1
(n+ 1)n+1. (Actually this is not induction
but telescoping sums, however it resembles induction). Surprisingly itworks! (on difficult exams, it won’t). We need to prove that
(n+ 1)n+1 − nn
nn × (n+ 1)n+1>
1
11 + 22 + . . .+ nn.
Rewrite this as(n+ 1)n+1 − nn
(n+ 1)n+1· 1 + 22 + . . .+ nn
nn> 1.
To compare nn with (n+1)n+1, we may use
(n+ 1
n
)n< e <
(n+ 1
n
)n+1
.
Therefore, as(n+ 1)n+1
nn= (n+1)
(n+ 1
n
)n= n
(n+ 1
n
)n+1
, we have
ne <(n+ 1)n+1
nn< (n+ 1)e.
So
(n+ 1)n+1 − nn
(n+ 1)n+1·1 + 22 + . . .+ nn
nn>
(1− 1
ne
)[1 +
1
ne+
1
n(n− 1)e2+ . . .
].
22
Since(1− 1
ne
)[1 +
1
ne+
1
n(n− 1)e2
]= 1− 1
n2e2+
1
n(n− 1)e2− 1
n2(n− 1)e3
= 1 +1
n2(n− 1)e2− 1
n2(n− 1)e3> 1.
And we are done.
10. If a1 ≥ a2 ≥ . . . ≥ an ≥ 0 then a21 − a22 + a23 ± . . .± a2n ≥ (a1 − a2 + a3 ±. . .± an)2
Solution: Again we want to use induction by deleting some of the vari-ables. Since we don’t really know the sign of an, it’s probably betterto start from a1. However deleting it produces an expression of oppo-site sign and applying induction to it would give an inequality also ofopposite sign, not helpful to us. So we should delete the first terms,a1, a2.
Indeed, by the induction assumption for n− 2 we can deduce that
a21 − a22 + a23 ± . . .± a2n ≥ a21 − a22 + (a3 − a4 + a5 ± . . .± an)2.
If we denote by t = a2 − a3 ± . . .± an then we must prove that
a21 − a22 + t2 ≥ (a1 − a2 + a3)2.
It resembles our inequality for n = 3 and it is indeed our inequality forn = 3 because t = (a3−a4)+ . . . = a3−(a4−a5)− . . . so 0 ≤ t ≤ a3 ≤ a2.Therefore it suffices to prove the basis, which consists of two separatecases: n = 2 and n = 3 (this is because we did an unusual inductionstep of 2 - if you want, we do induction on [n2 ]). For n = 2 we mustprove (a1 − a2)(a1 + a2) ≥ (a1 − a2)
2 which is clear. For n = 3 wemust prove a21 − a22 + a23 − (a1 − a2 + a3)
2 ≥ 0 which is equivalent to2(a1 − a2)(a2 − a3) ≥ 0, true!
11. Let x1 = 1, xn+1 =xnn
+n
xn, for n ≥ 1. Prove that the sequence (xn) is
increasing and that [x2n] = n for n ≥ 4
Solution: To perform the induction step, we will prove a stronger hy-
pothesis, that for n ≥ 3 we have√n ≤ xn ≤
n√n− 1
.
We may note that f(x) =x
n+n
xis decreasing for x ≤ n so it suffices to
prove that f(√n) ≤ n+ 1√
nand f
(n√n− 1
)≥√n+ 1. We have
xn+1 ≥ f(
n√n− 1
)=√n− 1 +
1√n− 1
=n√n− 1
>√n+ 1.
Replacing n+ 1 by n we get xn ≥n− 1√n− 2
. Therefore
xn+1 ≤ f(n− 1√n− 2
)=n2(n− 2) + (n− 1)2
n(n− 1)√n− 2
.
23
So we are left to prove thatn2(n− 2) + (n− 1)2
n(n− 1)√n− 2
<√n+ 2 which is
equivalent ton3 − n2 − 2n+ 1
n2 − n<√n2 − 4 which after squaring is equiv-
alent to 2n2(n − 3) + 4n − 1 ≥ 0, true for n ≥ 4. We thus deducen < x2n < n+ 1 so [x2n] = n.
12. Let n > 2. Find the least constant k such that for any a1, . . . , an > 0with product 1 we have
a1a2(a21 + a2)(a22 + a1)
+a2a3
(a22 + a3)(a23 + a2)+ . . .+
ana1(a2n + a1)(a21 + an)
≤ k.
Solution: We prove the answer is n− 2. If we put a1 = . . . = an−1 = xtending to 0 we get close to n− 2. Now to prove another part: Firstly
(a21 + a2)(a22 + a1) ≥ (a21 + a1)(a
22 + a2)
(this is easily deduced by opening the brackets). So we replace to prove
n∑i=1
1
(ai + 1)(ai+1 + 1)≤ n− 2.
For n = 3 it’s simple replacing a1 =x
y, a2 =
y
z, a3 =
z
x:
yz
(x+ y)(x+ z)+
zx
(y + z)(y + z)+
xy
(z + x)(z + y)<
<xy
xy + yz + zx+
yz
xy + yz + zx+
xz
xy + yz + zx= 1.
For n > 3 we prove by induction: If a1a2 ≤ 1 and a2a3 ≥ 1 (we mayassume this since the inequality is cyclic) we can replace a2 and a3 witha2a3 and prove the value decreases by at most 1 so we can use inductionstep. Indeed, the value decreases by
1
(a1 + 1)(a2 + 1)+
1
(a2 + 1)(a3 + 1)+
1
(a3 + 1)(a4 + 1)−
− 1
(a1 + 1)(a2a3 + 1)− 1
(a2a3 + 1)(a4 + 1)=
=a2(a3 − 1)
(a1 + 1)(a2 + 1)(a2a3 + 1)+
a3(a2 − 1)
(a4 + 1)(a3 + 1)(a2a3 + 1)+1− a2 + a3 + a2a3
(a2 + 1)(a3 + 1) + 1.
Now if a2 ≤ 1, a3 ≥ 1 then this quantity is at most
1 +a2a3
a2a3 + 1− (a2 + 1)(a3 + 1)− 1
(a2 + 1)(a3 + 1)≤
≤ 1 + 1− 1
a2a3 + 1− 1 +
1
a2a3 + a2 + a3 + 1≤ 1,
24
and analogously if a2 ≥ 1, a3 ≤ 1. Finally if a2, a3 > 1 then the quantityis at most
1 +a2a3
2(a2a3 + 1)+
a2a32(a2a3 + 1)
− a2a3 + a2 + a3(a2 + 1)(a3) + 1
=
= 1 +a2a3
a2a3 + 1− a2a3 + a2 + a3a2a3 + a2 + a3 + 1
≤ 1
again.
1.5 The AM-GM inequality
The inequality proven in the previous chapter:
a1 + a2 + . . .+ ann
≥ n√a1a2 . . . an
is the most important of all the inequalities used, and has enormously manyapplications.Here’s one of them:If ai, bi > 0 then
n∑i=1
aiai + bi
≥ n ·
n
√n∏i=1ai
n
√n∏i=1
(ai + bi)
and also
n∑i=1
biai + bi
≥ n ·
n
√n∏i=1bi
n
√n∏i=1
(ai + bi)
Summing these two inequalities and using the fact thatai
ai + bi+
biai + bi
= 1
we deduce
n ≥ n ·
n
√n∏i=1ai + n
√n∏i=1bi
n
√n∏i=1
(ai + bi)
or
n
√√√√ n∏i=1
(ai + bi) ≥ n
√√√√ n∏i=1
ai + n
√√√√ n∏i=1
bi
This is called Huygens Inequality.If we have the sequence a1, a1, . . . , a1, a2, . . . , a2, . . . , an, . . . , an, where ai re-peats pi times then according to AM-GM we have
p1a1 + p2a2 + . . .+ pnan ≥ (p1 + p2 + . . .+ pn) p1+p2+...+pn
√ap11 . . . apnn .
25
If we denote αi =pi
p1 + p2 + . . .+ pn, we obtain the following generalization
of AM-GM:
If α1, . . . , αn are positive numbers that sum to 1 then for any positive x1, x2, . . . , xnwe have α1x1 + . . .+ αnxn ≥ xα1
1 . . . xαnn
This is called Weighted AM-GM. Of course, in this setting αi need to berational - but in fact this condition is unnecessary.
Exercises
1. Show that for any a1, a2, . . . , an > 0 we have (a1+. . .+an)
(1
a1+ . . .+
1
an
)≥
n2.
Solution: We have a1 + a2 + . . .+ an ≥ n n√a1a2 . . . an and
1
a1+
1
a2+ . . .+
1
an≥ n n
√1
a1a2 . . . an
and by multiplying these two relations we get the result.
2. Let a1, a2, . . . , an > 0 with a1a2 . . . an = 1. Let sk = ak1 + ak2 + . . .+ akn.
a) Show that if k ≥ l ≥ 0 then sk ≥ sl;
b) Find all k > 0 for which sk ≥ s−1.
Solution: a) We have aki +k − ll≥ k
lali by weighted AM-GM. The
inequality now follows by summing, as sk ≥ n = s0 by AM-GM
b) We notice that sn−1 ≥ s−1 because
xn−12 + xn−13 + . . .+ xn−1n ≥ (n− 1)x2x3 . . . xn =n− 1
x1
and the inequality follows by summing. So using a) we get sk ≥ s−1for any k ≥ n − 1. If k < n − 1 then taking x1 = x2 = . . . = xn−1 =
x, xn =1
xn−1we have sk = (n−1)xk +
1
x(n−1)k, s−1 =
n− 1
x+xn−1 and
s−1 ≥ sk for all sufficiently big x. So k ∈ [n− 1,+∞).
3. Show that n√
(n+ 1)! > n√n! + 1
Solution: By Huygens Inequality
n√
(n+ 1)! = n√
(1 + 1)(2 + 1)(3 + 1) . . . (n+ 1) ≥ n√
1 · 2 . . . · n+1 =n√n!+1.
The equality cannot take place.
4. a) Show that
(1 +
1
n
)n<
(1 +
1
n+ 1
)n+1
;
b) Show that
(1 +
1
n
)n+1
>
(1 +
1
n+ 1
)n+2
.
26
Solution: a) By AM-GM we have n + 2 = 1 +
(1 +
1
n
)+
(1 +
1
n
)+
. . .+
(1 +
1
n
)(1+
1
nmeets n times) is bigger than n+1 n+1
√(1 +
1
n
)n,
which rising to power n+ 1 gives us exactly the conclusion.
b) By AM-GM we have n + 1 = 1 +n
n+ 1+ . . . +
n
n+ 1≥ (n +
2)n+2
√(n
n+ 1
)n+1
which after rising to power n + 2 yields the con-
clusion.
5. 2(a4 + b4) + 17 > 16ab.
Solution: 2(a4 + b4) + 17 ≥ 4a2b2 + 17 > 4a2b2 + 16 ≥ 16ab by AM-GM.
6. n n√a1a2 . . . an − (n− 1) n−1
√a1a2 . . . an−1 ≤ an, for ai > 0.
Solution: The conclusion from AM-GM applied to the n numbers an,n−1√a1a2 . . . an−1, n−1
√a1a2 . . . an−1, . . ., n−1
√a1a2 . . . an−1.
7. Show that n
(1− 1
n√n
)+ 1 > 1 +
1
2+ . . . +
1
n> n
(n√n+ 1− 1
), if
n ∈ N∗.
Solution: By AM-GM we have 1 +1
2+
2
3+ . . .+
n− 1
n≥ n 1
n√n
, or
n−(
1
2+
1
3+ . . .+
1
n
)≥ n 1
n√n,
thus n(1− 1n√n
) ≥ 1
2+
1
3+ . . .+
1
n, from where left inequality follows.
For the right inequality, we see that 2 +3
2+
4
3+ . . .+
n+ 1
n≥ n n√n+ 1,
so n+
(1 +
1
2+ . . .+
1
n
)≥ n n√n+ 1, and the rest is clear.
8. Let x1, x2, . . . , x2n be positive reals that sum to 1. Let
S = x21x22 . . . x
2n + x22x
23 . . . x
2n+1 + . . .+ x22nx
21 . . . x
2n−1.
Show that S <1
n2n.
Solution: We can notice that
S ≤ (x21 + x2n+1) . . . (x2n + x22n) ≤ (x1 + xn+1)
2 . . . (xn + x2n)2
≤(x1 + xn+1 + . . .+ xn + x2n
nn
)2
=1
n2n.
The inequality cannot hold in all cases.
9. If a, b, c > 0 then
√a
b+ c+
√b
a+ c+
√c
a+ b> 2.
27
Solution: This time we shall use the inequality between the geometric
and harmonic means:
√a
b+ c=
√a
(b+ c) · 1≥ 2
b+ c
a+ 1
=2a
a+ b+ c,
and the inequality now follows by summation.
10. For a, b, c > 0 we have
a√(a+ b)(a+ c)
+b√
(b+ a)(b+ c)+
c√(c+ a)(c+ b)
≤ 3
2.
Solution: By AM-GM we have
a√(a+ b)(a+ c)
=
√a
a+ b
a
a+ c≤ 1
2
(a
a+ b+
a
a+ c
)and summing with the analogous results we get the conclusion.
11. For x, y, z are positive numbers such that x+ x2 + y + y2 + z + z2 = 6,find the maximum of x2y + y2z + z2x
Solution: We see that
x3 + y3 + z3 + xy2 + yz2 + zx2 = (x3 + xy2) + (y3 + yz2) + (z3 + zx2)
≥ 2(x2y + y2z + z2x).
This means x2y + y2z + z2x ≤ (x+y+z)(x2+y2+z2)3 = a(6−a)
3 ≤ 3. Theequality can hold for x = y = z = 1. So the maximum is 3.
12. If a1, a2, . . . , an > 0 then
1
a1+
2
a1 + a2+ . . .+
n
a1 + a2 + . . .+ an< 2
(1
a1+
1
a2+ . . .+
1
an
).
Solution: We havek
a1 + a2 + . . .+ ak=
k
a1 +a22
+a22
+a33
+ . . .+akk
(so
we get 1 + 2 + . . .+ k =k(k + 1)
2terms at the denominator). In virtue
of exercise 1, it’s smaller than
4k
k2(k + 1)2
(1
a1+
2
a2+
2
a2+ . . .+
k
ak
)=
4
k(k + 1)2
(1
a1+
4
a2+
9
a3+ . . .+
k2
ak
).
By summing all these expressions, we see that this sum is smaller than
4
a1
(1
4+
1
18+ . . .
)+
4
a2
(4
18+ . . .
)+ . . . .
So we need to prove 4m2
(1
m(m+ 1)2+ . . .
)< 2. But we can prove
that1
m(m+ 1)2<
1
2
[1
m2− 1
(m+ 1)2
]and hence by telescoping
4m2
[1
m(m+ 1)2+ . . .
]< 4m2 · 1
2m2= 2
and we are done.
28
13. If a1, a2, . . . , an > 0 then
a1 +√a1a2 + 3
√a1a2a3 + . . .+ n
√a1a2 . . . an < 3(a1 + a2 + . . .+ an).
Solution: As in the above exercise, we assign ”weights” to ai. We have
k√a1a2 . . . ak =
1k√k!
k√a1 × 2a2 × . . .× kak ≤
1
k k√k!
(a1+2a2+. . .+kak).
After summation, the term alongside am will be
m
[1
m√m!m
+1
m+1√
(m+ 1)!(m+ 1)+ . . .+
1n√n!n
].
Now we prove that k√k! ≥ k + 1
3which will imply
m
[1
m√m!m
+1
m+1√
(m+ 1)!(m+ 1)+ . . .+
1n√n!n
]≤
≤ m
[3
m(m+ 1)+
3
(m+ 1)(m+ 2)+ . . .+
3
n(n+ 1)
]= m
(3
m− 3
m+ 1+
3
m+ 1− 3
m+ 2+ . . .+
3
n− 3
n+ 1
)= m
(3
m− 3
n+ 1
)= 3− 3m
n+ 1< 3
so our initial assertion. The inequality k√k! ≥ k + 1
3is proven by induc-
tion on k. For k ≤ 7 this is proven manually. The induction step followsfrom the fact that(
k + 2
3
)k+1
(k + 1
3
)k =k + 2
3
(k + 1
k
)k+1
<k + 2
3
(8
7
)8
< k + 1 =(k + 1)!
k!.
1.6 The quadratic trinomial
Consider a second-degree equation
ax2 + bx+ c = 0
Its discriminant isD = b2 − 4ac.
If D > 0, this equation has two different solutions, hence it changes sign.If D < 0, this equation has no solutions, so it has the same sign for all x. Thissign is the sign of a. Thus, we can enounce the following principle:
a) For a > 0, we have ax2 + bx+ c ≥ 0 for any x if and only if b2 ≤ 4ac.b) For a < 0, we have ax2 + bx+ c ≤ 0 for any x if and only if b2 ≤ 4ac.
29
We can also make the inequalities strict:
a) For a > 0, we have ax2 + bx+ c > 0 for any x if and only if b2 < 4ac.b) For a < 0, we have ax2 + bx+ c < 0 for any x if and only if b2 < 4ac.
This principle can be used for solving many problems or deducing theorems.
Consider the quadratic trinomial f(x) =
n∑i=1
(aix+ bi)2. It is always positive,
and hence must have non-positive discriminant. However, by opening thebrackets we see that
f(x) =n∑i=1
a2ix2 + 2
(n∑i=1
aibi
)x+
n∑i=1
b2i
and so the discriminant is
4
(n∑i=1
aibi
)2
− 4
(n∑i=1
a2i
)(n∑i=1
b2i
).
Therefore we have the following inequality (Cauchy-Buniakovsky-Schwartz):For any reals a1, a2, . . . , an, b1, b2, . . . , bn we have(
n∑i=1
a2i
)(n∑i=1
b2i
)≥
(n∑i=1
aibi
)2
.
If we take the equation f(x) = (a1x−b1)2−(a2x−b2)2− . . .−(anx−bn)2 thendoes takes non-positive values. Therefore if the leading coefficient is positive,it must have roots and so the discriminant must be positive, which leads toAczel’s Inequality:If a21 > a22 + . . .+ a2n then
(a1b1 − a2b2 − . . .− anbn)2 ≥ (a21 − a22 − . . .− a2n)(b21 − b22 − . . .− b2n).
If the dominant coefficient of the trinomial f(x) = ax2 + bx + c is positive,
then as f(x) can be written as a
(x+
b
2a
)2
+D
a2, it has its minimal value at
x = − b
2a, called the vertex of the trinomial (or parabola). We can also see
that f is increasing for x > − b
2aand decreasing otherwise. It can be therefore
deduced that a quadratic trinomial with positive dominant coefficient attainsits maximum at the extremities of an interval.For a negative, all properties above change sign.
The last principle deserves more a more detailed discussion. Let us firstlyconsider the case a > 0, then for all x ∈ [t1, t2] , we have
f(x)− f(t1) = a(x2 − t21) + b(x− t1) = (x− t1)(ax+ at1 + b),
f(x)− f(t2) = a(x2 − t22) + b(x− t2) = (x− t2)(ax+ at2 + b).
From this, we have that: if ax+at1 + b ≤ 0 then f(x) ≤ f(t1), and in the caseax + at1 + b ≥ 0, we have ax + at2 + b ≥ ax + at1 + b ≥ 0, and we get that
30
f(x) ≤ f(t2). Or in the other words, in any cases, the following inequality isvalid:
f(x) ≤ max {f(t1), f(t2)} ∀x ∈ [t1, t2] .
In the case a < 0, we proceed by the same way as the above for g(x) =−f(x) = −ax2 − bx− c, and we also have that
g(x) ≤ max {f(t1), f(t2)} ∀x ∈ [t1, t2] , or −f(x) ≤ max {−f(t1),−f(t2)}∀x ∈ [t1, t2] , which is equivalent to f(x) ≥ min {f(t1), f(t2)} ∀x ∈ [t1, t2] .
Finally, let us consider the case a = 0, then if b ≥ 0, it is trivial that bx1 + c ≤bx + c ≤ bx2 + c, or f(t1) ≤ f(x) ≤ f(t2). And if b ≤ 0, it is also trivial thatbt2 + c ≤ bx+ c ≤ bt1 + c, or f(t2) ≤ f(x) ≤ f(t1).Therefore, in any cases, wehave
min {f(t1), f(t2)} ≤ f(x) ≤ max {f(t1), f(t2)} ∀x ∈ [t1, t2] .
All these results can be summarized in the following proposition:
Proposition. Consider the following polynomial f(x) = ax2 + bx + c witha, b, c ∈ R and x ∈ [t1, t2] , then
1. If a > 0, then f(x) ≤ max {f(t1), f(t2)} ∀x ∈ [t1, t2] .
2. If a < 0, then f(x) ≥ min {f(t1), f(t2)} ∀x ∈ [t1, t2] .
3. If a = 0, then min {f(x1), f(x2)} ≤ f(x) ≤ max {f(x1), f(x2)} ∀x ∈[t1, t2] .
This theorem provides us a very interesting way to prove inequalities. Forexample, we need to prove that f(x) = ax2 + bx+ c ≤ m for all x ∈ [t1, t2] andwe know that a ≥ 0, then from the above theorem, it suffices to prove thatmax {f(t1), f(t2)} ≤ m. It means that we only need to consider the originalinequality in the cases x = t1 and x = t2 which is not hard to prove. For moreclearly, see the following examples
Example 1. Let a, b, c ∈ [1, 2] , show that a2 + b2 + c2 ≤ 3abc.
Solution: We need to prove that P (a, b, c) = a2− 3abc+ b2 + c2 ≤ 0. This is aquadratic polynomial in term of a with the highest coefficient it positive, there-fore, from the above theorem, we only have to prove that max {P (1, b, c), P (2, b, c)} ≤0. Now, we see that P (1, b, c), P (2, b, c) are also quadratic polynomials in termsof b, hence it suffices to prove the inequality max{max {P (1, 1, c), P (1, 2, c)} ,max {P (2, 1, c), P (2, 2, c)}} ≤0, which is equivalent to max {P (1, 1, c), P (1, 2, c), P (2, 1, c), P (2, 2, c)} ≤ 0.Again, we see that P (1, 1, c), P (1, 2, c), P (2, 1, c), P (2, 2, c) are also quadraticpolynomials in terms of c, hence it suffices to prove the inequality max{P (1, 1, 1),P (1, 1, 2), P (1, 2, 1), P (1, 2, 2), P (2, 1, 1), P (2, 1, 2), P (2, 2, 1), P (2, 2, 2)} ≤ 0.But our inequality is symmetric, therefore we only need to consider the fol-lowing cases
i) If a = b = c = 1, then a2 + b2 + c2 − 3abc = 0.
ii) If a = b = 1, c = 2, then a2 + b2 + c2 − 3abc = 0.
iii) If a = 1, b = c = 2, then a2 + b2 + c2 − 3abc = −3 < 0.
iv) If a = b = c = 2, then a2 + b2 + c2 − 3abc = −12 < 0.
Our proof is complete.
Example 2. If a, b, c ≥ 0 and a+b+c = 1. Prove that 1−4(ab+bc+ca)+9abc ≥0.
31
Solution: Put x = ab, then 0 ≤ x ≤ (a+ b)2
4=
(1− c)2
4. And our in-
equality becomes (9c − 4)ab + 1 − 4c(a + b) ≥ 0, or equivalently, f(x) =(9c − 4)x + 1 − 4c + 4c2 ≥ 0. This is a quadratic polynomial with zeroquadratic coefficient, therefore from the above theorem, it suffices to prove
that min
{f(0), f
((1− c)2
4
)}≥ 0. But it is true since f(0) = 1− 4c+ 4c2 =
(1− 2c)2 ≥ 0, and
f
((1− c)2
4
)= (9c− 4) · (1− c)2
4+ 1− 4c+ 4c2 =
1
4c(1− 3c)2 ≥ 0.
Remark. This is Schur’s inequality for third degree.
In the Example 2, if we put P (a, b, c) = 1 − 4(ab + bc + ca) + 9abc, then we
have showed that we only need to prove that min
{f(0), f
((a+ b)2
4
)}≥ 0.
But it is easy to see that
f(0) = P (a+ b, 0, c), f
((a+ b)2
4
)= P
(a+ b
2,a+ b
2, c
).
Therefore, our statement becomes min
{P (a+ b, 0, c), P
(a+ b
2,a+ b
2, c
)}≥
0. From this, we can obtain the following interesting idea: in order to prove theoriginal inequality, it suffices to consider the cases there is one variable equalto 0 or there are two variables equal. This is a very helpful idea in solvinginequalities.
Example 3. Let a, b, c, d be nonnegative real numbers such that a+b+c+d = 4.Show that
a2b2c2 + b2c2d2 + c2d2a2 + d2a2b2 + abc+ bcd+ cda+ dab ≤ 8.
Solution: Put x = ab, then 0 ≤ x ≤ (a+ b)2
4, our inequality becomes
f(x) = (c2 + d2)x2 + (c+ d− 2c2d2)x+ cd(a+ b) + (a+ b)2c2d2 − 8 ≤ 0,
This is a quadratic polynomial of x with the highest coefficient is nonnegative,
therefore, it suffices to prove that max
{f(0), f
((a+ b)2
4
)}≤ 0. It shows
that it suffices to consider the cases a = b or ab = 0. Similarly, we also obtainthat it suffices to consider the cases c = d or cd = 0. Combining both thesestatements, we obtain that, in order to prove the original inequality, it enoughto prove it in the following cases
Case 1. b = 0, d = 0, then the inequality is trivial since
a2b2c2 + b2c2d2 + c2d2a2 + d2a2b2 + abc+ bcd+ cda+ dab = 0.
Case 2. a = b, d = 0, then c = 4 − 2a ≥ 0, and our inequality becomesa4(4 − 2a)2 + a2(4 − 2a) ≤ 8, which is true since by AM-GM inequality, we
32
have a2(4−2a) ≤(a+ a+ 4− 2a
3
)3
=64
27, hence a4(4−2a)2+a2(4−2a)−8 ≤
642
272+
64
27− 8 = − 8
729< 0.
Case 3. a = b, c = d, then c = 2 − a ≥ 0, and our inequality becomes2a4(2 − a)2 + 2(2 − a)4a2 + 2a2(2 − a) + 2(2 − a)2a ≤ 8, or equivalent toa6−6a5+14a4−16a3+7a2+2a−2 ≤ 0, which is (a2−2a−1)(a2−2a+2)(a−1)2 ≤0. The last inequality is valid since a ≤ 2.
The property of always attaining the maximal (respectively minimal) valueat the extremities of the interval is not limited to quadratic polynomials - itis characteristic for convex (respectively) concave functions, but we will dealwith these concepts later.
Exercises
1. If x, y, z ∈ R then 3(x2y2 + 1)(z2 − y2 + 1) ≥ 2(x2y2z2 + xyz + 1)
Solution: Grouping by the powers of z we get a quadratic trinomial inz with positive leading coefficient: (x2y2 + 3)z2 − (3x2y2 + 2xy + 3)z +3x2y2 + 1 ≥ 0. Its discriminant can be computed to be −3(xy− 1)4 ≤ 0and from here we deduce the result.
2. If x, y, z ∈ [0, 1] then x2 + y2 + z2 ≤ x2y + y2z + z2x+ 1.
Solution: Let f(t) = t2(1− y)− z2t+ y2 + z2 − y2z − 1. We must provethat f(x) ≤ 0. If y = 1 then f(t) = −z2t+z2+1−z−1 = z2(1−t)−z ≤z2 − z ≤ 9. If y < 1 then f is a quadratic in t with positive dominantcoefficient and attains its maximum either at t = 0 or at t = 1. But
f(1) = 1− y − z2 + y2 + z2 − y2z − 1 = y2(1− z)− y ≥ y2 − y ≤ 0
and f(0) = y2 + z2 − y2z − 1 = (1− z)(y2 − 1) ≤ 0.
3. If a < b < c < d then (a+ b+ c+ d)2 > 8(ac+ bd)
Solution: This time, we can try two different approaches. The first is tolook at (a+b+c+d)2−8(ac+bd) as at a quadratic expression in, say, a.It is written as a2−2(3c−b−d)a+(b+c+d)2−8bd. It’s discriminant isD = 32(c− b)(c− d) < 0 so the expression is always positive (moreoverwe could restrict the condition to just b < c < d or just c between b andd).
The second is to write (a+ b+ c+ d)2 − 8(ac+ bd) as D = B2 − 4AC.It’s convenient to set B = a + b + c + d,A = 2, C = ac + bd. Then thequadratic function f(x) = Ax2 +Bx+C = (x+a)(x+c)+(x+b)(x+d)satisfies f(−a) = (b− a)(d− a) > 0 and f(−b) = (a− b)(c− b) < 0 so fmust have a root and hence its discriminant is positive.
4. Show that a2 + b2 + c2 − ab− bc− ca ≥ 3
4(a− b)2.
Solution: Regarding this as a quadratic trinomial in c, we write it as
c2 − (a+ b)c+(a+ b)2
4≥ 0 or
(c− a+ b
2
)2
≥ 0.
33
5. Let 0 < a < b and xi ∈ [a, b]. Show that
(x1 + x2 + . . .+ xn)
(1
x1+
1
x2+ . . .+
1
xn
)≤ (a+ b)2
4abn2.
Solution: If xi ∈ [a, b] then (xi− a)(xi− b) ≤ 0 so x2i + ab ≤ (a+ b)xi or
xi +ab
xi≤ a + b. We conclude that
n∑i=1
xi + abn∑i=1
1
xi≤ (a + b)n and we
use AM-GM to conclude the proof.
6. If ai ∈ [a,A], bi ∈ [b, B] where 0 < a < A, 0 < b < B then
(a21 + . . .+ a2n)(b21 + . . .+ b2n)
(a1b1 + . . .+ anbn)2≤ 1
4
(√AB
ab+
√ab
AB
).
Solution: As this inequality has a form similar to the Cauchy-Schwartz,but has a inverse sign, we might try to find a quadratic function with azero and therefore positive discriminant. We can take
f(x) =n∑i=1
(a2ix
2 −
(√AB
ab+
√ab
AB
)x+ b2i
)
=n∑i=1
a2i
(x− bi
ai
√AB
ab
)(x− bi
ai
√ab
AB
).
To have f(t) ≤ 0 it suffices to find a t s.t.biai
√ab
AB≤ t ≤ bi
ai
√AB
ab, so
the intervals
[biai
√ab
AB;biai
√AB
ab
]must have non-empty intersection, or
biai
√ab
AB≤ bjaj
√AB
ab
for any i, j. This is equivalent toajbiaibj
≤ AB
ab, and is true from the
condition.
7. If a1, a2, . . . , an ∈ [1, n− 1] then
n(a1 + a2 + . . .+ an)2 ≥ 4(n− 1)(a21 + a22 + . . .+ a2n).
Solution: We have (ai−1) [ai − (n− 1)] ≤ 0 so a2i +(n−1) ≤ nai. Thus(a21 + a22 + . . .+ a2n) + n(n− 1) ≤ n(a1 + a2 + . . .+ an). Thus
4(n− 1)(a21 + a22 + . . .+ a2n) ≤ 4n(n− 1)(a1 + a2 + . . .+ an)− 4n(n− 1)2.
But 4n(n− 1)(a1 + a2 + . . .+ an)− 4n(n− 1)2 ≤ n(a1 + a2 + . . .+ an)2
as this is equivalent to n [a1 + a2 + . . .+ an − 2(n− 1)]2 ≥ 0.
34
8. Show that a2 + b2 + c2 ≥√
3(a3b+ b3c+ c3a), where a, b, c ≥ 0.
Solution: If all of a, b, c are equal, equality holds. If not, since theinequality is cyclic, we may assume that b − a > 0. As the inequalityis obvious we can set b − a = 1. Let x = c − a. Then f(a) = (a2 +b2 + c2)2 − 3(a3b + b3c + c3a) can be written as (x2 − x + 1)a2 + (x3 −5x2 + 4x+ 1)a+x4−3x3 + 2x2 + 1. Its discriminant can be evaluated as−3(x3−x2−2x−1)2 so is non-positive and thus f is always non-negative,QED.
9. Let a, b, c ∈ [0, 1] . Determine the greatest value of P = a+ b+ c− ab−bc− ca.Solution: We have that P (a, b, c) = a+ b+ c−ab− bc− ca is a quadraticpolynomial of a with the highest coefficient is 0, therefore, to find themaximum of P , we only need to consider the cases a = 0 or a = 1.Similarly, P is also a quadratic polynomial of b and c, therefore it sufficesto consider the cases b = 0 or b = 1, and c = 0 or c = 1. Combiningall cases with notice that the expression P is symmetric, we see that itsuffices to consider the following cases
Case 1. If a = b = c = 0, then P = 0.
Case 2. If a = b = 0, c = 1, then P = 1.
Case 3. If a = 0, b = c = 1, then P = 1.
Case 4. If a = b = c = 1, then P = 0.
Therefore, we obtain that maxP = 1.with equality holds when a = b =0, c = 1 or a = 0, b = c = 1 and their cyclic permutations.
10. Let a, b, c ∈[
1
3, 3
], then show that
a
a+ b+
b
b+ c+
c
c+ a≥ 7
5.
Solution: Since the inequality is cyclic, we may assume that c = max {a, b, c} ,
then since a, b, c ∈[
1
3, 3
], we have that
c
9≤ a ≤ c. We will change the
inequailty into a quadratic polynomial of a in order to use our theorem.
Put x =7
5− b
b+ c, then our inequality becomes
a
a+ b+
c
c+ a≥ x,
which is equivalent to a2 + ac + ac + bc ≥ x(a2 + ac + ab + bc), or(1− x)a2 + [(2− x)c− xb] a+ (1− x)bc ≥ 0.
If 1 − x ≤ 0 or eqivalently c ≥ 3
2b, we have that the last inequality is a
quadratic polynomial of a with the highest coefficient is not greater than
0, therefore, it suffices to consider the cases a =c
9or a = c to prove the
original inequality. If a = c9 , then we have that
a
a+ b+
b
b+ c+
c
c+ a=
c
9b+ c+
b
b+ c+
9
10=
7
5+
(3b− c)2
2(9b+ c)(b+ c)≥ 7
5,
and if a = c, then we have thata
a+ b+
b
b+ c+
c
c+ a=
c
c+ b+
b
b+ c+
1
2=
3
2>
7
5. So our inequality is proved in this case.
35
Now, let us consider the case 1 − x ≥ 0 or b ≥ 2
3c, then we have that
(2 − x)c − xb ≥ (2 − x)c − xc = 2(1 − x)c ≥ 0. Therefore, since a ≥ c
9,
we have
(1− x)a2 + [(2− x)c− xb] a+ (1− x)bc ≥
≥ (1− x)( c
9
)2+ [(2− x)c− xb]
( c9
)+ (1− x)bc.
And it suffices to prove that (1 − x)( c
9
)2+ [(2− x)c− xb]
( c9
)+ (1 −
x)bc ≥ 0. Or in another word, it means that we only need to prove the
original inequality in the case a =c
9, but this case has been already
proved in case 1. Our proof is now complete.
11. Let a, b, c ∈ [1, 2] , then show that (a+ b+ c)
(1
a+
1
b+
1
c
)≤ 10.
Solution: We can rewrite the original inequality as P (a, b, c) = (a+ b+c)(ab + bc + ca) − 10abc ≤ 0 which is a quadratic polynomial of a withthe highest coefficient is (b + c) > 0, therefore, it suffices to considerthe cases a = 1 or a = 2. Similarly, by the same manner, we also seethat P is also a quadratic polynomial of b and c, and thus, it sufficesto consider the cases b = 1 or b = 2, and c = 1 or c = 2. Combiningall cases with notice that the inequality is symmetric, we obtain that itsuffices to consider the following cases
Case 1. If a = b = c = 1, then P (1, 1, 1) = −1 < 0.
Case 2. If a = b = 1, c = 2, then P (1, 1, 2) = 0.
Case 3. If a = 1, b = c = 2, then P (1, 2, 2) = 0.
Case 4. If a = b = c = 2, then P (2, 2, 2) = −1 < 0.
Therefore, our inequality is proved.
12. Let a, b, c be nonnegative real numbers such that a + b + c = 3, thena2 + b2 + c2 + abc ≥ 4.
Solution: Put x = ab, then 0 ≤ x ≤ (a+ b)2
4, and our inequality becomes
f(x) = (c − 2)x + 2c2 − 6c + 5 ≥ 0 which is a quadratic polynomial ofx with the highest coefficient is 0, therefore, it suffices to prove that
min
{f(0), f
((a+ b)2
4
)}≥ 0. Now, putting P (a, b, c) = a2 + b2 + c2 +
abc, then we see that
f(0) = P (a+ b, 0, c), f
((a+ b)2
4
)= P
(a+ b
2,a+ b
2, c
).
which means that we have to prove min
{P (a+ b, 0, c), P
(a+ b
2,a+ b
2, c
)}≥
0, or in the other words, we only need to consider the original inequalityin the cases there is one variable equal to 0 and there are 2 variables
36
equal. Since the inequality is symmetric, therefore from the above state-ments, we have 2 cases
Case 1. If a = 0, then b = 3−c, and our inequality becomes (3−c)2+c2 ≥4, or equivalently 2c2 − 6c + 5 ≥ 0, which is true since by AM-GMinequality, we have 2c2 + 5 ≥ 2
√10c ≥ 6c.
Case 2. If a = b, then c = 3 − 2a ≥ 0, and our inequality becomes2a2 + (3 − 2a)2 + a2(3 − 2a) ≥ 4, or equivalently (5 − 2a)(a − 1)2 ≥ 0which is true since a ≤ 3
2 .This ends our proof.
13. Let a, b, c, d ∈ [0, 1] . Prove that a2b+b2c+c2d+d2a−ab2−bc2−cd2−da2 ≤8
27.
Solution: Since the inequality is cyclic, we may assume that a = max {a, b, c, d} ,then from a, b, c, d ∈ [0, 1] , we have that 0 ≤ d ≤ a. Now, notice that wecan rewrite our inequality as
f(d) = (a− c)d2 + (c2 − a2)d+ a2b+ b2c− ab2 − bc2 − 8
27≤ 0
which is a quadratic polynomial of d with the highest coefficient is non-negative, therefore, it suffices to prove that max {f(0), f(a)} ≤ 0. Wehave
f(0) = a2b+ b2c− ab2 − bc2 − 8
27= b(a− c)(a− b+ c)− 8
27
≤[b+ (a− c) + (a− b+ c)
3
]3− 8
27=
8
27(a3 − 1) ≤ 0,
and
f(a) = (a− c)a2 + (c2 − a2)a+ a2b+ b2c− ab2 − bc2 − 8
27
= (a− b)c2 + (b2 − a2)c+ a2b− ab2 − 8
27
= (b− c)(a− c)(a− b)− 8
27≤ b(a− c)(a− b)− 8
27
≤ ab(a− b)− 8
27≤ b(1− b)− 8
27≤[b+ (1− b)
2
]2− 8
27= − 5
108.
Therefore, our inequality is proved.
14. If a, b, c are the side-lengths of a triangle, then
3
(a
b+b
c+c
a
)≥ 2
(b
a+c
b+a
c
)+ 1.
Solution: Without loss of generality, we may assume that a ≥ b ≥ c,then b ≤ a ≤ b+ c. The inequality is equivalent to f(a) = (3c− 2b)a2 +(3b2 − 3bc− 2c2)a+ 3bc2 − 2b2c ≥ 0. If 3c ≤ 2b, then f(a) is a quadraticpolynomial of a with the highest coefficient is 3c − 2b ≤ 0, thereforeaccording to our theorem, it suffices to prove that min {f(b), f(b+ c)} ≥
37
0. But this inequality is true because f(b) = b(b−c)2 ≥ 0, and f(b+c) =c(b − c)2 ≥ 0. Now, let us consider the case 3c ≥ 2b, then by AM-GMinequality, we have that a2 + b2 ≥ 2ab, therefore
f(a) ≥ (3c− 2b)(2ab− b2
)+ (3b2 − 3bc− 2c2)a+ 3bc2 − 2b2c
= a(b− c)(2c− b) + b(2b− 3c)(b− c)≥ b(b− c)(2c− b) + b(2b− 3c)(b− c) = b(b− c)2 ≥ 0.
Remark. We still have to justify why we can assume that a ≥ b ≥ cbecause the inequality is not symmetric, it is only cyclic. Put P (a, b, c) =
3
(a
b+b
c+c
a
)−2
(b
a+c
b+a
c
)−3, then since the inequality is cyclic,
we may assume that a = max {a, b, c} . If b ≥ c, then the statement istrivial. If c ≥ b, then we have that
P (a, b, c)− P (a, c, b) = 5
(a
b+b
c+c
a− b
a− c
b− a
c
)=
5(a− b)(a− c)(c− b)abc
≥ 0.
Hence, it suffices to prove that P (a, c, b) ≥ 0 which is equivalent toP (a, b′, c′) ≥ 0 with a ≥ b′ ≥ c′. Therefore, we only need to consider thecase a ≥ b ≥ c in order to prove the original inequality. We may usesimilar methods to explain the reason why we can assume that a ≥ b ≥ cor c ≥ b ≥ a in proving cyclic inequalities.
15. If a, b, c are the side-lengths of a triangle, then
3
(a2
b2+b2
c2+c2
a2
)≥(a2 + b2 + c2
)( 1
a2+
1
b2+
1
c2
).
Solution: Put x = a2, y = b2, z = c2 and assume that x ≥ y ≥ z (wecan argue this like in the previous problem), then we have that y ≤ x ≤(√y +√z)2
since a, b, c are the side-lengths of a triangle. Our inequality
is equivalent to 3
(x
y+y
z+z
x
)≥ (x+ y + z)
(1
x+
1
y+
1
z
),which can
be rewritten as
f(x) = (2z − y)x2 + (2y2 − 3yz − z2)x+ 2yz2 − y2z ≥ 0.
If y ≥ 2z, then f(x) is a quadratic polynomial of x with the highest coeffi-
cient is 2z−y ≤ 0, therefore, it suffices to prove that min{f(y), f
((√y +√z)2)} ≥
0 which is true because f(y) = y(y − z)2 ≥ 0, and
f((√
y +√z)2)
=
= (2z − y)(√y +√z)4
+ (2y2 − 3yz − z2)(√y +√z)2
+ 2yz2 − y2z= (2c2 − b2)(b+ c)4 + (2b4 − 3b2c2 − c4)(b+ c)2 + 2b2c4 − b4c2
= b6 − 6b4c2 − 2b3c3 + 9b2c4 + 6bc5 + c6
= (b3 − 3bc2 − c3)2 ≥ 0.
38
Now, let use consider the case 2z ≥ y, then by AM-GM inequality, wehave that x2 + y2 ≥ 2xy. Therefore
f(x) = (2z − y)x2 + (2y2 − 3yz − z2)x+ 2yz2 − y2z≥ (2z − y)(2xy − y2) + (2y2 − 3yz − z2)x+ 2yz2 − y2z= xz(y − z) + y(y − z)(y − 2z) ≥ yz(y − z) + y(y − z)(y − 2z)
= y(y − z)2 ≥ 0.
Our proof is complete.
16. If a, b, c, d are nonnegative real numbers such that a+ b+ c+d = 1, then
abc+ bcd+ cda+ dab ≤ 1
27+
176
27abcd.
Solution: Put x = ab, then 0 ≤ x ≤ (a+b)2
4 , our inequality becomes
f(x) =
(c+ d− 176
27cd
)x + cd(a + b) − 1
27≤ 0. This is a quadratic
polynomial of x with the highest coefficient is 0, therefore, it suffices to
prove that max
{f(0), f
((a+ b)2
4
)}≤ 0. It shows that it suffices to
consider the cases a = b or ab = 0. Similarly, we also obtain that itsuffices to consider the cases c = d or cd = 0. Combining both thesestatements, we obtain that, in order to prove the orginal inequality, itenough to prove it in the following cases
Case 1. If b = d = 0, then the inequality is trivial since abc+bcd+cda+
dab− 176
27abcd = 0.
Case 2. If a = b, d = 0, then c = 1 − 2a ≥ 0, the inequality becomes
a2(1 − 3a) ≤ 1
27which is true since by AM-GM inequality, we have
a2(1− 3a) ≤[a+ a+ (1− 2a)
3
]3=
1
27.
Case 3. If a = b, c = d, then c = 1−2a2 , the inequality becomes a2(1 −
2a) + 12a(1 − 2a)2 ≤ 1
27 + 4427a
2(1 − 2a)2, which is equivalent to (4a −1)2(22a2 − 11a + 2) ≥ 0. The last inequality is valid since by AM-GMinequality, we have 22a2 + 2 ≥ 4
√11a ≥ 11a. This ends our proof.
17. If a, b, c, d are nonnegative real numbers such that a+ b+ c+d = 4, then
(1 + 3a)(1 + 3b)(1 + 3c)(1 + 3d) ≤ 125 + 131abcd.
Solution: Proceeding similar to the previous problem, we see that itenough to prove it in the following cases
Case 1. If b = d = 0, then the inequality is trivial since
(1 + 3a)(1 + 3b)(1 + 3c)(1 + 3d) = (1 + 3a)(1 + 3c) ≤ 1
4(2 + 3a+ 3c)2
= 49 < 125 = 125 + 131abcd.
39
Case 2. If a = b, d = 0, then c = 4− 2a ≥ 0, and the inequality becomes(1 + 3a)2(13 − 6a) ≤ 125 which is true since by AM-GM inequality, wehave that
(1 + 3a)2(13− 6a) ≤[
(1 + 3a) + (1 + 3a) + (13− 6a)
3
]3= 125.
Case 3. If a = b, c = d, then c = 2− a ≥ 0, and the inequality becomes(1 + 3a)2(7− 3a)2 ≤ 125 + 131a2(2− a)2, which is equivalent to (25a2−50a+ 38)(a− 1)2 ≥ 0 which is true. Our proof is complete.
18. Let a, b, c, d be nonnegative real numbers with sum 4, prove that 3(a2 +b2 + c2 + d2) + 4abcd ≥ 16.
Solution: Again, proceeding similar to the previous problem, we see thatit enough to prove it in the following cases
Case 1. If b = d = 0, then the inequality is trivial since
3(a2 + b2 + c2 + d2) + 4abcd = 3(a2 + c2) ≥ 3
2(a+ c)2 = 24 > 16.
Case 2. If a = b, d = 0, then c = 4 − 2a and the inequality becomes3[2a2 + (4− 2a)2
]≥ 16, which is true since by Cauchy Schwartz in-
equality (see at the next chapter), we have that
2a2 + (4− 2a)2 ≥ 1
3[a+ a+ (4− 2a)]2 =
16
3.
Case 3. If a = b, c = d, then c = 2 − a ≥ 0 and the inequality becomes6a2+6(2−a)2+4a2(2−a)2 ≥ 16, or equivalently (a2−2a+2)(a−1)2 ≥ 0which is trivial. The proof is complete.
19. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Provethat
1
6− ab+
1
6− bc+
1
6− ca≤ 3
5.
Solution: By expanding, the inequality is equivalent to 108 + 13abc(a+b+ c)− 48(ab+ bc+ ca)− 3a2b2c2 ≥ 0, or 36 + 13abc− a2b2c2 − 16(ab+
bc+ ca) ≥ 0. Put x = ab, then 0 ≤ x ≤ (a+ b)2
4, our inequality becomes
f(x) = −c2x2 + (13c − 16)x + 36 − 16c(3 − c) ≥ 0. This is a quadraticpolynomial of x with the highest coefficient is −c2 ≤ 0, therefore, it
suffices to prove that min{f(0), f
((a+b)2
4
)}≥ 0. It shows that it suffices
to consider the cases a = b or ab = 0.
Case 1. If b = 0, then a = 3−c, and the inequality becomes 36−16c(3−c) ≥ 0, or equivalently 4(3− 2c)2 ≥ 0.
Case 2. If a = b, then c = 3 − 2a ≥ 0, and the inequality becomes36 + 13a2(3− 2a)− a4(3− 2a)2− 16a2− 32a(3− 2a) ≥ 0, or equivalently−4a6 + 12a5− 9a4− 26a3 + 87a2− 96a+ 36 ≥ 0, or (3− 2a)(a+ 2)(2a2−3a+ 6)(a− 1)2 ≥ 0. Our proof is complete.
40
20. Let a, b, c, d be nonnegative real numbers, no three of which are zero.Show that√
a
a+ b+ c+
√b
b+ c+ d+
√c
c+ d+ a+
√d
d+ a+ b≤ 4√
3.
Solution: By Cauchy Schwarz inequality, we have(∑cyc
√a
a+ b+ c
)2
≤
≤
[∑cyc
(a+ b+ d)(a+ c+ d)
][∑cyc
a
(a+ b+ c)(a+ b+ d)(a+ c+ d)
]
=2[2(a+ b+ c+ d)2 + (a+ c)(b+ d)][(a+ c)(b+ d) + ac+ bd]
(a+ b+ c)(b+ c+ d)(c+ d+ a)(d+ a+ b).
We need to prove that
P (a, b, c, d) = 8(a+ b+ c)(b+ c+ d)(c+ d+ a)(d+ a+ b)
−[2(a+ b+ c+ d)2 + (a+ c)(b+ d)][(a+ c)(b+ d) + ac+ bd] ≥ 0.
It is easy to see that P (a, b, c, d) is a quadratic polynomial of x = ac withthe highest coefficient is 0, therefore, in order to prove P (a, b, c, d) ≥ 0,it suffices to prove it in the case a = c or ac = 0. Similarly, P (a, b, c, d)is also a quadratic polynomial of t = bd with the highest coefficient is 0,therefore, in order to prove P (a, b, c, d) ≥ 0, it suffices to prove it in thecase b = d or bd = 0. Combining all cases, we see that it is enough toprove the inequality in the following cases
Case 1. c = d = 0, then the inequality becomes 8ab(a+ b)2 ≥ 3ab(2a2 +5ab+ 2b2), or equivalently ab(2a2 + ab+ 2b2) ≥ 0.
Case 2. a = c, d = 0, then the inequality becomes 16a(2a+ b)(a+ b)2 ≥6a(a+ b)(a+ 2b)(4a+ b), or equivalently 2a(a+ b)(4a2− 3ab+ 2b2) ≥ 0.
Case 3. a = c, b = d, then the inequality becomes 8(2a+ b)2(a+ 2b)2 ≥12(2a2+5ab+2b2)(a2+4ab+b2), or equivalently 4(a+2b)(2a+b)(a−b)2 ≥0.Our proof is now complete. Equality holds if and only if a = b = c = d.
1.7 Cauchy-Schwartz Inequality
The inequality
(a21 + a22 + . . .+ a2n)(b21 + b22 + . . .+ b2n) ≥ (a1b1 + a2b2 + . . .+ anbn)2
has incredibly many applications.Especially useful is the following form of the inequality:CBS Lemma: For yi > 0,
x21y1
+ . . .+x2nyn≥ (x1 + x2 + . . .+ xn)2
y1 + y2 + . . .+ yn
41
It can be easily noticed that it’s equivalent to CBS inequality, and it can beproven also by induction on n.
With this lemma, inequalities like the following become straightforward:
Example 1. If a, b, c > 0 then
a
a+ 2b+
b
b+ 2c+
c
c+ 2a≥ 1.
Indeed, multiplying the denominator and numerator of the first fraction witha, of the second with b, and of the third with c, we get
a2
a2 + 2ab+
b2
b2 + 2bc+
c2
c2 + 2ca≥ (a+ b+ c)2
a2 + 2ab+ b2 + 2bc+ c2 + 2ca= 1.
The importance of this inequality is revealed by the multitude of exercises ithelps solve.
Exercises
1. If x, y, z ≥ 0 show that√
(x+ y)(x+ z) ≥ x+√yz.
Solution: Immediate from Cauchy-Schwartz for√x,√y and
√x,√z.
2. Show that n(x21 + x22 + . . .+ x2n) ≥ (x1 + x2 + . . .+ xn)2
Solution: In the form (1 + 1 + . . . + 1)(x21 + x22 + x23 + . . . + x2n) ≥(x1 + . . .+ xn)2, Cauchy-Schwartz is clearly seen.
3. For a, b, c > 0, show thata
a+ 2b+ c+
b
b+ 2c+ a+
c
c+ 2a+ b≥ 3
4.
Solution: Amplify the first fraction by a, the second by b, the third byc and apply CBS Lemma.
4. For a, b, c, d, e, f > 0 show thata
b+ c+
b
c+ d+
c
d+ e+
d
e+ f+
e
f + a+
f
a+ b≥ 3.
Solution: We again amplify by the corresponding numerators and applyCBS Lemma, obtaining
a
b+ c+
b
c+ d+
c
d+ e+
d
e+ f+
e
f + a+
f
a+ b≥
≥ (a+ b+ c+ d+ e+ f)2
a(b+ c) + b(c+ d) + c(d+ e) + d(e+ f) + e(f + a) + f(a+ b).
It remains to show that
a(b+c)+b(c+d)+c(d+e)+d(e+f)+e(f+a)+f(a+b) ≤ (a+ b+ c+ d+ e+ f)2
3.
However this can be written as (a+d)(b+e)+(b+e)(c+f)+(c+f)(a+d)
and we apply the inequality uv + vw + wu ≤ (u+ v + w)2
3.
42
5. If1
x+
1
y+
1
z= 2 then
√x+ y + z ≥
√x− 1 +
√y − 1 +
√z − 1.
Solution: By CBS,(√x− 1 +
√y − 1 +
√z − 1
)2≤
≤ (x+ y + z)
(x− 1
x+y − 1
y+z − 1
z
)= x+ y + z.
6. If abc = 1, a, b, c > 0 thena
1 + b+ c+
b
1 + c+ a+
c
1 + a+ b≥ 1.
Solution: Amplifying the fractions by a, b, c we get
a2
a+ ab+ ac+
b2
b+ ab+ bc+
c2
c+ ac+ bc≥ (a+ b+ c)2
a+ b+ c+ 2(ab+ bc+ ca)
≥ (a+ b+ c)2
(a+ b+ c)2= 1.
The last inequality follows from the fact that a2 + b2 + c2 ≥ a+ b+ c asa2 + 1 + b2 + 1 + c2 + 1 ≥ 2a+ 2b+ 2c and a+ b+ c ≥ 3.
7. Let n ≥ 3, a1, a2, . . . , an ≥ 0 with a21 + a22 + . . .+ a2n = 1. Prove that
a1a22 + 1
+a2
a23 + 1+ . . .+
ana21 + 1
≥ 4
5(a1√a1 + a2
√a2 + . . .+ an
√an)2 .
Solution: Amplifying the fractions by a2i and applying the CBS Lemmawe get
a1a22 + 1
+a2
a23 + 1+ . . .+
ana21 + 1
≥(a1√a1 + a2
√a2 + . . .+ an
√an)2
a21a22 + a21 + . . .+ a2na
21 + a2n
.
So we are left to show that a21a22 + . . .+a2na
21 ≤
1
4which we have already
proven in exercise 7 at induction.
8. Show that
x
x+√
(x+ y)(x+ z)+
y
y +√
(y + z)(y + x)+
z
z +√
(z + x)(z + y)≤ 1.
Solution: We have√
(x+ y)(x+ z) =√
(x+ y)(z + x) ≥√xz +
√xy
hence
x
x+√
(x+ y)(x+ z)≤ x
x+√xy +
√xz
=
√x√
x+√y +√z.
Summing with the analogously deduced inequalities we get the result.
43
9. If a, b, c ≥ 0 thena
1 + a+
b
1 + b+
c
1 + c+
1
1 + a+ b+ c≥ 1.
Solution:
a2
a2 + a+
b2
b2 + b+
c2
c2 + c+
1
a+ b+ c+ 1≥
≥ (a+ b+ c+ 1)2
a2 + a+ b2 + b+ c2 + c+ a+ b+ c+ 1= 1.
10. If m ≥ n ≥ p ≥ q are integers numbers with m + q = n + p anda1, a2, . . . , ak are positive reals then(
k∑i=1
ami
)(k∑i=1
aqi
)≥
(k∑i=1
ani
)(k∑i=1
api
).
Solution: Let sr =
k∑i=1
ari . We must prove that smsq ≥ snsp. For
n = p = n− 1 = q+ 1 we have sn+1sn−1 ≥ s2n, which is true in virtue of
CBS inequality. Rewriting this assn+1
sn≥ snsn−1
. Therefore rn =sn+1
snis increasing in n. Hence rm−1rm−2 . . . rn ≥ rp−1rp−2 . . . rq (recall that
m− n = p− q) which means thatsmsn≥ spsq
or smsq ≥ snsp.
11. If n, k > l ∈ N, a1, a2, . . . , an ≥ 0 thenk
√ak1 + . . .+ akn
n≥ l
√al1 + . . .+ aln
n.
Solution: We shall use induction on k − l. If k = l this is obvious.Rephrasing the problem using the notation in the previous exercise weneed to show that slkn
k−l ≥ skl . Now remember that we proved that
ri =si+1
si≥ ri−1 =
sisi−1
. From here we deduce that
rlk ≥ rl−1rl−2 . . . r0 or
(sk+1
sk
)l≥ sln
or
(sk+1
sk
)ln ≥ sl,
and this relation performs the induction step on k − l: multiplying itslkn
k−l ≥ skl we get slk+1nk−l+1 ≥ sk+1
l .
12. (Minkowski) If ai, bi are reals then
n∑i=1
√a2i + b2i ≥
√√√√( n∑i=1
ai
)2
+
(n∑i=1
bi
)2
.
Solution: If n = 2 the inequality is equivalent to√a2 + b2 +
√c2 + d2 ≥
√(a+ c)2 + (b+ d)2,
which after squaring and cancelling common terms becomes√(a2 + b2)(c2 + d2) ≥ ac+ bd,
which is CBS. The general case follows by induction on n.
44
13. If a, b, c > 0 add up to 1, thena+ 1
c(2− b)+
b+ 1
a(2− c)+
c+ 1
b(2− a)≥ 36
5.
Solution:
1
2− a=
1
2 (a+ b+ c)− a=
1
2b+ 2c+ a≤ 1
52
(2
b+
2
c+
1
a
),
where the last inequality is because
(2
b+
2
c+
1
a
)(2b+ 2c+ a) ≥ 52
(Cauchy-Schwartz). Similarly,
1
2− b≤ 1
52
(2
c+
2
a+
1
b
),
1
2− c≤ 1
52
(2
a+
2
b+
1
c
).
Summing up results in
1
2− a+
1
2− b+
1
2− c≤
≤ 1
52
(2
b+
2
c+
1
a
)+
1
52
(2
c+
2
a+
1
b
)+
1
52
(2
a+
2
b+
1
c
)=
1
52
(5
a+
5
b+
5
c
)=
1
5
(1
a+
1
b+
1
c
).
Thus, using a+ 1 = 2− (1− a) = 2− (b+ c) = (2− b)− c and similarlyb+ 1 = (2− c)− a and c+ 1 = (2− a)− b, we get
a+ 1
c (2− b)+
b+ 1
a (2− c)+
c+ 1
b (2− a)=
=(2− b)− cc (2− b)
+(2− c)− aa (2− c)
+(2− a)− bb (2− a)
=
(1
c− 1
2− b
)+
(1
a− 1
2− c
)+
(1
b− 1
2− a
)=
(1
a+
1
b+
1
c
)−(
1
2− a+
1
2− b+
1
2− c
)≥
(1
a+
1
b+
1
c
)− 1
5
(1
a+
1
b+
1
c
)=
4
5
(1
a+
1
b+
1
c
).
So it remains to show that 45
(1
a+
1
b+
1
c
)≥ 36
5, i. e. that
1
a+
1
b+
1
c≥ 9.
But this is clear, since Cauchy-Schwartz gives (a+ b+ c)
(1
a+
1
b+
1
c
)≥
9, and a+ b+ c = 1. This completes the proof.
14. Let a, b, c, x, y, z be reals such that
(a+ b+ c)(x+ y + z) = 3 and (a2 + b2 + c2)(x2 + y2 + z2) = 4.
Show that ax+ by + cz ≥ 0
Solution: Set u = ax + by + xz. As (a, b, c) and (x, y, z) are unrelatedin the condition, we can also take its ”brothers”: v = ay + bz + cx,w = az + bx+ cy. Then u+ v + w = (a+ b+ c)(x+ y + z) = 3.
u2 + v2 + w2 = (a2 + b2 + c2)(x2 + y2 + z2) + 2(ab+ bc+ ca)(xy + yz + zx)
= 4 + 2(ab+ bc+ ca)(xy + yz + xz).
45
Assume now that u < 0, then v+w ≥ 3 so a(y+z)+b(x+z)+c(x+y) > 3.Applying CBS we get
(a2+b2+c2)[2(x2 + y2 + z2) + 2(xy + yz + zx)
]≥ 9
4(a2+b2+c2)(x2+y2+z2).
This means that xy+ yz+ zx ≥ x2 + y2 + z2
8and analogously ab+ bc+
ca ≥ a2 + b2 + c2
8. This estimates are too rough, however they tell us
that ab + bc + ca, xy + yz + zx are positive, and so we can apply CBSin the following form:[
a2 + b2 + c2 + 2(ab+ bc+ ca)] [x2 + y2 + z2 + 2(xy + yz + zx)
]≥
≥[√
(a2 + b2 + c2)(x2 + y2 + z2) + 2√
(ab+ bc+ ca)(xy + yz + zx)]2
or 9 ≥[2 + 2
√(ab+ bc+ ca)(xy + yz + zx)
]2which implies
(ab+ bc+ ca)(xy + yz + zx) ≤ 1
4.
It suffices to derive a contradiction now: we have
u2 + v2 + w2 = 4 + 2(ab+ bc+ ca)(xy + yz + xz) ≤ 9
2,
however from the other side u2 + v2 + w2 > v2 + w2 ≥ (v + w)2
2>
9
2.
15. If a, b, c, x, y, z > 0 then
a
b+ c(y + z) +
b
a+ c(x+ z) +
c
a+ b(x+ y) ≥ 3
xy + yz + zx
x+ y + z.
Solution: We add (y+ z) + (x+ z) + (x+ y) to both sides to get a morecomfortable form of the LHS:
(a+b+c)
(y + z
b+ c+z + x
c+ a+x+ y
a+ b
)=
2(x+ y + z)2 + 3(xy + yz + zx)
x+ y + z.
Now from CBS we have
(a+ b+ c)
(y + z
b+ c+z + x
c+ a+x+ y
a+ b
)=
=1
2[(b+ c) + (c+ a) + (a+ b)]
(y + z
b+ c+z + x
c+ a+x+ y
a+ b
)≥ 1
2
(√y + z +
√z + x+
√x+ y
)2.
So it suffices to prove that(√y + z +
√z + x+
√x+ y
)2 ≥ 4(x+ y + z) + 6xy + yz + zx
x+ y + z,
46
or
2(x+ y + z) + 2∑cyc
√(x+ y)(x+ z) ≥ 4(x+ y + z) + 6
xy + yz + zx
x+ y + z
or√(x+ y)(x+ z)+
√(y + z)(y + x)+
√(z + x)(z + y) ≥ (x+y+z)+3
xy + yz + zx
x+ y + z.
However√(x+ y)(x+ z) +
√(y + z)(y + x) +
√(z + x)(z + y) =
=√x2 + (xy + yz + zx) +
√y2 + (xy + yz + zx) +
√z2 + (xy + yz + zx)
≥√
(x+ y + z)2 + 9(xy + yz + zx)
from Minkowski. The fact that√(x+ y + z)2 + 9(xy + yz + zx) ≥ (x+ y + z) + 3
xy + yz + zx
x+ y + z
follows by squaring.
16. If a1, a2, . . . , an are positive numbers that sum to 1, then(a1
a22 + a2+
a2a23 + a3
+ . . .+an
a21 + a1
)(a1a2+a2a3+. . .+ana1) ≥
n
n+ 1.
Solution: We may note thatai
a2i+1 + ai+1=
aiai+1
− aiai+1 + 1
, thus
a1a22 + a2
+a2
a23 + a3+ . . .+
ana21 + a1
=
=
(a1a2
+a2a3
+ . . .+ana1
)−(
a1a2 + 1
+ . . .+an
a1 + 1
).
But
(a1a2
+a2a3
+ . . .+ana1
)(a1a2 + . . .+ana1) ≥ 1 so it suffices to prove
that
a1a2
+a2a3
+ . . .+ana1≥ (n+ 1)
(a1
a2 + 1+ . . .+
ana1 + 1
).
However we may note thata1a2
+ . . .+ana1≥ n but as
(a2 + 1)
(1
n2a2+ 1
)≥(n+ 1
n
)2
(CBS)
we deducea1
a2 + 1≤ 1
(n+ 1)2a1a2
+n2
(n+ 1)2a1. Summing with the anal-
ogous relations we get the result.
47
1.8 Young Inequality
In the AM-GM inequality put m terms equal a and n terms equal b. Wededuce
ma+ nb ≥ (m+ n)am
m+n bn
m+n .
Now setting p =m+ n
m, q =
m+ n
n, x = a
mm+n , y = b
nm+n we rewrote the
inequality asxp
p+yq
q≥ pq for any a, b > 0, p, q ∈ Q with 1
p + 1q = 1. We can
extend this inequality to any positive real p, q: taking sequences of rationals
pn, qn with1
pn+
1
qn= 1 and pn tending to p, qn tending to q we deduce
xpn
pn+yqn
qn≥ xy. Passing to limit we have
Theorem. If p, q > 0 with 1p + 1
q > 0 then for any positive reals x, y > 0
xp
p+yq
q≥ xy.
This is called Young’s inequality.
Exercise: a) Prove that Young Inequality changes sign when one of p, q is
negative. (Hint: if, for example, q < 0, use normal Young Inequality for bq
1−q ,
(ab)qq−1 and weights 1− q, 1− q
q).
b) Prove the generalization of Young Inequality:
If p1, p2, . . . , pn, x1, x2, . . . , xn > 0 and p1 + p2 + . . .+ pn = 1 then
p1x1 + . . .+ pnxn ≥ xp11 . . . xpnn
(hint: use induction).
Take now some A,B, a1, a2, . . . , an, b1, b2, . . . , bn positive reals. Then
aibiAB≤ 1
p
(aiA
)p+
1
q
(biB
)q.
Summing this over all i we have
a1b1 + . . .+ anbnAB
≤ 1
p
n∑i=1
api
A
p
+1
q
n∑i=1
bpi
B
q
.
Taking now A =
(n∑i=1
api
)1p
, B =
(n∑i=1
nbqi
)1q
we deduce
Theorem. If p, q > 0 and 1p+1
q = 1 then for any positive reals a1, . . . , an, b1, . . . , bnthe following inequality holds:
48
n∑i=1
aibi ≤
(n∑i=1
api
)1p(∑i=1
nbqi
)1q
This is Holder’s Inequality. Like Young’s Inequality, when one of p, q isnegative it changes sign. [Note: for p = q = 2 this is Cauchy-Schwartz].Take p > 1 and apply Holder for the numbers ai, (ai + bi)
p−1 with weights
p,p
p− 1. We deduce
n∑i=1
ai(ai + bi)p−1 ≤
(n∑i=1
api
)1p[
n∑i=1
(ai + bi)p
]p−1p
.
Writing the analogous inequality for bi, (ai + bi)p−1 and summing we obtain
Minkovski’s Inequality:(n∑i=1
(ai + bi)p
)1p
≤
(n∑i=1
api
)1p
+
(n∑i=1
bpi
)1p
.
When 0 < p < 1 it changes sign.
Exercises
1. Show that for any x ≥ y ≥ 0,
(ax1 + . . .+ axn)(a−x1 + . . .+ a−xn ) ≥ (ay1 + . . .+ ayn)(a−y1 + . . .+ a−yn ).
Solution: By Holder,
(ax1 + . . .+ axn)yx (1 + 1 + . . .+ 1)
x−yx ≥ (ay1 + . . .+ ayn),
and analogously
(a−x1 + . . .+ a−xn )yx (1 + 1 + . . .+ 1)
x−yx ≥ (a−y1 + . . .+ a−yn ).
The inequality follows from multiplying these two and using the factthat
(ax1 + . . .+ axn)(a−x1 + . . .+ a−xn ) ≥ n2.
2. For a, b, c > 0√a− ab+ b2 +
√a2 − ac+ c2 ≥
√b2 + bc+ c2.
Solution: we write it as√(b
2− a)2
+
(b
2
)2
+
(b
2
)2
+
(b
2
)2
+
√(a− c
2
)2+( c
2
)2+( c
2
)2+( c
2
)2≥
≥
√(b
2− c
2
)2
+
(b
2+c
2
)2
+
(b
2+c
2
)2
+
(b
2+c
2
)2
,
and it now follows from Minkowski.
49
3. (a2 + ab+ b2)(b2 + bc+ c2)(c2 + ca+ a2) ≥ (ab+ bc+ ca)3
Solution: Write it as
(a2 + ab+ b2)(ac+ a2 + c2)(c2 + b2 + bc) ≥ (ab+ bc+ ca)3.
By CBS
(a2 + ab+ b2)(ac+ a2 + c2) ≥(a32 c
12 + a
32 b
12 + bc
)2
and
(a32 c
12 + a
32 b
12 + bc
)2
(c2 + b2 + bc) ≥ (ab+ bc+ ca)3 by Holder for
p =3
2, q = 3.
Note: by the very same method we can prove by induction the general-ized version of Holder for more that two sequences of variables.
1.9 Advanced techniques with Cauchy-Buniakowski-Schwarz and Holder Inequalities
In the previous parts, we have presented the Cauchy-Schwartz inequality andthe Holder’s inequality with their basic techniques in use. These inequalitiesare so important, however, that they deserve an additional chapter for finerapplications.
Firstly, let us recall the Cauchy-Schwartz inequality and the Holder’s inequal-ity.
Theorem 1. (Cauchy-Schwartz inequality) For any real numbers (a1, a2, . . .,an) and (b1, b2, . . ., bn), we have
(a1b1 + a2b2 + . . .+ anbn)2 ≤ (a21 + a22 + . . .+ a2n)(b21 + b22 + . . .+ b2n).
The equality holds if and only if ai : aj = bi : bj ∀i, j ∈ {1, 2, . . . , n} .
Corollary 1. (CBS lemma) For any real numbers (a1, a2, . . ., an) and (b1, b2,. . ., bn) with bi > 0 ∀i = 1, 2, . . . , n, we have
a21b1
+a22b2
+ . . .+a2nbn≥(a1 + a2 + . . .+ an)2
b1 + b2 + . . .+ bn.
The equality holds if and only if ai : aj = bi : bj ∀i, j ∈ {1, 2, . . . , n} .
Theorem 2. (Generalized Holder’s inequality) Given positive reals xij (i =1,m, j = 1, n). Then, for all ω1, . . ., ωn ≥ 0 satisfy ω1 + . . .+ ωn = 1, we have
n∏i=1
m∑j=1
xij
ωj
≥m∑j=1
(n∏i=1
xωj
ij
).
50
A special case of Holder’s inequality which we used to apply is(m∑i=1
ani
)(m∑i=1
bni
)n−1≥
(m∑i=1
aibn−1i
)n,
for all ai, bi ≥ 0. The equality holds in this inequality whena1b1
= . . . =ambm
.
In most advanced inequalities it is not immediately clear how to apply to CBSor Holder. One needs to do some more work to ”prepare” it for their use. Inthe following, we will describe several common techniques that help one findthe way to apply Cauchy-Schwartz and Holder to problems. These techniquesare not restricted only to Cauchy-Schwartz and Holder, they work for otherinequalities as well, but in this paragraph we will only focus on these twoparticular inequalities.
Balancing coefficients
Grouping the coefficients to use classical inequalities is not always easy, espe-cially in non-symmetric inequalities. In these cases, the coefficients of similarterms are normally not equal to each other and therefore there is confusionabout not only the use basic inequalities properly but also taking care of thecase of equality so that this case is valid throughout the process of solution.One way to deal with this is to use additional variables (”weights”) in orderto apply some inequality with unknown parameters, and the solve a system ofequations to find assign the corresponding values to the weights. Knowing anequality case or what should be in the right-hand side of the inequality usuallygives information that is helpful in finding them. This method is called thebalancing coefficients technique. It is mainly applied in conjunction with twoinequalities the AM-GM inequality and Cauchy Schwartz/Holder inequality.In some sense, Holder as deduced from Young’s inequality is a consequence ofthis method for AM-GM: in the inequality ap
p + bq
q ≥ ab, p and q are ”weights”
with respect to which weighted AM-GM is applied (i.e. ap appears ”1p times”
and bq appears ”1q times”).
We will focus on applying the technique for Cauchy-Schwarts/Holder inequal-ities in the examples.
Example 1. Given nonnegative real numbers x, y, z satisfying 2x+ 3y+ z = 1.Determine the minimum value of P = x3 + y3 + z3.
Solution: According to the Holder’s inequality, we have that for all a, b, c ≥ 0,
(x3 + y3 + z3)(a3 + b3 + c3)2 ≥ (xa2 + yb2 + zc2)3.
It follows that
P = x3 + y3 + z3 ≥ (xa2 + yb2 + zc2)3
(a3 + b3 + c3)2.
Now, let us choose a, b, c, so that we can apply the given hypothesis 2x+ 3y+z = 1 to the above inequality, and therefore, naturally, we will choose a, b, c,
51
such thata2
2=b2
3=c2
1= 1, or a =
√2, b =
√3, c = 1. The equality holds
whenx
a=y
b=z
c, or
x
a=y
b=z
c=
2x+ 3y + z
2a+ 3b+ c=
1
2a+ 3b+ c, or
x =a
2a+ 3b+ c, y =
b
2a+ 3b+ c, z =
c
2a+ 3b+ c.
From now, we have
P ≥ (xa2 + yb2 + zc2)3
(a3 + b3 + c3)2=
(2x+ 3y + z)3[(√2)3
+(√
3)3
+ 13]2 =
1(2√
2 + 3√
3 + 1)2 .
Since the equality can be attained, the minimum value of P is1(
2√
2 + 3√
3 + 1)2 .
Example 2. Given nonnegative real numbers x, y such that x3 + y3 = 1. Findthe maximum value of P = x+ 2y.
Solution: This is in a sense a converse to the previous example. By Holder’sinequality, we have that for all a, b ≥ 0
(x3 + y3)(a3 + b3)2 ≥ (xa2 + yb2)3,
hence
xa2 + yb2 ≤ 3
√(x3 + y3)(a3 + b3)2.
Naturally we will choose a, b so that xa2 + yb2 ≡ P, which gives us a = 1, b =√
2. The equality holds whenx
a=y
b, or
x3
a3=y3
b3=x3 + y3
a3 + b3=
1
a3 + b3, or
x3 =a3
a3 + b3, y3 =
b3
a3 + b3.
Then
P ≤ 3
√(x3 + y3)
[13 +
(√2)3]2
=3
√(2√
2 + 1)3.
Again equality is attained and we obtain that the maximum of P is3
√(2√
2 + 1)3.
Example 3. Let a1, a2, . . . , an be positive real numbers. Prove that
n∑k=1
kk∑j=1
aj
≤ 2
n∑k=1
1
ak.
Solution: By the Cauchy Schwartz inequality,
(a1+a2+ . . .+ ak)
(b21a1
+b22a2
+ . . .+b2kak
)≥ (b1 + b2 + . . .+ bk)
2,
52
where bi are arbitrary positive real numbers for all i = 1, 2, . . . , n. Therefore
kk∑j=1
aj
≤ k
(b1 + b2 + . . .+ bk)2
(b21a1
+b22a2
+ . . .+b2kak
),
from which we deduce thatn∑k=1
kk∑j=1
aj
≤n∑i=1
ciai,
where
ck =kb2k
(b1 + b2 + . . .+ bk)2+
(k + 1)b2k(b1 + b2 + . . .+ bk+1)2
+. . .+nb2k
(b1 + b2 + . . .+ bk+1)2,
for all k = 1, 2, . . . , n. Choosing now bk = k, we get
ck =k3(k∑i=1
i
)2 +k2(k + 1)(k+1∑i=1
i
)2 + . . .+k2n(n∑i=1
i
)2
= 4k2(
1
k(k + 1)2+
1
(k + 1)(k + 2)2+ . . .+
1
n(n+ 1)2
)= 4k2
(1
k(k + 1)+ . . .+
1
n(n+ 1)− 1
(k + 1)2− . . .− 1
(n+ 1)2
)< 4k2
(1
2
(1
k2+
1
(k + 1)2
)+ . . .+
1
2
(1
n2+
1
(n+ 1)2
)− 1
(k + 1)2− . . .− 1
(n+ 1)2
)= 4k2
(1
2k2+
1
2(n+ 1)2+
1
(k + 1)2+ . . .+
1
n2− 1
(k + 1)2− . . .− 1
(n+ 1)2
)= 4k2
(1
2k2− 1
2(n+ 1)2
)< 2,
for all k = 1, 2, . . . , n. In this case, it follows that
n∑k=1
kk∑j=1
aj
≤ 2
n∑i=1
1
ai,
which ends the proof.
Example 4. Given positive real numbers a1, a2, . . . , an. Prove that
4(a21 + a22 + . . .+ a2n) ≥ a21 +
(a1 + a2
2
)2
+ . . .+
(a1 + a2 + . . .+ an
n
)2
.
Solution: By the Cauchy Schwartz inequality, for all k = 1, 2, . . . , n, bi > 0, wehave (
a21b1
+a22b2
+ . . .+a2kbk
)(b1 + b2 + . . .+ bk) ≥ (a1 + a2 + . . .+ ak)
2,
53
It follows that(a1 + a2 + . . .+ ak
k
)2
≤ b1 + b2 + . . .+ bkk2
·(a21b1
+a22b2
+ . . .+a2kbk
),
orn∑k=1
(a1 + a2 + . . .+ ak
k
)2
≤n∑k=1
cka2k,
where
ck =b1 + b2 + . . .+ bk
k2bk+b1 + b2 + . . .+ bk+1
(k + 1)2bk+ . . .+
b1 + b2 + . . .+ bnn2bk
.
Now, choosing bk =√k −√k − 1, we obtain
ck =b1 + b2 + . . .+ bk
k2bk+b1 + b2 + . . .+ bk+1
(k + 1)2bk+ . . .+
b1 + b2 + . . .+ bnn2bk
=1√
k −√k − 1
(1
k3/2+
1
(k + 1)3/2+ . . .+
1
n3/2
)
≤ 1√k −√k − 1
2
1√k − 1
2
− 1√k + 1
2
+ . . .+ 2
1√n− 1
2
− 1√n+ 1
2
=
1√k −√k − 1
1√k − 1
2
− 1√n+ 1
2
<2√
k − 12
(√k −√k − 1
)=
2√
2(√
k +√k − 1
)√
2k − 1≤ 4.
The inequality is proved.
This last example, by far the hardest, requires some explanation - namely,how to arrive to the choice bk =
√k −√k − 1? There is no universal method
to choose these coefficients, and hard problems require a certain degree ofcreativity from the solver. In this case, there are several reasons that point tothe correct solution.One way would be to look at the equality case. We don’t know it (and infact it is never achieved unless all variables are zero), but we can produce an”asymptotic” case, i.e. an examples where the two expression are close to eachother. The choice to take all ai equal fails. Then, because the inequality isnot symmetric, one can try to choose ai as a function of i, and the easiestone is ai = ik. This is good, because the sum of ik is approximately ik+1
k+1 fork > −1 (this is done by tricks as the integral approximation or the Bernoulli
polynomials), so that a1 + . . .+ an would be approximately nk+1
k+1 .
One therefore gets 4∑i2k ∼ 4
2k+1n2k+1 (for 2k > −1) and we compare it to∑
( ik
k+1)2 ∼ 1(k+1)2(2k+1)
n2k+1. Therefore, in order for them to be equal we
need (k + 1)2 = 14 so k = −1
2 . Note that we cannot do any better because weneed 2k > −1 - in fact we are already on the border and violate this conditionfor k = −1
2 . This case is still very close to the equality anyway, as checkeddirectly.
54
So now if we want the method to work, we better choose bi that ”approxi-mately” respect the inequality case, i.e. that are ”approximately proportional”to ai = 1√
i. Such a choice is bi =
√i.
This one isn’t too nice however - the reason being that ck involves the sum1 + 1√
2+ . . .+ 1√
kwhich is approximately 2
√k but however is not precise. So
perhaps a nicer choice would be the one that makes the partial sum b1+. . .+bkactually equal to
√k (forgetting the unnecessary constant of 2) i.e. bk =√
k −√k − 1. These numbers are still approximately proportional to 1√
kas
√k −√k − 1 = k−(k−1)√
k+√k−1 = 1√
k+√k−1 ∼
12√k, and this is how the weights are
found.
Another idea is to apply the ”power guess” to the coefficients bk: one could tryto set bk = kα and then solve for α using the same estimates to get α = −1
2 .
Make it nonnegative Suppose one has to prove an inequality of the forma1b1
+a2b2
+ . . .+anbn≥ k, bi > 0 ∀n = 1, 2, . . . , n
It is similar to Cauchy Schwartz and Holder. However these inequalities maynot be applicable if the quantities ai are not positive. In some cases, it ispossible to resolve this inconvenience. The suggestion is to add up ai
bito mi so
that ai +mibi ≥ 0 and as small as possible (it is great if we choose mi so thatthe inequality ai +mibi ≥ 0 has equality case), then the inequality becomes
a′1b1
+a′2b2
+ . . .+a′nbn≥ k +m1 +m2 + . . .+mn.
where a′i ≥ 0, bi > 0 ∀i = 1, 2, . . . , n.
In this case, the Cauchy-Schwartz or Holder inequality can be applied. In thecase that ai ≥ 0 ∀n = 1, 2, . . . , n, one can subtract a positive quantity mi
from aibi
such that ai −mibi ≥ 0 and as small as possible, then we can applythe Cauchy Schwartz-Holder inequality more effectively than if we apply itdirectly (in some cases).
Example 5. Let a, b, c, d be real numbers such that a2 + b2 + c2 +d2 = 1. Provethat
1
1− ab+
1
1− bc+
1
1− cd+
1
1− da≤ 16
3.
Solution: Rewrite the inequality in the form(2− 1
1− ab
)+
(2− 1
1− bc
)+
(2− 1
1− cd
)+
(2− 1
1− da
)≥ 8
3,
or1− 2ab
1− ab+
1− 2bc
1− bc+
1− 2cd
1− cd+
1− 2da
1− da≥ 8
3.
Now, since 1− 2ab = a2 + b2 + c2 + d2− 2ab = (a− b)2 + c2 + d2 ≥ 0, applying
55
the Cauchy Schwartz inequality, we have
1− 2ab
1− ab+
1− 2bc
1− bc+
1− 2cd
1− cd+
1− 2da
1− da≥
≥ [(1− 2ab) + (1− 2bc) + (1− 2cd) + (1− 2da)]2
(1− 2ab)(1− ab) + (1− 2bc)(1− bc) + (1− 2cd)(1− cd) + (1− 2da)(1− da)
=4[2− (a+ c)(b+ d)]2
4− 3(a+ c)(b+ d) + 2(a2 + c2)(b2 + d2).
It suffices to prove that
3[2− (a+ c)(b+ d)]2 ≥ 2[4− 3(a+ c)(b+ d) + 2(a2 + c2)(b2 + d2)],
or
3(a+ c)2(b+ d)2 − 6(a+ c)(b+ d) + 4− 4(a2 + c2)(b2 + d2) ≥ 0,
3[1− (a+ c)(b+ d)]2 + 1− 4(a2 + c2)(b2 + d2) ≥ 0.
By the AM-GM inequality, we have
4(a2 + c2)(b2 + d2) ≤ (a2 + b2 + c2 + d2)2 = 1.
The inequality is proved. Equality holds if and only if a = b = c = d = ±12 .
Example 6. For any real numbers a, b, c, prove that
a2 − bca2 + 2b2 + 3c2
+b2 − ca
b2 + 2c2 + 3a2+
c2 − abc2 + 2a2 + 3b2
≥ 0.
Solution: Rewrite the inequality as
∑cyc
4(a2 − bc)a2 + 2b2 + 3c2
≥ 0,
or ∑cyc
[4(a2 − bc)
a2 + 2b2 + 3c2+ 1
]≥ 3,
2∑cyc
(b− c)2
a2 + 2b2 + 3c2+∑cyc
5a2 + c2
a2 + 2b2 + 3c2≥ 3.
Since 2∑cyc
(b− c)2
a2 + 2b2 + 3c2≥ 0, it suffices to show that
∑cyc
5a2 + c2
a2 + 2b2 + 3c2≥ 3,
but it follows from the Cauchy Schwartz inequality because(∑cyc
5a2 + c2
a2 + 2b2 + 3c2
)(∑cyc
(5a2 + c2)(a2 + 2b2 + 3c2)
)≥
[∑cyc
(5a2 + c2)
]2
= 36
(∑cyc
a2
)2
.
56
and
12
(∑cyc
a2
)2
−∑cyc
(5a2 + c2)(a2 + 2b2 + 3c2) = 4∑cyc
a4 − 4∑cyc
a2b2 ≥ 0.
Equality holds if and only if a = b = c.
Example 7. Let a, b, c be nonnegative real numbers, no two of which are zero.Prove that the following inequality holds
1
a2 + bc+
1
b2 + ca+
1
c2 + ab≥ 3(a+ b+ c)2
2(a2 + b2 + c2)(ab+ bc+ ca).
Solution: The inequality is equivalent to each of the following inequalities∑cyc
a2 + b2 + c2
a2 + bc≥ 3(a+ b+ c)2
2(ab+ bc+ ca),
∑cyc
(a2 + b2 + c2
a2 + bc− 1
)≥ 3(a+ b+ c)2
2(ab+ bc+ ca)− 3,
∑cyc
b2 + c2 − bca2 + bc
≥ 3(a2 + b2 + c2)
2(ab+ bc+ ca).
Now, by Cauchy Scwhartz inequality, we have that
∑cyc
b2 + c2 − bca2 + bc
≥
(2∑cyc
a2 −∑cyc
ab
)2
∑cyc
(a2 + bc)(b2 + c2 − bc)
=
(2∑cyc
a2 −∑cyc
ab
)2
(∑cyc
ab
)2
+
(∑cyc
ab
)(∑cyc
a2
)− 4abc
∑cyc
a
.
It suffices for us to prove that(2∑cyc
a2 −∑cyc
ab
)2
(∑cyc
ab
)2
+
(∑cyc
ab
)(∑cyc
a2
)− 4abc
∑cyc
a
≥ 3(a2 + b2 + c2)
2(ab+ bc+ ca).
To prove this inequality, we will use the pqr technique (refer to the appropriateparagraph for a general description of the method). Because the inequality ishomogeneous, we can assume that a+b+c = 1. and set q = ab+bc+ca, r = abc.Then this inequality becomes
(2− 5q)2
q2 + q(1− 2q)− 4r≥ 3(1− 2q)
2q,
57
or(2− 5q)2
q − q2 − 4r≥ 3(1− 2q)
2q.
If 1 ≥ 4q, then we have
(2− 5q)2
q − q2 − 4r− 3(1− 2q)
2q≥ (2− 5q)2
q − q2− 3(1− 2q)
2q=
(5− 11q)(1− 4q)
2q(1− q)≥ 0.
If 4q ≥ 1, then from Schur’s inequality for fourth degree, we see that r ≥(4q−1)(1−q)
6 . Hence,
(2− 5q)2
q − q2 − 4r≥ (2− 5q)2
q − q2 − 4 · (4q−1)(1−q)6
=3(2− 5q)
1− q.
It remains to prove that
3(2− 5q)
1− q≥ 3(1− 2q)
2q,
which can be easily simplified to
3(4q − 1)(1− 3q)
2q(1− q)≥ 0.
Sometimes one has to prove the inequalitya1b1
+a2b2
+ . . . +anbn≤ k, bi > 0
∀n = 1, 2, . . . , n. We can prove this inequality by adding aibi
to mi such thatai +mibi is a complete square of a expression, then we can apply the Cauchy-Schwarts to each of these terms to obtain an upper bound.Example 8. Let a, b, c be real numbers. Prove that the following inequalityholds
a2 − bc4a2 + 4b2 + c2
+b2 − ca
4b2 + 4c2 + a2+
c2 − ab4c2 + 4a2 + b2
≥ 0.
Solution: Rewrite the inequality in the form
4(bc− a2)4a2 + 4b2 + c2
+4(ca− b2)
4b2 + 4c2 + a2+
4(ab− c2)4c2 + 4a2 + b2
≤ 0,
or[4(bc− a2)
4a2 + 4b2 + c2+ 1
]+
[4(ca− b2)
4b2 + 4c2 + a2+ 1
]+
[4(ab− c2)
4c2 + 4a2 + b2+ 1
]≤ 3,
or(2b+ c)2
4a2 + 4b2 + c2+
(2c+ a)2
4b2 + 4c2 + a2+
(2a+ b)2
4c2 + 4a2 + b2≤ 3.
Now, using Cauchy Schwartz inequality, we have
(2b+ c)2
4a2 + 4b2 + c2=
(b+ b+ c)2
(a2 + 2b2) + (a2 + 2b2) + (c2 + 2a2)
≤ b2
a2 + 2b2+
b2
a2 + 2b2+
c2
c2 + 2a2
=2b2
a2 + 2b2+
c2
c2 + 2a2.
58
Similarly, we have
(2c+ a)2
4b2 + 4c2 + a2≤ 2c2
b2 + 2c2+
a2
a2 + 2b2,
(2a+ b)2
4c2 + 4a2 + b2≤ 2a2
c2 + 2a2+
b2
b2 + 2c2.
Adding up these inequalities, we get the result.
Make it symmetric It is usually the case that a symmetric inequality iseasier than cyclic inequality. This technique is based on this intuitive idea.We will use the Cauchy Schwartz-Holder inequality to make the inequalitysymmetric - and then it will be easier to prove.
Example 9. Let a, b, c be nonnegative real numbers, no two of which are zero.Prove that √
a
a+ b+
√b
b+ c+
√c
c+ a≤ 3√
2.
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
√a
a+ b
)2
≤
[∑cyc
(a+ c)
][∑cyc
a
(a+ b)(a+ c)
]
=
4
(∑cyc
a
)(∑cyc
ab
)(a+ b)(b+ c)(c+ a)
.
Moreover, by the AM-GM inequality,
(a+ b)(b+ c)(c+ a) =
(∑cyc
a
)(∑cyc
ab
)− abc ≥ 8
9
(∑cyc
a
)(∑cyc
ab
).
Hence (∑cyc
√a
a+ b
)2
≤ 9
2.
Equality holds if and only if a = b = c.
Example 10. Let a, b, c be positive real numbers. Prove that√a
4a+ 4b+ c+
√b
4b+ 4c+ a+
√c
4c+ 4a+ b≤ 1.
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
√a
4a+ 4b+ c
)2
≤
[∑cyc
(4a+ b+ 4c)
][∑cyc
a
(4a+ 4b+ c)(4a+ b+ 4c)
]
=
9
(∑cyc
a
)(∑cyc
a2 + 8∑cyc
ab
)(4a+ 4b+ c)(4b+ 4c+ a)(4c+ 4a+ b)
.
59
It suffices to prove that
9
(∑cyc
a
)(∑cyc
a2 + 8∑cyc
ab
)≤ (4a+ 4b+ c)(4b+ 4c+ a)(4c+ 4a+ b),
or
7∑cyc
a3 + 3∑cyc
ab(a+ b) ≥ 39abc.
The last inequality is trivial by the AM-GM inequality. Equality holds if andonly if a = b = c.
Substitutions This technique is simply the application of the ubiquitoussubstitution method to the Cauchy-Schwartz and Holder inequalities.
Example 11. Let a, b, c be positive real numbers. Prove that
a3
a3 + abc+ b3+
b3
b3 + abc+ c3+
c3
c3 + abc+ a3≥ 1.
Solution: Set x =b
a, y =
a
c, z =
c
b, then x, y, z > 0, xyz = 1 and the inequality
becomes ∑cyc
1
x3 +x
y+ 1≥ 1,
or ∑cyc
1
x3 + x2z + 1≥ 1,
which can be rewritten as ∑cyc
yz
x2 + yz + zx≥ 1.
By the Cauchy Schwartz inequality, we have
∑cyc
yz
x2 + yz + zx≥
(∑cyc
yz
)2
∑cyc
yz(x2 + yz + zx)= 1.
Example 12. Let a, b, c be positive real numbers. Prove that√a2
a2 + 7ab+ b2+
√b2
b2 + 7bc+ c2+
√c2
c2 + 7ca+ a2≥ 1.
Solution: Set x =b
a, y =
c
b, z =
a
c, then x, y, z > 0, xyz = 1 and the inequality
becomes ∑cyc
1√x2 + 7x+ 1
≥ 1.
60
Again, since x, y, z > 0, xyz = 1 then there exist m,n, p > 0 such that x =n2p2
m4, y =
p2m2
n4, z =
m2n2
p4, the inequality is transformed into
∑cyc
m4√m8 + 7m4n2p2 + n4p4
≥ 1
By Holder’s inequality, we have(∑cyc
m4√m8 + 7m4n2p2 + n4p4
)2 [∑cyc
m(m8 + 7m4n2p2 + n4p4)
]≥
(∑cyc
m3
)3
.
It suffices to prove that(∑cyc
m3
)3
≥∑cyc
m(m8 + 7m4n2p2 + n4p4),
or ∑sym
(5m6n3 + 2m3n3p3 − 7m5n2p2) +∑sym
(m6n3 −m4n4p) ≥ 0.
The last inequality is obviously true from the AM-GM inequality.
Example 13. Let a, b, c be positive real numbers. Prove that
1
a(a+ b)+
1
b(b+ c)+
1
c(c+ a)≥ 3
23√a2b2c2
.
Solution: Due to the homogeneity, we may assume abc = 1, then there existx, y, z > 0 such that a = x
y , b = zx , c = y
z . The inequality becomes
∑cyc
y2
x2 + yz≥ 3
2.
By the Cauchy Schwartz inequality, we have
∑cyc
y2
x2 + yz≥
(∑cyc
y2
)2
∑cyc
x2y2 +∑cyc
y3z.
Moreover (∑cyc
y2
)2
− 3∑cyc
x2y2 =1
2
∑cyc
(x2 − y2)2 ≥ 0,
and (∑cyc
y2
)2
− 3∑cyc
y3z =1
2
∑cyc
(x2 − z2 − 2xy + yz + zx)2 ≥ 0.
61
Therefore
∑cyc
y2
x2 + yz≥
(∑cyc
y2
)2
∑cyc
x2y2 +∑cyc
y3z≥
(∑cyc
y2
)2
13
(∑cyc
y2
)2
+ 13
(∑cyc
y2
)2 =3
2.
Equality holds if and only if a = b = c.
The CYH technique This is the most advanced technique that we wantto present in this paper. It may not be applicable to such a wide range ofinequalities as others - however it solves some very hard problems. Here is, forexample, an inequality of Jack Garfunkel, famous for creating many difficultinequalities.
Example 14. Let a, b, c be nonnegative real numbers, no two of which are zero.Prove that
a√a+ b
+b√b+ c
+c√c+ a
≤ 5
4
√a+ b+ c.
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
a√a+ b
)2
≤
[∑cyc
a(5a+ b+ 9c)
][∑cyc
a
(a+ b)(5a+ b+ 9c)
]
= 5
(∑cyc
a
)2 [∑cyc
a
(a+ b)(5a+ b+ 9c)
].
It suffices to prove that(∑cyc
a
)[∑cyc
a
(a+ b)(5a+ b+ 9c)
]≤ 5
16.
This holds since since
5
16−
(∑cyc
a
)[∑cyc
a
(a+ b)(5a+ b+ 9c)
]=A+B
C.
where
A =∑cyc
ab(a+ b)(a+ 9b)(a− 3b)2 ≥ 0,
B = 243∑cyc
a3b2c+ 835∑cyc
a2b3c+ 232∑cyc
a4bc+ 1230a2b2c2 ≥ 0,
C = 16(a+ b)(b+ c)(c+ a)(5a+ b+ 9c)(5b+ c+ 9a)(5c+ a+ 9b) > 0.
Equality holds if and only if a3 = b
1 = c0 or any cyclic permutations (of course,
the expression c0 in the equality simply means c = 0).
62
Of course, this solution has a strategy behind it:
Similar to the symmetric technique, we want to apply the Cauchy Schwartzinequality like this
(∑cyc
a√a+ b
)2
≤
[∑cyc
a(ma+ nb+ pc)
][∑cyc
a
(a+ b)(ma+ nb+ pc)
].
where m,n, p are nonnegative real numbers that we will choose later.
Now, note that the original inequality has equality for a = 3, b = 1, c = 0 thenwe must choose m,n, p so that this step, the equality also holds for this point.Moreover, due to the equality of Cauchy Schwartz inequality, this step, wehave equality if and only if
√a(ma+ nb+ pc)√
a
(a+ b)(ma+ nb+ pc)
=
√b(mb+ nc+ pa)√
b
(b+ c)(mb+ nc+ pa)
=
√c(mc+ na+ pb)√
c
(c+ a)(mc+ na+ pb)
.
We must choose m,n, p such that this equation has a root (a, b, c) = (3, 1, 0),that is√
3 · (3 ·m+ 1 · n+ 0 · p)√3
(3 + 1)(3 ·m+ 1 · n+ 0 · p)
=
√1 · (1 ·m+ 0 · n+ 3 · p)√
1
(1 + 0)(1 ·m+ 0 · n+ 3 · p)
=
√0 · (0 ·m+ 3 · n+ 1 · p)√
0
(0 + 3)(0 ·m+ 3n+ 1 · p)
,
or
2(3m+ n) = m+ 3p,
or
5m+ 2n = 3p.
Moreover, note that if the expression∑cyc
a(ma + nb + pc) = m∑cyc
a2 + (n +
p)∑cyc
ab has the form k
(∑cyc
a
)2
, then after using the Cauchy Schwartz in-
equality, it will be simpler thus it might be easier to proceed. Hence, let uschoose m,n, p such that n + p = 2m. Now, due to the homogeneity, we canchoose m = 5, n = 1, p = 9. That is the reason why we have the above solutionfor this very hard inequality.
Unfortunately, the second part of the proof does not have such a generalmethod behind it. Still, this method of choosing m,n, p can reduce the in-equality to something one may have more freedom to tackle. For example,with the the same Jack Garfunkel’s inequality, we can proceed in a slightlydifferent way :
From the same Cauchy Schwartz inequality (this time for (m,n, p) = (3, 3, 1)),
63
we have that(∑cyc
a√a+ b
)2
≤
[∑cyc
a(3a+ 3b+ c)
a+ b
](∑cyc
a
3a+ 3b+ c
)
=
[2∑cyc
a+∑cyc
a(a+ b+ c)
a+ b
](∑cyc
a
3a+ 3b+ c
)
=
(∑cyc
a
)(∑cyc
a
3a+ 3b+ c
)(∑cyc
a
a+ b+ 2
),
hence it suffices for us to prove that(∑cyc
a
3a+ 3b+ c
)(∑cyc
a
a+ b+ 2
)≤ 25
16.
The detailed proof of the last inequality can be found in the following book:Old And New Inequalities 2, Vo Quoc Ba Can - Cosmin Pohoata, GIL pub-lishing house, 2008.
Example 15. Let a, b, c be nonnegative real numbers, no two of which are zero.Prove that
(ab+ bc+ ca)
(1
(b+ c)2+
1
(c+ a)2+
1
(a+ b)2
)≥ 9
4.
Solution: The appearance of∑cyc
1
(b+ c)2suggests the Cauchy Schwartz in-
equality (since it is the sum of squares). Let us use the above idea to attackthis inequality.
By the Cauchy Schwartz inequality, we have[∑cyc
(ma+ nb+ nc)2
][∑cyc
1
(b+ c)2
]≥
(∑cyc
ma+ nb+ nc
b+ c
)2
.
The equality holds if and only if
ma+ nb+ nc1
b+ c
=mb+ nc+ na
1
c+ a
=mc+ na+ nb
1
a+ b
.
We notice that the original inequality has an equality case for a = b = 1, c = 0,hence the above solution must be satisfied at this point, that is
m+ n
1=m+ n
1=
2n12
⇔ m = 3n⇒ m = 3, n = 1.
And now, we have the solution as follows:
64
By the Cauchy Schwartz inequality, we have(11∑cyc
a2 + 14∑cyc
ab
)[∑cyc
1
(b+ c)2
]=
[∑cyc
(3a+ b+ c)2
][∑cyc
1
(b+ c)2
]
≥
(∑cyc
3a+ b+ c
b+ c
)2
= 9
(1 +
∑cyc
a
b+ c
)2
.
It suffices to prove that
4
(1 +
∑cyc
a
b+ c
)2
≥11∑cyca2 + 14
∑cycab∑
cycab
.
Due to the homogeneity, we may assume a+ b+ c = 1, setting q =∑cyc
ab, r =
abc, then by Schur’s inequality for third degree, we obtain r ≥ max
{0,
4q − 1
9
}.
The inequality becomes
4
(1 + q
q − r− 2
)2
≥ 11− 8q
q.
If 1 ≥ 4q, then
4
(1 + q
q − r− 2
)2
−11− 8q
q≥ 4
(1 + q
q− 2
)2
−11− 8q
q=
(4− 3q)(1− 4q)
q2≥ 0.
If 4q ≥ 1, then
4
(1 + q
q − r− 2
)2
− 11− 8q
q≥ 4
1 + q
q − 4q − 1
9
− 2
2
− 11− 8q
q
=(1− 3q)(4q − 1)(11− 17q)
q(5q + 1)2≥ 0.
Equality holds if and only if a = b = c or a = b, c = 0 and any cyclicpermutations.
Example 16. Let a, b, c, d be positive real numbers such that
(a+ b+ c+ d)
(1
a+
1
b+
1
c+
1
d
)= 20.
Prove that
(a2 + b2 + c2 + d2)
(1
a2+
1
b2+
1
c2+
1
d2
)≥ 36.
65
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
1
a2
)[∑cyc
(b+ c+ d− a)2
]≥
(∑cyc
b+ c+ d− aa
)2
= 144.
Moreover, we see that
4∑cyc
a2 =∑cyc
(b+ c+ d− a)2.
Sometimes, the above way to choose m,n, p, . . . cannot be used, and we needto another way to choose them. A nice idea to apply the Cauchy Schwartzinequality is by taken note at some well-known inequalities or taken note atthe special form of the given inequalities. Here are some examples.
Example 17. Let a, b, c be positive real numbers. Prove that for all k ≥ 2, wehave ∑
cyc
√a2 + kab+ b2 ≤
√4∑cyc
a2 + (3k + 2)∑cyc
ab.
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
√a2 + kab+ b2
)2
≤
[∑cyc
(a+ b)
](∑cyc
a2 + kab+ b2
a+ b
)
= 2
(∑cyc
a
)(∑cyc
a2 + kab+ b2
a+ b
).
We need to prove that
2∑cyc
a2 + kab+ b2
a+ b≤
4∑cyc
a2 + (3k + 2)∑cyc
ab∑cyc
a,
This inequality is equivalent to each of the following
2∑cyc
(a+ b) + 2(k − 2)∑cyc
ab
a+ b≤ 4
∑cyc
a+
3(k − 2)∑cyc
ab∑cyc
a,
2∑cyc
ab
a+ b≤
3∑cyc
ab∑cyc
a,
2∑cyc
ab(a+ b+ c)
a+ b≤ 3
∑cyc
ab,
66
2abc∑cyc
1
a+ b≤∑cyc
ab.
And the last inequality follows from the Cauchy Schwartz inequality
2abc∑cyc
1
a+ b≤ 2abc
∑cyc
(1
4a+
1
4b
)=∑cyc
ab.
Equality holds if and only if a = b = c.
Example 18. Let a, b, c be positive real numbers such that a+ b+ c = 1. Provethat
a√a+ 2b
+b√
b+ 2c+
c√c+ 2a
≤√
3
2.
Solution: At this point, we can easily show that
a
2a+ 4b+ c+
b
2b+ 4c+ a+
c
2c+ 4a+ b≤ 1
2,
with equality holding when abc = 0. (this is left as an exercise)Thus, using the Cauchy Schwartz inequality, we have that(∑
cyc
a√a+ 2b
)2
≤
(∑cyc
a
2a+ 4b+ c
)[∑cyc
a(2a+ 4b+ c)
a+ 2b
]
≤ 1
2
∑cyc
a(2a+ 4b+ c)
a+ 2b
=1
2
(2∑cyc
a+∑cyc
ca
a+ 2b
)
≤ 1
2
(2∑cyc
a+∑cyc
ca
a
)=
3
2
∑cyc
a =3
2.
Exercises
1. Let x, y ≥ 0 such that x3 + y3 = 1. Prove that
√x+ 2
√y ≤ 6
√(1 + 2
5√
2)5.
Solution: By Holder’s inequality, we have that for all a, b ≥ 0,(a6 + b6
)5 (x3 + y3
)≥(a5√x+ b5
√y)6.
Hence,
a5√x+ b5
√y ≤ 6
√(a6 + b6)5 (x3 + y3) =
6
√(a6 + b6)5.
Choosing a = 1, b = 5√
2, then we have
√x+ 2
√y ≤ 6
√[16 +
(5√
2)6]5
=6
√(1 + 2
5√
2)5.
67
2. Let a, b, c ≥ 0 such that a2 + b2 + c2 = 1. Find the minimum value of
P = a3 + 3b3 + 2c3.
Solution: By Holder’s inequality, we have that for all m,n, p ≥ 0,
(a3 + 3b3 + 2c3)2(m3 + n3 + p3) ≥(a2m+ b2n
3√
9 + c2p3√
4)3.
Choosing m = 1, n = 13√9, p = 1
3√4, then we have
(a3 + 3b3 + 2c3)2 ≥ (a2 + b2 + c2)3
1 +1
9+
1
4
=36
49,
or
P = a3 + 3b3 + 2c3 ≥ 6
7.
We have equality when
a
m=b 3√
3
n=c 3√
2
p,
ora2
m2=b2 3√
9
n2=c2 3√
4
p2=
a2 + b2 + c2
m2 +n2
3√
9+
p2
3√
4
=36
49,
or
a =6
7, b =
6
7 3√
9, c =
6
7 3√
4.
Hence
minP =6
7.
3. Given a, b, c ≥ 0 satisfying a+ b+ c = 3. Determine the minimum valueof
P = a4 + 2b4 + 4c4.
Solution: By Holder’s inequality, we have that for all m,n, p ≥ 0,
(a4 + 2b4 + 3c4)(m4 + n4 + p4)3 ≥(am3+bn3
4√
2+cp34√
3)4.
Choosing m = 1, n = 112√2
, b = 112√3
, then we have
P = a4 + 2b4 + 3c4 ≥ (a+ b+ c)4[14 +
(1
12√
2
)4
+
(1
12√
3
)4]3
=1(
1 +13√
2+
13√
3
)3 .
68
We have equality when
a
m=b 4√
2
n=c 4√
3
p=
a+ b+ c
m+n4√
2+
p4√
3
=1
1 +13√
2+
13√
3
,
or
a =1
1 +13√
2+
13√
3
, b =1
3√
2
(1 +
13√
2+
13√
3
) , c =1
3√
3
(1 +
13√
2+
13√
3
) .Hence
minP =1(
1 +13√
2+
13√
3
)3 .
4. Let x1, x2, . . . , xn be real numbers. Prove that
x21+(x1+x2)2+. . .+(x1+x2+. . .+xn)2 ≤ 1
4 sin2 π
2(2n+ 1)
·(x21+x22+. . .+x2n).
Solution: By Cauchy Schwartz inequality, we have that for all ci > 0,
n∑k=1
(k∑i=1
xi
)2
≤n∑k=1
[(k∑i=1
ci
)(k∑i=1
x2ici
)]
=
n∑k=1
[Sk
(k∑i=1
x2ici
)]=
n∑k=1
Si + . . .+ Snci
x2i .
Choosing ci such that
S1 + S2+ . . .+Snc1
=S2 + . . .+ Sn
c2= . . . =
Sncn,
then we obtain ci = siniπ
2n+ 1− sin
(i− 1)π
2n+ 1, and
S1 + S2+ . . .+Snc1
=S2 + . . .+ Sn
c2= . . . =
Sncn
=1
4 sin2 π
2(2n+ 1)
.
Thusn∑k=1
(k∑i=1
xi
)2
≤ 1
4 sin2 π
2(2n+ 1)
n∑k=1
x2i .
5. Let a, b, c ≥ 0 such that a+ b+ c = 1. Determine the minimum value of
P =
√a2 +
1
b2+
√b2 +
1
c2+
√c2 +
1
a2.
69
Solution: We guess that the minimum attains equality when a = b = c =1
3. Hence, by Cauchy Schwartz Inequality, we have that for all m,n ≥ 0,√(
a2 +1
b2
)(m2 + n2) ≥ ma+
n
b,√(
b2 +1
c2
)(m2 + n2) ≥ mb+
n
c,√(
c2 +1
a2
)(m2 + n2) ≥ mc+
n
a.
Therefore
P√m2 + n2 ≥ m(a+ b+ c) + n
(1
a+
1
b+
1
c
)≥ m(a+ b+ c) +
9n
a+ b+ c= m+ 9n,
or
P ≥ m+ 9n√m2 + n2
.
Choosing m,n > 0 such that
a = b = c =1
3,
a
m=
1
bn,
b
m=
1
cn,
c
m=
1
an.
Then n = 9m, and we may choose m = 1, n = 9 to obtain
P ≥ 1 + 9 · 9√12 + 92
=√
82.
Thus minP =√
82.
6. Let a, b, c > 0 such that a2 + b2 + c2 = 3. Prove that
1
2− a+
1
2− b+
1
2− c≥ 3.
Solution: The inequality is equivalent to(2
2− a− 1
)+
(2
2− b− 1
)+
(2
2− c− 1
)≥ 3,
ora
2− a+
b
2− b+
c
2− c≥ 3.
By the Cauchy Schwartz inequality and AM-GM inequality, we have
a
2− a+
b
2− b+
c
2− c=
a4
2a3 − a4+
b4
2b3 − b4+
c4
2c3 − c4
≥ (a2 + b2 + c2)2
2a3 + 2b3 + 2c2 − a4 − b4 − c4
≥ (a2 + b2 + c2)2
(a4 + a2) + (b4 + b2) + (c4 + c2)− a4 − b4 − c4
= a2 + b2 + c2 = 3.
70
7. Let a, b, c be the side lengths of a triangle. Prove that
a(a− b)a2 + 2bc
+b(b− c)b2 + 2ca
+c(c− a)
c2 + 2ab≥ 0.
Solution: The inequality is equivalent to∑cyc
[a(a− b)a2 + 2bc
+ 1
]≥ 3,
or ∑cyc
2a2 − ab+ 2bc
a2 + 2bc≥ 3.
Since a, b, c are the side lengths of a triangle, we have c ≥ b− a, then
2a2 − ab+ 2bc ≥ 2a2 − ab+ 2b(b− a) = 2(a− b)2 + ab ≥ 0.
Hence, by the Cauchy Schwartz inequality, we have
∑cyc
2a2 − ab+ 2bc
a2 + 2bc≥
[∑cyc
(2a2 − ab+ 2bc)
]2∑cyc
(2a2 − ab+ 2bc)(a2 + 2bc)
It suffices to prove that[∑cyc
(2a2 − ab+ 2bc)
]2≥ 3
∑cyc
(2a2 − ab+ 2bc)(a2 + 2bc),
or equivalently,
7∑cyc
a3b+ 4∑cyc
ab3 ≥ 2∑cyc
a4 + 3∑cyc
a2b2 + 6∑cyc
a2bc.
Again, since a, b, c are the side lengths of a triangle, there exist x, y, z > 0such that a = y + z, b = z + x, c = x+ y. The inequality becomes
2∑cyc
x4 + 2∑cyc
xy(x2 + y2) + 3∑cyc
xy3 ≥ 6∑cyc
x2y2 + 3∑cyc
x2yz,
But it follows from the AM-GM inequality since
2∑cyc
x4 ≥2∑cyc
x2y2, 2∑cyc
xy(x2 + y2) ≥ 4∑cyc
x2y2, 3∑cyc
xy3 ≥ 3∑cyc
x2yz.
Equality holds if and only if a = b = c.
8. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat
2a2 − bcb2 − bc+ c2
+2b2 − ca
c2 − ca+ a2+
2c2 − aba2 − ab+ b2
≥ 3.
71
Solution: Rewrite the inequality as
∑cyc
[2a2 − bc
b2 − bc+ c2+ 1
]≥ 6,
or ∑cyc
2a2 + (b− c)2
b2 − bc+ c2≥ 6.
By the Cauchy Schwartz inequality, we have
∑cyc
2a2 + (b− c)2
b2 − bc+ c2≥
[∑cyc
[2a2 + (b− c)2]
]2∑cyc
[2a2 + (b− c)2](b2 − bc+ c2).
It suffices to prove that[∑cyc
[2a2 + (b− c)2]
]2≥ 6
∑cyc
[2a2 + (b− c)2](b2 − bc+ c2),
or
2∑cyc
a4 + 2abc∑cyc
a+∑cyc
ab(a2 + b2) ≥ 6∑cyc
a2b2,
that is
2∑cyc
a2(a− b)(a− c) + 3∑cyc
ab(a− b)2 ≥ 0.
which is true by Schur’s inequality for fourth degree. Equality holds ifand only if a = b = c or a = b, c = 0 and its cyclic permutations.
9. Let a, b, c be nonnegative real numbers, not all are zero. Prove that
3a2 − bc2a2 + b2 + c2
+3b2 − ca
2b2 + c2 + a2+
3c2 − ab2c2 + a2 + b2
≤ 3
2.
Solution: Rewrite the inequality as
∑cyc
[3− 2(3a2 − bc)
2a2 + b2 + c2
]≥ 6,
or ∑cyc
3b2 + 2bc+ 3c2
2a2 + b2 + c2≥ 6,
that is
3∑cyc
(b− c)2
2a2 + b2 + c2+ 8
∑cyc
bc
2a2 + b2 + c2≥ 6.
72
If (a−b)2+(b−c)2+(c−a)2 = 0, it is trivial. If (a−b)2+(b−c)2+(c−a)2 >0, then by the Cauchy Schwartz inequality, we have
∑cyc
(b− c)2
2a2 + b2 + c2≥
[∑cyc
(b− c)2]2
∑cyc
(b− c)2(2a2 + b2 + c2)
=
4
(∑cyc
a2 −∑cyc
ab
)2
[∑cyc
(b− c)2](∑
cyc
a2
)+∑cyc
a2(b− c)2
=
2
(∑cyc
a2 −∑cyc
ab
)2
(∑cyc
a2 −∑cyc
ab
)(∑cyc
a2
)+∑cyc
b2c2 −∑cyc
a2bc
,
and
∑cyc
bc
2a2 + b2 + c2≥
(∑cyc
bc
)2
∑cyc
bc(2a2 + b2 + c2)=
(∑cyc
bc
)2
(∑cyc
bc
)(∑cyc
a2
)+∑cyc
a2bc
.
It suffices to prove that
6
(∑cyc
a2 −∑cyc
ab
)2
(∑cyc
a2 −∑cyc
ab
)(∑cyc
a2
)+∑cyc
b2c2 −∑cyc
a2bc
+
8
(∑cyc
bc
)2
(∑cyc
bc
)(∑cyc
a2
)+∑cyc
a2bc
≥ 6.
Due to the homogeneity, we may assume a + b + c = 1. Putting q =∑cyc
bc, r = abc, then the inequality becomes
3(1− 3q)2
(1− 3q)(1− 2q) + q2 − 3r+
4q2
q(1− 2q) + r≥ 3,
or(3r + q − 4q2)2
(1− 5q + 7q2 − 3r)(q − 2q2 + r)≥ 0,
which is true.
10. Let a, b, c be the side lengths of a triangle. Prove that
a(b+ c)
a2 + 2bc+b(c+ a)
b2 + 2ca+c(a+ b)
c2 + 2ab≤ 2.
73
Solution: The inequality is equivalent to∑cyc
[1− a(b+ c)
a2 + 2bc
]≥ 1,
or ∑cyc
a2 − ab− ac+ 2bc
a2 + 2bc≥ 1
We will show that a2 − ab − ac + 2bc ≥ 0, or a(a − b) + (2b − a)c ≥ 0.Indeed, if 2b ≥ a, we have
a(a− b) + (2b− a)c ≥ a(a− b) + (2b− a)(b− a) = 2(a− b)2 ≥ 0.
If a ≥ 2b, we have
a(a− b) + (2b− a)c ≥ a(a− b) + (2b− a)(a+ b) = 2b2 ≥ 0.
Now, by the Cauchy Schwartz inequality, we have
∑cyc
a2 − ab− ac+ 2bc
a2 + 2bc≥
[∑cyc
(a2 − ab− ac+ 2bc
)]2∑cyc
(a2 − ab− ac+ 2bc
)(a2 + 2bc)
=
(∑cyc
a2
)2
∑cyc
(a2 − ab− ac+ 2bc
)(a2 + 2bc)
.
It suffices to show that(∑cyc
a2
)2
≥∑cyc
(a2 − ab− ac+ 2bc
)(a2 + 2bc),
which can be easily simplified to∑cyc
ab(a− b)2 ≥ 0.
which is obviously true. Equality holds if and only if a = b = c ora = b, c = 0 and its cyclic permutations.
11. Let a, b, c be the side lengths of a triangle. Prove that
a
b+ c+
b
c+ a+
c
a+ b+ab+ bc+ ca
a2 + b2 + c2≤ 5
2.
Solution: The inequality is equivalent to
∑cyc
(1− a
b+ c
)≥ 1
2+
∑cyc
ab∑cyc
a2,
74
or
∑cyc
b+ c− ab+ c
≥
(∑cyc
a
)2
2∑cyc
a2.
By the Cauchy Schwartz inequality, we have
2
(∑cyc
a2
)(∑cyc
b+ c− ab+ c
)=
[∑cyc
(b+ c)(b+ c− a)
](∑cyc
b+ c− ab+ c
)
≥
[∑cyc
(b+ c− a)
]2=
(∑cyc
a
)2
.
Equality holds if and only if a = b = c or a = b, c = 0 or any cyclicpermutations.
12. Let a, b, c be the side lengths of a triangle. Prove that
a
3a− b+ c+
b
3b− c+ a+
c
3c− a+ b≥ 1.
Solution: We have4a
3a− b+ c=
a+ b− c3a− b+ c
+ 1. Hence, it suffices to
prove thata+ b− c3a− b+ c
+b+ c− a3b− c+ a
+c+ a− b3c− a+ b
≥ 1,
which is obviously true since by Cauchy Schwartz inequality, we have
a+ b− c3a− b+ c
+b+ c− a3b− c+ a
+c+ a− b3c− a+ b
≥ (a+ b+ c)2∑cyc
(a+ b− c)(3a− b+ c)
=(a+ b+ c)2∑
cyc
a2 + 2∑cyc
ab= 1.
13. Let x1, x2, . . . , xn be positive real numbers such thatn∑i=1
xi = 1. Prove
that the following inequality holds
n∑i=1
√x2i + x2i+1 ≤ 2− 1
√2
2+
n∑i=1
x2ixi+1
.
Solution: The original inequality is equivalent to
n∑i=1
(xi + xi+1 −
√x2i + x2i+1
)≥ 1√
2
2+
n∑i=1
x2ixi+1
,
75
orn∑i=1
x2ix2ixi+1
+ xi +xixi+1
√x2i + x2i+1
≥ 1
√2 + 2
n∑i=1
x2ixi+1
.
By the Cauchy Schwartz inequality, we have that
n∑i=1
x2ix2ixi+1
+ xi +xixi+1
√x2i + x2i+1
≥
(n∑i=1
xi
)2
n∑i=1
(x2ixi+1
+ xi +xixi+1
√x2i + x2i+1
)=
1n∑i=1
(x2ixi+1
+ xi +xixi+1
√x2i + x2i+1
) .It suffices to prove that
n∑i=1
(xixi+1
√x2i + x2i+1 −
x2ixi+1
)≤√
2− 1,
orn∑i=1
xixi+1√x2i + x2i+1 + xi
≤√
2− 1.
Since√x2i + x2i+1 ≥
xi + xi+1√2
, it remains to show that
n∑i=1
xixi+1(1 +√
2)xi + xi+1
≤ 1−√
2
2,
or
n∑i=1
xixi+1(1 +√
2)xi + xi+1
≤n∑i=1
(3− 2
√2
2xi +
√2− 1
2xi+1
),
that isn∑i=1
(√2− 1
)(xi − xi+1)
2
2[(
1 +√
2)xi + xi+1
] ≥ 0.
The proof is complete. Equality holds if and only if x1 = x2 = . . . =
xn =1
n.
14. Let a, b, c be positive real numbers. Prove that
a4
a2 + ab+ b2+
b4
b2 + bc+ c2+
c4
c2 + ca+ a2≥ a3 + b3 + c3
a+ b+ c.
Solution: Rewrite the inequality in the form∑cyc
(a4
a2 + ab+ b2+ ab− a2
)≥ a3 + b3 + c3
a+ b+ c+ab+bc+ca−a2−b2−c2,
76
or ∑cyc
ab3
a2 + ab+ b2≥ 3abc
a+ b+ c,
or ∑cyc
b2
a2 + b2 + ab
ab
≥ 3abc
a+ b+ c.
By the Cauchy Schwartz inequality, we have∑cyc
b2
a2 + b2 + ab
ab
≥ (a+ b+ c)2∑cyc
a2 + b2 + ab
ab
=abc(a+ b+ c)
ab+ bc+ ca.
It suffices to prove that
abc(a+ b+ c)
ab+ bc+ ca≥ 3abc
a+ b+ c,
or(a+ b+ c)2 ≥ 3(ab+ bc+ ca),
which is obviously true by the AM-GM inequality.
15. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
ab√ab+ bc
+bc√
bc+ ca+
ca√ca+ ab
≤ 1√2.
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
ab√ab+ bc
)2
=
[∑cyc
√a+ b ·
√a2b
(a+ b)(a+ c)
]2
≤
[∑cyc
(a+ b)
][∑cyc
a2b
(a+ b)(a+ c)
]
= 2∑cyc
a2b
(a+ b)(a+ c),
hence it suffices to prove that
4∑cyc
a2b
(a+ b)(a+ c)≤ a+ b+ c,
which is equivalent to
4a2b(b+ c) + 4b2c(c+ a) + 4c2a(a+ b) ≤ (a+ b)(b+ c)(c+ a)(a+ b+ c),
this last one being true, because it can be written as
ab(a− b)2 + bc(b− c)2 + ca(c− a)2 ≥ 0.
77
16. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat √
a2
4a2 + ab+ 4b2+
√b2
4b2 + bc+ 4c2+
√c2
4c2 + ca+ 4a2≤ 1.
Solution: By the Cauchy Schwartz inequality, we have(∑cyc
√a2
4a2 + ab+ 4b2
)2
≤
≤
[∑cyc
(4a2 + ac+ 4c2)
][∑cyc
a2
(4a2 + ab+ 4b2)(4a2 + ac+ 4c2)
].
It suffices to prove that[∑cyc
(4a2 + ac+ 4c2)
][∑cyc
a2
(4a2 + ab+ 4b2)(4a2 + ac+ 4c2)
]≤ 1.
By expanding, we can see the inequality is equivalent to
8∑cyc
a3b3 + 8∑cyc
a4bc+ 3abc∑cyc
ab(a+ b) ≥ 66a2b2c2.
which follows from the AM-GM inequality.
17. Let a, b, c be positive real numbers. Prove that
1
a√a+ b
+1
b√b+ c
+1
c√c+ a
≥ 3√2abc
.
Solution: Due to the homogeneity, we may assume abc = 1, then there
exist x, y, z > 0 such that a =x
y, b =
z
x, c =
y
z. The inequality becomes
∑cyc
y√y√
x(x2 + yz)≥ 3√
2
By the Cauchy Schwartz inequality,
∑cyc
y√y√
x(x2 + yz)≥
(∑cyc
y
)2
∑cyc
√xy(x2 + yz)
,
and∑cyc
√xy(x2 + yz) ≤
√√√√(∑cyc
xy
)(∑cyc
x2 +∑cyc
xy
)
=1√2
√√√√(2∑cyc
xy
)(∑cyc
x2 +∑cyc
xy
)
≤ 1
2√
2
(∑cyc
x2 + 3∑cyc
xy
)≤ 2
3√
2
(∑cyc
x
)2
.
78
Hence ∑cyc
y√y√
x(x2 + yz)≥ 3√
2.
Equality holds if and only if a = b = c.
18. Let a, b, c be positive real numbers such that abc = 1. Prove that
1
a2 − a+ 1+
1
b2 − b+ 1+
1
c2 − c+ 1≤ 3.
Solution: First of all, we will show that for any x, y, z > 0 which satis-fying xyz = 1
1
x2 + x+ 1+
1
y2 + y + 1+
1
z2 + z + 1≥ 1.
Indeed, since x, y, z > 0 and xyz = 1 then there exist m,n, p > 0 such
that x =np
m2, y =
pm
n2, z =
mn
p2. The inequality is transformed into
∑cyc
m4
m4 +m2np+ n2p2≥ 1.
From the Cauchy Schwartz inequality and the known∑cyc
n2p2 ≥∑cyc
m2np,
we have
∑cyc
m4
m4 +m2np+ n2p2≥
(∑cyc
m2
)2
∑cyc
(m4 +m2np+ n2p2)≥
(∑cyc
m2
)2
∑cyc
m4 + 2∑cyc
n2p2= 1.
Back to the original problem, from the statement above, we have∑cyc
11
a4+
1
a2+ 1≥ 1,
or ∑cyc
a4
a4 + a2 + 1≥ 1,
∑cyc
2(a2 + 1)
a4 + a2 + 1≤ 4,
∑cyc
(a2 + a+ 1) + (a2 − a+ 1)
(a2 + a+ 1)(a2 − a+ 1)≤ 4,
∑cyc
1
a2 + a+ 1+∑cyc
1
a2 − a+ 1≤ 4.
79
Using the above statement again, we have∑cyc
1
a2 + a+ 1≥ 1, and so
∑cyc
1
a2 − a+ 1≤ 3.
Equality holds if and only if a = b = c = 1.
19. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat
(b+ c)2
a2 + bc+
(c+ a)2
b2 + ca+
(a+ b)2
c2 + ab≥ 6.
Solution: By the Cauchy Schwartz inequality, we obtain[∑cyc
(b+ c)2
a2 + bc
][∑cyc
(b+ c)2(a2 + bc)
]≥
[∑cyc
(b+ c)2
]2
= 4
(∑cyc
a2 +∑cyc
ab
)2
.
Hence, it suffices to show that
2
(∑cyc
a2 +∑cyc
ab
)2
≥ 3∑cyc
(b+ c)2(a2 + bc),
or equivalently,
2∑cyc
a4 + 2abc∑cyc
a+∑cyc
ab(a2 + b2)− 6∑cyc
a2b2 ≥ 0.
This inequality follows by summing Schur’s inequality of fourth degree
2∑cyc
a4 + 2abc∑cyc
a ≥ 2∑cyc
ab(a2 + b2)
to the inequality
3∑cyc
ab(a2 + b2) ≥ 6∑cyc
a2b2.
which follows from the AM-GM inequality. Equality holds if a = b = cor a = b, c = 0 and its cyclic permutations.
20. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat
1√4a2 + bc
+1√
4b2 + ca+
1√4c2 + ab
≥ 2√ab+ bc+ ca
.
Solution: By Holder’s inequality, we have(∑cyc
1√4a2 + bc
)2 [∑cyc
(b+ c)3(4a2 + bc)
]≥
[∑cyc
(b+ c)
]3
= 8
(∑cyc
a
)3
.
80
It suffices to prove that
2
(∑cyc
a
)3(∑cyc
ab
)≥∑cyc
(b+ c)3(4a2 + bc),
or ∑cyc
ab(a3 + b3)−∑cyc
a2b2(a+ b) + 14abc∑cyc
a2 ≥ 0,
or ∑cyc
ab(a− b)2(a+ b) + 14abc∑cyc
a2 ≥ 0,
which is trivial. Equality holds if and only if a = b, c = 0 and its cyclicpermutations.
21. Let a, b, c be nonnegative real numbers, not all are zero. Prove that
(a+ b)2
a2 + 2b2 + 3c2+
(b+ c)2
b2 + 2c2 + 3a2+
(c+ a)2
c2 + 2a2 + 3b2≥ 3
2.
Solution: By the Cauchy Schwartz inequality, we have
∑cyc
(a+ b)2
a2 + 2b2 + 3c2≥
[∑cyc
(a+ b)(2a+ b)
]2∑cyc
(2a+ b)2(a2 + 2b2 + 3c2)
=
9
(∑cyc
a2 +∑cyc
ab
)2
∑cyc
(2a+ b)2(a2 + 2b2 + 3c2).
It suffices to prove that
6
(∑cyc
a2 +∑cyc
ab
)2
≥∑cyc
(2a+ b)2(a2 + 2b2 + 3c2),
which can be easily simplified to
4∑cyc
a3b+ 2∑cyc
ab3 − 3∑cyc
a2b2 + 6abc∑cyc
a ≥ 0,
or
2∑cyc
ab(a− b)2 + 2∑cyc
a3b+∑cyc
a2b2 + 6abc∑cyc
a ≥ 0,
which is obviously true. Equality holds if and only ifa
1=b
0=c
0and its
cyclic permutations.
81
22. Let a, b, c be nonnegative real numbers satisfying a + b + c = 1, andmoreover from which at least two are nonzero. Prove that
a√
4b2 + c2 + b√
4c2 + a2 + c√
4a2 + b2 ≤ 3
4.
Solution: With the help of the Cauchy Schwartz inequality, we get(∑cyc
a√
4b2 + c2
)2
≤
[∑cyc
a(2b+ c)
][∑cyc
a(4b2 + c2)
2b+ c
]
= 3
(∑cyc
ab
)[∑cyc
a(4b2 + c2)
2b+ c
],
hence it suffices to prove that
3
16(ab+ bc+ ca)≥ a(4b2 + c2)
2b+ c+b(4c2 + a2)
2c+ a+c(4a2 + b2)
2a+ b,
or equivalently,
3
16(ab+ bc+ ca)+ 4abc
(1
2b+ c+
1
2c+ a+
1
2a+ b
)≥ 3(ab+ bc+ ca)
Using again the Cauchy Schwartz inequality we see that
1
2b+ c+
1
2c+ a+
1
2a+ b≥ 3
a+ b+ c= 3,
and so, we are left to show that
1
16(ab+ bc+ ca)+ 4abc ≥ ab+ bc+ ca.
Setting q = ab + bc + ca, notice that 0 ≤ q ≤ 1
3. From the Schur’s
inequality, applied for the fourth degree, we have
a4 + b4 + c4 + abc(a+ b+ c) ≥ ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2),
and thus, in terms of q, we get
abc ≥ (4q − 1)(1− q)6
.
Therefore
1
16(ab+ bc+ ca)+ 4abc− ab− bc− ca =
1
16x− q + 4abc
≥ 1
16q− q +
2
3(4q − 1)(1− q)
=(3− 8q)(1− 4q)2
48q≥ 0.
82
23. Let n > 2 and x1, x2, . . . , xn > 0 such that(n∑k=1
xk
)(n∑k=1
1
xk
)= n2 + 1.
Prove that (n∑k=1
x2k
)(n∑k=1
1
x2k
)> n2 + 4 +
2
n(n− 1).
Solution: Put s =
n∑k=1
xk. Then by Cauchy-Schwartz we have that
[n∑k=1
(s− n
2xk
)2]( n∑k=1
1
x2k
)≥
(n∑k=1
s− n2xk
xk
)2
=
[(n∑k=1
xk
)(n∑k=1
1
xk
)− n2
2
]2
=
(n2 + 1− n2
2
)2
=(n2 + 2)2
4.
Now, notice that
n∑k=1
(s− n
2xk
)2= ns2 − ns
n∑k=1
xk +n2
4
n∑k=1
x2k =n2
4
n∑k=1
x2k.
Hence
n2
4
(n∑k=1
x2k
)(n∑k=1
1
x2k
)≥ (n2 + 2)2
4,
It follows that (n∑k=1
x2k
)(n∑k=1
1
x2k
)≥ (n2 + 2)2
n2.
It suffices to prove that
(n2 + 2)2
n2> n2 + 4 +
2
n(n− 1),
which is true because
(n2 + 2)2
n2− n2 − 4− 2
n(n− 1)=
2(n− 2)
n2(n− 1)> 0.
Therefore, our inequality is proved.
83
1.10 The principle of extremality and monotonicity
To prove an inequality of the form f(x1, x2, . . . , xn) ≥ (≤)0 it suffices to proveit for the multiset of x1, x2, . . . , xn which minimizes (maximizes) f on the givendomain.
This is actually a very powerful idea, because there are often methods todetermine when a set of variables minimize f . However, the bulk of suchexamples comes from analysis, which is too advanced for the purposes of thischapter (but will be dealt with later).
There are still simpler instances of this method, for which we can find thepoints of minimal/maximal values using elementary methods.
For example, to prove the inequality
a2 + b2 + c2 ≥ ab+ bc+ ca
we can take the function
f(a) = a2 + b2 + c2 − ab− ac− bc
as a quadratic function in a and we must prove f(a) ≥ 0. It suffices to prove
it for a being the vertex of the parabola a =b+ c
2which minimizes f . But
f
(b+ c
2
)=
3
4(b− c)2 ≥ 0.
If the function is monotone the extremal values are always attained on theends on an interval. Therefore if for each of xi the domain is an interval, itsuffices to check the inequality for xi being the extremities of the intervals.We can actually consider just one of the extremities: If f is increasing and theinequality is f ≥ 0 or f is decreasing and the inequality is f ≤ 0 we can takejust the smallest extremity, otherwise just the largest extremity.
For example, consider the inequality x2 − xy + y2 ≤ 4 for x, y ∈ [0, 2]. Thefunction
f(x, y) = x2 − xy + y2
is increasing in both x and y, because its a quadratic trinomial and x >y
2which is the vertex of g(x) = x2 − xy + y2, y >
x
2which is the vertex of
h(y) = y2 − yx+ x2.
Therefore the inequality must be proven only for the extremities of [1, 2]. How-ever, as the function is increasing, we can consider only the largest extremities,which are x = y = 2, for which we get the value 4.
84
Exercises
1. If a, b, c ∈ [0, 1] then 1 ≤ a+ b+ c+ 3(1− a)(1− b)(1− c) ≤ 3.
Solution: If we fix b, c then E = a + b + c + 3(1 − a)(1 − b)(1 − c) is alinear function in a so attains its extremal value at the extremes, so weneed to consider only the case when a ∈ {0, 1}. Analogously it sufficesto consider that b, c ∈ {0, 1}. This cases are easy to handle to obtain thedesired result.
2. If x1, x2, x3, . . . , xn ∈ [0, 1] then
x1 + x2 + x3 + . . .+ xn − x1x2 − x2x3 − . . .− xnx1 ≤⌊n
2
⌋.
Solution: Again this function is linear in all xi, so it suffices to look at thecases xi ∈ {0, 1}. Now assume that there are k ones among x1, x2, . . . , xnand n−k zeroes. Then x1+x2+. . .+xn = k and x1x2+x2x3+. . .+xnx1is the number of pairs of consecutive ones in the sequence. How smallcan the number of such pairs be? As there are n−k zeroes and each zeroenters in two pairs of consecutive numbers, there are at most 2(n − k)pairs of consecutive numbers at least one of which is zero, so there areat least n− 2(n− k) = 2k − n pairs of consecutive ones. So our sum isat most x1 + x2 + . . .+ xn = k and at most k − (2k − n) = n− k, so atmost max{k, n− k} ≤
⌊n2
⌋.
3. If a, b, c ∈ [0, 1]; p, q, r ∈[0, 12]; a+ b+ c = p+ q + r = 1 then
abc ≤ 1
8(pa+ qb+ rc).
Solution: we have r = 1− p− q so the expression
abc− 1
8(pa+ qb+ rc) = abc− 1
8[pa+ qb+ (1− p− q)c]
is linear in q for fixed p, and so takes its minimal value at the extremities.
The extremities are1
2− p (which yields r =
1
2) and
1
2. So it suffices to
look only at this case, which gives us that one of q, r is1
2.
Analogously we can suppose one of p, q is1
2and one of q, r is
1
2. It’s
easy to see that in this case two of p, q, r are1
2and one is 0. WLOG
assume p = q =1
2. Then we have to prove 16abc ≤ a + b. If we let
x = a+ b ∈ [0, 1] then 16abc ≤ 16
(a+ b
2
)2
c = 4x2(1− x). So we need
to show that 4x2(1− x) ≤ x or 4x(1− x) ≤ 1 or 0 ≤ (1− 2x)2, which istrue.
85
1.11 Breaking the inequality
Most complicated inequalities can be broken into simpler ones, by summingwhich we get the original inequality. We have already seen an example wehave broken the inequality
a2 + b2 + c2 ≥ ab+ bc+ ca
intoa2 + b2
2≥ ab, b
2 + c2
2≥ bc, and
c2 + a2
2≥ ca.
There are different tips for breaking inequalities.If we have a homogeneous inequality of degree one, like the one
a2
b+ c+
b2
a+ c+
c2
b+ a≥ a+ b+ c
2, (a, b, c > 0),
we can try to compare each term from the left sides with a part of the rightside:
a2
b+ c≥ αa+ βb+ γc.
We can see that α+β+γ = 12 as for a = b = c = 1 we get equality. Moreover,
sincea2
b+ cincreases with respect to a and decreases with respect to b, c we
must have α ≥ 0, β, γ ≤ 0. Moreover we must have β = γ because of thesymmetry of b, c. So, we must have an inequality of the form
a2
b+ c≥(
1
2+ 2x
)a− x(b+ c)
or
a2 ≥(
1
2+ 2x
)a(b+ c)− x(b+ c)2 or a2−
(1
2+ 2x
)a(b+ c) + x(b+ c)2 ≥ 0.
Since equality occurs when a = b = c, this quadratic is probably a perfectsquare that vanishes when 2a = b+ c, thus it must be (a− b+c
2 )2, and indeed
for x =1
4it is, thus we get the inequality
a2
b+ c≥ a− b+ c
4
with analogous ones which break up the original inequality. The most usedinequalities are xy ≥ 0 if x, y have the same sign, and especially the inequalityx2 ≥ 0. since every inequality can be transformed to E ≥ 0, a common way isbreaking E into sum of products of squares with positive numbers.
For example, the above inequality
a2 + b2 + c2 ≥ ab+ bc+ ca
is equivalent toa2 + b2 + c2 − ab− bc− ca ≥ 0
86
which breaks into
1
2(a− b)2 +
1
2(b− c)2 +
1
2(c− a)2 ≥ 0,
clearly true.Whereas most inequalities break by sum, some of them can be broken byproduct. Look at this problem from Saint-Petersburg 2005:(
x2 +3
4
)(y2 +
3
4
)(z2 +
3
4
)≥√
(x+ y)(y + z)(z + x).
Although the inequality is in three variables, it can be easily broken down intothree inequalities in just two variables:√(
x2 +3
4
)(y2 +
3
4
)≥√x+ y
and analogous.This inequalities are actually true: keeping in mind that x = y = z = 1
2 givethe equality, we can deduce(
x2 +1
4+
1
2
)(1
4+ y2 +
1
2
)≥(x+ y + 1
2
)2
but (x+ y + 1)2 ≥ 4(x+ y) by AM-GM, and from here the conclusion.
Some inequalities in many variables, like Cauchy–Schwartz, can be brokendown into simpler ones according to pairs of indices. For example, the in-equality
(a21 + . . .+ a2n)(b21 + . . .+ b2n)− (a1b1 + . . .+ anbn)2 ≥ 0,
if opened the brackets, can be grouped by pairs i, j of indices:∑1≤i<j≤n
(a2i b
2j + a2jb
2i − 2aibiajbj
)=
∑1≤i<j≤n
(aibj − ajbi)2 ≥ 0.
That’s how we established Lagrange’s Identity:
(a21 + . . .+ a2n)(b21 + . . .+ b2n)− (a1b1 + . . .+ anbn)2 =∑
1≤i<j≤n(aibj − ajbi)2.
Analogously one can prove the following inequality:If x1, x2, . . . , xn > 0 then
(x1 + x2 + . . .+ xn)
(1
x1+ . . .+
1
xn
)≥ n2.
Indeed, opening the brackets the inequality is equivalent to∑1≤i<j≤n
(xixj
+xjxi− 2
)≥ 0
which follows from AM-GM for two variables.This method of extracting squares leads to another way of proving inequali-ties, conventionally called the SOS (Sum of Squares) method, which shall bepresented in a section to come.
87
Exercises
1. If a, b, c > 0 then (a+ b)(b+ c)(c+ a) ≥ 8abc
Solution: The inequality breaks by product into
a+ b ≥ 2√ab, b+ c ≥ 2
√bc, c+ a ≥ 2
√ca.
2. (a+ b− c)2 + (a+ c− b)2 + (b+ c− a)2 +3
4≥ a+ b+ c.
Solution: we break it by sum into
(a+b−c)2+1
4≥ a+b−c, (a+c−b)2+1
4≥ a+c−b, (b+c−a)2+
1
4≥ b+c−a.
3. If a, b, c, d > 0 then a4b+ b4c+ c4d+ d4a ≥ abcd(a+ b+ c+ d)
Solution: We wish to break the inequality into
ma4b+ nb4c+ pc4d+ qd4a ≥ a2bcd,
where m+n+ p+ q = 1. Combined with the analogous cyclic relations,this would yield the conclusion.
Now according to weighted AM-GM we have
ma4b+ nb4c+ pc4d+ qd4a ≥ a4m+qb4n+mc4p+nd4q+p.
Thus we need 4m+ q = 2, 4n+m = 1, 4p+n = 1, 4q+p = 1. We thendeduce q = 2− 4m, p = 1− 4q = 1− 4(2− 4m) = 16m− 7, n = 1− 4p =1 − 4(16m − 7) = 29 − 64m. Then 4n + m = 1 or 116 − 255m = 1 som = 115
255 = 2351 . So we compute then n = 7
51 , p = 1151 , q = 10
51 and we’vegot the breaking.
4. If a, b, c > 0 then
(a+b+c)
(a
(2a+ b+ c)(b+ c)+
b
(2b+ c+ a)(c+ a)+
c
(2c+ b+ a)(b+ a)
)≥ 9
8.
Solution: We may notice that
a
(2a+ b+ c)(b+ c)=
1
2(b+ c)− 1
2(a+ b) + 2(a+ c)
≥ 1
2(b+ c)− 1
8(a+ b)− 1
8(a+ c).
By summing with the analogous terms we have
a
(2a+ b+ c)(b+ c)+
b
(2b+ c+ a)(c+ a)+
c
(2c+ b+ a)(b+ a)≥
≥ 1
4
(1
a+ b+
1
b+ c+
1
c+ a
)≥ 1
4
9
2(a+ b+ c)=
9
8(a+ b+ c).
88
5. Let a, b, c ≥ 0 with a2+b2+c2 = 1. Show thata
1− a2+
b
1− b2+
c
1− c2≥
3√
3
2.
Solution: we notice the inequalityx
1− x2≥ 3√
3
2x2, which is equivalent
to 3√
3x(1− x2) ≤ 4 or 27x2 · 1−x22 · 1−x22 ≤ 1, which follows by AM-GM
for x2, 1−x2
2 , 1−x2
2 . This inequality solves the problem.
6. If a, b, c ≥ 0 and a2 + b2 + c2 = 1 then a1−a4 + b
1−b4 + c1−c4 ≥
5 4√54 .
Solution: Again we break the inequality according to the relation
x
1− x4≥ 5 4√
5
4x2.
This relation is equivalent to5 4√
5
4x(1− x4) ≤ 1, or raised to the fourth
power to x4 · 1−x44 · 1−x44 · 1−x44 · 1−x44 ≤ 55, which follows by AM-GM.
7. Let x, y, z > 0. Then
(x2+y2−z2)(y2+z2−x2)(x2+z2−y2) ≤ (x+y−z)2(y+z−x)2(x+z−y)2.
Solution: It suffices to consider the case when all x2+y2−z2, x2+z2−y2,y2 + z2 − x2 are positive, because at most one of them can be negative(the pairwise sums are positive, so we can’t have two negative numbersamong them), and when one of them is negative, LHS is negative andRHS is positive.
Now we shall break the inequality into
(x2 + y2 − z2)(x2 + z2 − y2) ≤ (x+ y − z)2(x+ z − y)2
and the analogous ones.
This is equivalent to x4 − (y2 − z2)2 ≤[x2 − (y − z)2
]2or
x4 − (y + z)2(y − z)2 ≤ x4 − 2(y − z)2x2 + (y − z)4,
or that (y − z)4 + (y + z)2(y − z)2 − 2(y − z)2x2 ≥ 0 or
(y − z)2[(y + z)2 + (y − z)2 − 2x2
]≥ 0
or 2(y − z)2(y2 + z2 − x2) ≥ 0, which is true.
8. For a, b, c > 0 we havea2
b2+b2
c2+c2
a2≥ a
b+b
c+c
a.
Solution: As in example 3, we have a breaking that can be computed by
constructing a system of equations:2
3
a2
b2+
1
6
b2
c2+
1
6
c2
a2≥ a
b.
89
9. For x, y, z > 0,
1
3x+ y+
1
3y + z+
1
3z + x≥ 1
2x+ y + z+
1
2y + z + x+
1
2z + x+ y.
Solution: We seek an inequality of form
m
3x+ y+
n
3y + z+
p
3z + x≥ 1
2x+ y + z
where m+ n+ p = 1.
We have
m
3x+ y+
n
3y + z+
p
3z + x=
m2
3mx+my+
n2
3ny + nz+
p2
3pz + xp
≥ 1
(3m+ p)x+ (3n+m)y + (3p+m)z.
By solving the system we get m = 47 , n = 1
7 , p = 27 .
10. Let 0 ≤ x ≤ y, a1, a2, . . . , an, b1, b2, . . . , bn > 0, n ≥ 2.
Then(n∑i=1
a1+xi b1−xi
)(n∑i=1
a1−xi b1+xi
)≤
(n∑i=1
a1+yi b1−yi
)(n∑i=1
a1−yi b1+yi
).
Solution: Like CBS, we break the inequality into pairs of indices. Indeedthe LHS is
n∑i=1
a2i b2i +
∑1≤i<j≤n
(a1+xi b1−xi a1−xj b1+xj + a1−xi b1+xi a1+xj b1−xj )
and the RHS is
n∑i=1
a2i b2i +
∑1≤i<j≤n
(a1+yi b1−yi a1−yj b1+yj + a1−yi b1+yi a1+yj b1−yj ).
So we have to show that
a1+xi b1−xi a1−xj b1+xj +a1−xi b1+xi a1+xj b ≤ a1+yi b1−yi a1−yj b1+yj +a1−yi b1+yi a1+yj b1−yj .
By dividing by aibiajbj we transform this to a nicer form(aibjbiaj
)x+
(biajaibj
)x≤(aibjbiaj
)y+
(biajaibj
)y,
of ux+1
ux≤ uy +
1
uy, or
u2x + 1
ux≤ u2y + 1
uyor (ux+y−1)(xy−x−1) ≥ 0,
which is true.
90
11. If x1, . . . , xn > 0 such that x1x2 . . . xn = 1 and p, q > 0 such thatq
p≥ n(n− 2), then
√px21 + q +
√px22 + q + . . .+
√px2n + q ≤
√p+ q(x1 + x2 + . . .+ xn).
Solution: We try to break the inequality into√mx21 + n ≤ ax1 +
√m+ n− an− 1
(x2 + x3 + . . .+ xn).
As the LHS depends only on x1, we apply AM-GM to the RHS to get
the inequality ax1 +
√p+ q − an− 1
(x2 + x3 + . . .+ xn) ≥ ax+
√p+ q − a
x1
n−1(we let x = x1 for simplicity).
Now square to get
a2x2+2a(√p+ q − a
)x1− 1
n−1+(p+ q − 2
√p+ qa+ a2
)x− 2n−1 ≥ px2+q,
or (a2−p)x2+2a (√p+ q − a)x
1− 1n−1+
(p+ q − 2
√p+ qa+ a2
)x− 2n−1 ≥
q. We apply now weighted AM-GM to get
(a2 − p)x2 + 2a(√p+ q − a
)x1− 1
n−1 +(p+ q − 2
√p+ qa+ a2
)x− 2n−1 ≥
≥[(a2 − p) + 2a
(√p+ q − a
)+(p+ q − 2
√p+ qa+ a2
)]·
·x2(a2−p)+
(1− 1
n−1)2a(√p+q−a)+
(− 2n−1
)(p+q−2
√p+qa+a2)
(a2−p)+2a(√p+q−a)+(p+q−2
√p+qa+a2)
= qx
2nn−1
√p+qa−2
(pnn−1 + q
n−1)
q .
So now for the desired inequality to take place, we must take
a =
n
n− 1p+
1
n− 1q
n√p+ q
=√p+ q
np+ q
n(p+ q).
We can see that√p < a <
√p+ q so for this value the desired breaking
holds and the conclusion follows.
12. If a, b, c > 0 and abc = 1 then1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a≤ 1.
Solution: Let’s try to find a breaking of the form1
1 + a+ b≤ ct
at + bt + ct,
or at+bt+ct ≤ ct+ct(a+b), so at+bt ≤ ct(a+b) so a2tbt+b2tat ≤ a+b.
This is achieved for t =1
3.
13. If a, b, c > 0 thena√
a2 + 8bc+
b√b2 + 8ac
+c√
c2 + 8ab≥ 1
91
Solution: Again let’s break the inequality intoa√
a2 + 8bc≥ as
as + bs + cs
or a2s(a2 + 8bc) ≤ a2(as + bs + cs)2. We must have
a2s + 8a2s−2bc ≤ a2s + b2s + c2s + 2bscs + 2as(bs + cs)
so 8a2s−2bc ≤ b2s + c2s + 2bscs + 2as(bs + cs).
As b2s + c2s ≥ 2bscs and bs + cs ≥ 2bs2 c
s2 , it suffices to have
8a2s−2bc ≤ 4bscs + 4asbs2 c
s2 .
If we let d =√bc this amount to 2a2s−2d2 ≤ d2s + asds. However by
AM-GM d2s+asds ≥ 2as2d
3s2 , and for s =
4
3we get the desired relation.
1.12 Separating the squares
Sometimes it’s hard to break a inequality into sum of squares, and we must use
some special tricks. One very good trick is the easy principle a− b =a2 − b2
a+ b,
which helps us deal with inequalities involving radicals. Look at√a2 + 2b2 +
√b2 + 2a2 ≥ 2
√a2 + ab+ b2.
The first idea is squaring, selecting the radical, and squaring again, whichproduces a terrible mess of computations.
For another way, we first observe that if
a = b,√a2 + 2b2 =
√b2 + 2a2 =
√a2 + ab+ b2,
so we want to extract a− b as a factor. To do this, use our trick:√a2 + 2b2 −
√a2 + b2 + ab =
a2 + 2b2 − a2 − b2 − ab√a2 + 2b2 +
√a2 + b2 + ab
=b(b− a)√
a2 + 2b2 +√a2 + b2 + ab
.
We can factor√2a2 + b2 −
√a2 + b2 + ab =
a(a− b)√2a2 + b2 +
√a2 + b2 + ab
,
and so the inequality is equivalent to
(b− a)
(b√
a2 + 2b2 +√a2 + b2 + ab
− a√2a2 + b2 +
√a2 + b2 + ab
)≥ 0.
We have selected just b − a, which is not a square, and so cannot help us:in fact it can take both positive and negative values. But, if we think thatthe original expression is symmetric in a, b, we can expect to factor it alsosymmetrically in a, b. As b − a is not symmetric but (b − a)2 is, we can be
92
pretty sure that we can extract b−a once again (this is just intuitive reasoning,but it leads us to the correct result). First, let’s clear the denominators:
(b− a)b(√
a2 + 2b2 +√a2 + b2 + ab
)− a
(√2a2 + b2 +
√a2 + b2 + ab
)(√
2a2 + b2 +√a2 + b2 + ab
)(√a2 + 2b2 +
√a2 + b2 + ab
) ≥ 0.
The denominator is clearly positive, so let’s look at the numerator which canbe written as
(b− a)√a2 + b2 + ab+
√2a2b2 + b4 −
√2a2b2 + a4.
We have no problem in extracting b− a from√2a2b2 + b4 −
√2a2b2 + a4
using our trick:√2a2b2 + b4 −
√2a2b2 + a4 =
(b− a)(b4 + b3a+ b2a2 + ba3 + a4)√2a2b2 + b4 +
√2a2b2 + a4
.
That’s why the numerator is
(b− a)2(√
a2 + b2 + ab+b4 + b3a+ b2a2 + ba3 + a4√
2a2b2 + b4 +√
2a2b2 + a4
),
thus positive, which finishes the proof. [Note: actually this inequality can be
easily handled by Minkowski’s Inequality for p =1
2].
When we have such inequalities, symmetric in a few variables, and that becomeequalities when all the variables are equal, its useful to select out the squaresof the differences ((a− b)2, (b− c)2), perhaps using the trick above. But thesedifferences help us even sometimes when the inequalities do not transform intoequalities when all the variables are equal:Consider the inequalityIf a, b, c are positive numbers which form a triangle (that if a < b+c, b < c+a,c < a+ b), then
2ab+ 2bc+ 2ca > a2 + b2 + c2.
We don’t see an obvious method of writing it as a sum of squares, but sincewe see a2, b2, c2, 2ab, 2bc and 2ca We try to get the squares
(a+ b)2, (b+ c)2, (c+ a)2 or (a− b)2, (b− c)2, (c− a)2.
We transform the inequality into
0 > a2 + b2 + c2 − 2ab− 2bc− 2ca
or intoa2 + b2 + c2 ≥ (a− b)2 + (b− c)2 + (c− a)2
which breaks into the inequalities
a2 ≥ (b− c)2, b2 ≥ (c− a)2, c2 ≥ (a− b)2.
93
They are true: since b < a+c we get a > b−c; since c < a+b we get a > c−b,hence |a| ≥ |b− c| and a2 ≥ (b− c)2, the other inequalities being analogous.When trying to extract the squares we must reason logically and usually thisleads to a solution. For example, let’s try to break the inequality
a3 + b3 + c3 + 3abc− a2b− b2a− b2c− c2b− a2c− c2a ≥ 0.
If we write it in the form
P1(a, b, c)(a− b)2 + P2(a, b, c)(b− c)2 + P3(a, b, c)(c− a)2
then P1, P2, P3 must be linear, thus we must have P1(a, b, c) = ax + by + cz,and from the symmetry in a, b ce suppose x = y. Further, from the totalsymmetry we may assume
P2(a, b, c) = x(b+ c) + za, P3(a, b, c) = x(a+ c) + zb.
To compute, x, z, set a = b. Then the relation reduces to
c3−2c2a+ca2 = c(a−c)2 = (P2(a, a, c)+P3(a, a, c))(a−c)2 = (2(x+z)a+2xc)(c−a)2.
This yields us x =1
2, z = −1
2and so the inequality is equivalent to
(b+ c− a)(b− c)2 + (a+ c− b)(a− c)2 + (a+ b− c)(a− b)2.
Actually one of these terms may be negative, but the problem is simpler tosolve in this form.Indeed, as it’s symmetric, we may assume that a ≥ b ≥ c. If b+ c ≥ a, we aredone. Otherwise, we must prove that
(a− b− c)(b− c)2 ≤ (a+ c− b)(a− c)2 + (a+ b− c)(a− b)2.
However a−b−c < a+c−b and b−c ≤ a−c and from here the result follows.
The method of extracting squares of differences is conventionally called SOS(Sum of Squares) method, and is usually used to solve hard inequalities.
Sometimes it’s hard to see the square. In this case it’s good to find an in-
termediary expression. For example, if we want to express√ab − 2ab
a+ bwith
respect to (a− b)2, we might remember that
a+ b
2−√ab =
(√a−√b)2
2=
(a− b)2
2(√
a+√b)2
anda+ b
2− 2ab
a+ b=
(a− b)2
a+ b
hence√ab− 2ab
a+ b=
(a− b)2
2
(1
a+ b− 1
a+ b+ 2√ab
)≥ 0.
94
Exercises
1. If a2 + b2 = 1 and a, b > 0 then a+ b+1
ab≥ 2 +
√2
Solution: The inequality can be rewiteen as√
2− a− b ≤ a2 + b2
ab− 2 or
as√
2(a2 + b2)−a−b ≤ a
b+b
a−2 or as
(a− b)2
a+ b+√
2≤
(√a
b−√b
a
)2
=
(a− b)2
abwhich transforms to ab ≤ a+ b+
√2 which is obviously true.
2. If a, b, c > 0 and a2 + b2 + c2 = 1 then a+ b+ c+1
abc≥ 4√
3
Solution: We write the conclusion as√
3− (a+ b+ c) ≤ 1
abc− 3√
3. It’s
clear that both sides are positive (from AM-GM). This is equivalent to
3− (a+ b+ c)2√3 + a+ b+ c
≤ 1− 27a2b2c2
abc(1 + 3
√3abc
) .Now we remember that a2 + b2 + c2 = 1 and we use the identities
3(a2 + b2 + c2)− (a+ b+ c)2 = (a− b)2 + (b− c)2 + (c− a)2
and (x+y+z)3−27xyz =∑cyc
(x−y)2x+ y + 7z
2(for x = a2, y = b2, z =
c2) to transform our inequality into∑cyc
(a− b)2(a+ b)2 a2+b2+7c2
2
abc(1 + 3
√3abc
) ≥
∑cyc
(a− b)2
√3 + a+ b+ c
.
However (a + b)2 ≥ 4ab,a2 + b2 + 7c2
2=
1 + 6c2
2≥√
6c from AM-
GM, and abc ≤ 13√3, implying that LHS is at least
∑cyc
2√
6(a − b)2 ≥
∑cyc
(a− b)2√3
, which is, in turn, greater than RHS.
3. Let a, b, c be nonnegative numbers. Show that
a+ b+ c
3− 3√abc ≤ max
{(√a−√b)2, (
√a−√c)2, (
√b−√c)2}.
Solution: Set a = x6, b = y6, c = z6. WLOG suppose x ≥ y ≥ z. Wemake use of the identity
u3 + v3 + w3 − 3uvw =1
2(u+ v + w)
[(u− v)2 + (v − w)2 + (w − u)2
],
rewriting the conclusion as
1
6(x2 + y2 + z2)
[(x2 − y2)2 + (y2 − z2)2 + (x2 − z2)2
]≤ (x3 − z3)2,
95
or
(x2 + y2 + z2)(x+ y)2(x− y)2 + (x2 + y2 + z2)(y + z)2(y − z)2 +
+(x2 + y2 + z2)(x+ z)2(x− z)2 ≤ 6(x2 + z2 + zx)(z − x)2.
However we have
x2 + y2 + z2 ≤ 2(x2 + z2 + xz),
(x− y)2 + (y − z)2 ≤ (x− z)2,(y − z)2 ≤ (x− z)2,
Therefore we have
6(x2 + xz + z2)2(x− z)2 − [(x2 + y2 + z2)(x+ y)2(x− y)2 +
+(x2 + y2 + z2)(y + z)2(y − z)2 + (x2 + y2 + z2)(x+ z)2(x− z)2] ≥≥ 2(x2 + zx+ x2)
[(2x2 + xz + 2z2)(x− z)2 − (y + z)2(y − z)2 − (x+ y)2(x− y)2
].
Therefore it suffices to show that
(2x2 + 2z2 + xz)(x− z)2 ≥ (x+ y)2(x− y)2 + (y + z)2(y − z)2.
Now set u = x − y, v = y − z, u + v = x − z. The inequality to showtransforms then into[2(z + u+ v)2 + 2z2 + z(z + u+ v)
](u+v)2 ≥ (2z+2u+v)2u2+(2z+v)2v2,
or
z2[5(u+ v)2 − 4u2 − 4v2
]+[5(u+ v)3 − 4(2u+ v)u2 − 4v3
]z +
+2(u+ v)4 − (2u+ v)2v2 − v4 ≥ 0.
It is now pretty clear that 5(u+v)2−4u2−4v2, 5(u+v)3−4(2u+v)u2−4v3,2(u+ v)4 − (2u+ v)2v2 − v4 are positive, so the proof is finished.
4. Let a, b, c > 0. Show that
4(ab+ bc+ ca)
(1
(a+ b)2+
1
(b+ c)2+
1
(c+ a)2
)≥ 9.
Solution: We have
4(ab+bc+ca) = (a+b)2+(b+c)2+(c+a)2−(a−b)2−(b−c)2−(c−a)2
so
4(ab+ bc+ ca)
[1
(a+ b)2+
1
(b+ c)2+
1
(c+ a)2
]− 9 =
= −[(a− b)2 + (b− c)2 + (c− a)2
] [ 1
(a+ b)2+
1
(b+ c)2+
1
(c+ a)2
]+
+[(a+ b)2 + (b+ c)2 + (c+ a)2
] [ 1
(a+ b)2+
1
(b+ c)2+
1
(c+ a)2
]− 9
= −[(a− b)2 + (b− c)2 + (c− a)2
] [ 1
(a+ b)2+
1
(b+ c)2+
1
(c+ a)2
]+
+(a− b)2(a+ b+ 2c)2
(a+ c)2(b+ c)2+
(b− c)2(b+ c+ 2a)2
(a+ b)2(a+ c)2+
(a− c)2(a+ 2b+ c)2
(a+ b)2(b+ c)2
96
(we used here the identity (x + y + z)
(1
x+
1
y+
1
z
)=
(x− y)2
xy+
(x− z)2
xz+
(y − z)2
yz.)
As(a+ b+ 2c)2
(a+ c)2(b+ c)2=
1
(a+ c)2+
1
(b+ c)2+
2
(a+ c)(b+ c), we finally
write
4(ab+ bc+ ca)
[1
(a+ b)2+
1
(b+ c)2+
1
(c+ a)2
]− 9 =
= (a− b)2[
2
(a+ c)(b+ c)− 1
(a+ b)2
]+ (a− c)2
[2
(a+ b)(b+ c)− 1
(a+ c)2
]+
+(b− c)2[
2
(a+ b)(a+ c)− 1
(b+ c)2
].
Assume that a ≥ b ≥ c. Then we have2
(a+ b)(b+ c)− 1
(a+ c)2≥ 0 and
2
(a+ c)(b+ c)− 1
(a+ b)2≥ 0. If
2
(a+ b)(a+ c)− 1
(b+ c)2≥ 0 then we
are done. So assume that2
(a+ b)(a+ c)− 1
(b+ c)2< 0. Then we rewrite
the inequality as
(a− b)2[
2
(a+ c)(b+ c)− 1
(a+ b)2
]+ (a− c)2
[2
(a+ b)(b+ c)− 1
(a+ c)2
]≥
≥ (b− c)2[
1
(b+ c)2− 2
(a+ b)(a+ c)
].
We shall prove that
(a−c)2[
2
(a+ b)(b+ c)− 1
(a+ c)2
]≥ (b−c)2
[1
(b+ c)2− 2
(a+ b)(a+ c)
].
After clearing denominators and cancelling common terms we reducethis to
(a− c)2
a+ c(2a2+3ac+2c2−b2−ab−bc) ≥ (b− c)2
b+ c(a2+ab+ac−2b2−2c2−3bc).
However (a−c)2a+c ≥
(b−c)2b+c as this is equivalent to
(a− b)[(a+ b)(b+ c)− 4c2
]≥ 0
and 2a2 + 3ac + 2c2 − b2 − ab − bc ≥ a2 + ab + ac − 2b2 − 2c2 − 3bc asthis is equivalent to a2 + b2 + 2c2 − 2ab+ 2ac+ 2bc ≥ 0, clearly true.
5. If a, b, c > 0 then show that (a3 + b3 + c3)2 ≥ (a4 + b4 + c4)(ab+ bc+ ca).
Solution: We need to write the difference
(a3 + b3 + c3)2 − (a4 + b4 + c4)(ab+ bc+ ca)
in SOS style?.
97
We might remember that (a3 + b3 + c3)2 − (a4 + b4 + c4)(a2 + b2 + c2)can be written in this form according to Lagrange Identity, and clearly
(a4 + b4 + c4)(a2 + b2 + c2 − ab− bc− ca),
can be written in that form, too.
So
(a3 + b3 + c3)− (a4 + b4 + c4)(ab+ bc+ ca) =
= (a3 + b3 + c3)2 − (a4 + b4 + c4)(a2 + b2 + c2) +
+(a4 + b4 + c4)(a2 + b2 + c2 − ab− bc− ca)
= −(a− b)2a2b2 − (b− c)2b2c2 − (c− a)2c2a2 +
+(a4 + b4 + c4)
[(a− b)2 + (b− c)2 + (c− a)2
2
]=
1
2[(a− b)2(a4 + b4 − 2a2b2 + c4) + (b− c)2(b4 + c4 − 2b2c2 + a4) +
+(c− a)2(c4 + a4 − 2a2c2 + b4)] ≥ 0.
6. If a, b, c > 0 then 9(a4 + b4 + c4)2 ≥ (a5 + b5 + c5)(a+ b+ c)3.
Solution: We use the same method as in the previous problem. Keepingin mind that
(a4+b4+c4)2−(a5+b5+c5)(a3+b3+c3) = −a3b3(a−b)2−b3c3(b−c)2−c3a3(a−c)2
and
9(a3 + b3 + c3)− (a+ b+ c)3 =
= 2(a3 + b3 + c3 − 3abc) + 3(a3 + b3 − a2b− ab2) + 3(b3 + c3 − b2c− bc2) +
+3(a3 + c3 − a2c− ac2)= (a− b)2(a+ b+ c+ 3(a+ b)) + (b− c)2(a+ b+ c+ 3(b+ c)) +
+(c− a)2(a+ b+ c+ 3(a+ c)),
we write 9(a4 + b4 + c4)2 − (a5 + b5 + c5)(a+ b+ c)3 as
(a− b)2[(a5 + b5 + c5)(4a+ 4b+ c)− 9a3b3] +
+(b− c)2[(a5 + b5 + c5)(4b+ 4c+ a)− 9b3c3] +
+(c− a)2[(a5 + b5 + c5)(4a+ 4c+ b)− 9a3c3].
As (a5+b5+c5)(4a+4b+c) ≥ 4(a5+b5)(a+b) = 4(a6+b6+a5b+ab5) ≥16a3b3, we conclude that all terms accompanying the squares are non-negative so we are done.
7. If a, b, c > 0 then1
a+
1
b+
1
c+
9
a+ b+ c≥ 4
(1
a+ b+
1
b+ c+
1
c+ a
).
Solution: We have
1
a+
1
b+
1
c+
9
a+ b+ c− 4
(1
a+ b+
1
b+ c+
1
c+ a
)=
=1
2
[(a− b)2
ab(a+ b)+
(a− c)2
ac(a+ c)+
(b− c)2
bc(b+ c)
]+
9
a+ b+ c− 2
a+ b− 2
b+ c− 2
a+ c.
98
Now knowing the identity
(x+ y + z)
(1
x
1
y+
1
z
)− 9 =
(x− y)2
xy+
(x− z)2
xz+
(y − z)2
yz,
we obtain
9
a+ b+ c− 2
a+ b− 2
b+ c− 2
a+ c=
=1
a+ b+ c
[(b− c)2
(a+ b)(a+ c)+
(a− b)2
(c+ a)(c+ b)+
(a− c)2
(b+ a)(b+ c)
],
so finally
1
a+
1
b+
1
c+
9
a+ b+ c− 4
(1
a+ b+
1
b+ c+
1
c+ a
)=
=1
2
[(a− b)2
[1
ab(a+ b)− 4
(a+ c)(b+ c)(a+ b+ c)
]+
+(a− c)2[
1
ac(a+ c)− 2
(a+ b)(b+ c)(a+ b+ c)
]+
+(c− b)2[
1
cb(c+ b)− 2
(a+ c)(b+ a)(a+ b+ c)
] ].
In what cases do we have the inequality1
ab(a+ b)≤ 2
(a+ c)(b+ c)(a+ b+ c)?
This is equivalent to ab(a + b) ≥ c(a2 + b2 + 3ab) + 2c2(a + b) + c3 soc < ab
a+b and c is the smallest of a, b, c.
Now assume that a ≥ b ≥ c and write the inequality as
(a− b)2[
2
(a+ c)(b+ c)(a+ b+ c)− 1
ab(a+ b)
]≤
≤ (a− c)2[
1
ac(a+ c)− 2
(a+ b)(b+ c)(a+ b+ c)
]+
+(c− b)2[
1
cb(c+ b)− 2
(a+ c)(b+ a)(a+ b+ c)
].
However we may note that (a− c) ≥ (a− b) and
1
ac(a+ c)− 2
(a+ b)(b+ c)(a+ b+ c)≥ 2
(a+ c)(b+ c)(a+ b+ c)− 1
ab(a+ b).
This is equivalent to
1
ab(a+ b)+
1
ac(a+ c)≥ 2
[1
(a+ c)(b+ c)(a+ b+ c)+
1
(a+ b)(b+ c)(a+ b+ c)
].
However we may prove that
1
ab(a+ b)+
1
ac(a+ c)≥ 2
a√bc(a+ b)(a+ c)
≥ 8
a(b+ c)(2a+ b+ c)
99
and1
(a+ c)(b+ c)(a+ b+ c)+
1
(a+ b)(b+ c)(a+ b+ c)<
<1
a(b+ c)(a+ b+ c)+
1
a(b+ c)(a+ b+ c)=
2
a(b+ c)(a+ b+ c).
So it suffices to prove that8
a(b+ c)(2a+ b+ c)≥ 4
a(b+ c)(a+ b+ c)which is equivalent to 2(a+ b+ c) ≥ 2a+ b+ c, true!
8. Let a, b, c > 0 and a+ b+ c = 1. Prove that√a+ abc+
√b+ abc+
√c+ abc ≥ 2
3
(√a+√b+√c).
Solution: Let’s substitute
a =x2
x2 + y2 + z2, b =
y2
x2 + y2 + z2, c =
z2
x2 + y2 + z2.
We have reduced the inequality to proving
3∑cyc
x√
(x2 + y2)(x2 + z2) ≥ 2(x+ y + z)(x2 + y2 + z2).
Lemma: 2(x3 + y3 + z3)−∑sym
x2y =∑sym
(x+ y)(x− y)2.
Proof : Immediately from x3 + y3 − xy(x+ y) = (x+ y)(x− y)2.
Now we see that
3∑cyc
x(x2 + y2) + (x2 + z2)
2−RHS =
1
2
∑cyc
(x+ y)(x− y)2
using Lemma. However
3∑cyc
x(x2 + y2) + (x2 + z2)
2− LHS =
3
2
∑cyc
(√x2 + y2 −
√x2 + z2
)2.
So we are left to prove that∑cyc
(x+ y)(x− y)2 ≥ 3∑cyc
x(√
x2 + y2 −√x2 + z2
)2.
We prove that
(x+y)(x−y)2 ≥ 3z(√
z2 + y2 −√z2 + x2
)2=
3z(x− y)2(x+ y)2(√z2 + y2 +
√z2 + x2
)2 .Thus we have to prove
(√z2 + y2 +
√z2 + x2
)2≥ 3z(x+ y) or
2z2 + x2 + y2 + 2√z2 + x2
√z2 + y2 ≥ 3z(x+ y).
However√z2 + x2
√z2 + y2 ≥ z2 + xy by CBS. Thus we reduce it to
4z2 + x2 + y2 + 2xy ≥ 3z(x+ y)
or 4z2 + (x + y)2 ≥ 3z(x + y). However 4z2 + (x + y)2 ≥ 4z(x + y) byAM-GM and so we are done.
100
1.13 The Dual Principle
This is very simple idea, and yet it prove to have quite a few applications - andnot only in inequalities, but also in computational geometry. In particular, itis useful in solving algebraic inequalities, as we will see in the examples.
The positive reals a, b, c are sides of a triangle if and only if there are positivereals x, y, z with a = y + z, b = x+ z, c = x+ y.
Indeed, if a = y+ z, b = x+ z, c = x+ y, then a = y+ z < 2x+ y+ z = b+ c
and analogously b < c + a, c < a + b. Conversely, setting x =b+ c− a
2,
y =a+ c− b
2, z =
a+ b− c2
we satisfy the requirements.
Let’s look at one already met inequality through the prism of dual principle:
2(ab+ bc+ ca) ≥ a2 + b2 + c2
is equivalent, by dual principle, to
2 [(x+ y)(y + z) + (x+ y)(x+ z) + (y + z)(x+ z)] ≥ (x+y)2+(z+x)2+(y+z)2 ≥ 0
or to
2(x2 + y2 + z2) + 3(xy + yz + zx) ≥ 2(x2 + y2 + z2) + 2(xy + yz + zx)
which is obvious.
101
Exercises
1. If a, b, c are sides of a triangle then
a(b− c)2 + b(c− a)2 + c(a− b)2 + 4abc > a3 + b3 + c3.
Solution: After substituting a = y+z, b = x+z, c = x+y by cancellingcommon terms the inequality becomes xyz > 0.
2. If a, b, c form a triangle thena
2a+ b+ c+
b
2b+ c+ a+
c
2c+ a+ b>
2
3.
Solution: After applying the CBS Lemma we get
a
2a+ b+ c+
b
2b+ c+ a+
c
2c+ a+ b≥ (a+ b+ c)2
2(a2 + b2 + c2) + 2(ab+ bc+ ca)
so we left to prove that 3(a+b+c)2 > 2[2(a2 + b2 + c2) + 2(ab+ bc+ ca)
]or 2(ab+ bc+ ca) > a2 + b2 + c2 which we have already proven.
3. If a, b, c are the sides of a triangle and ax+by+cz = 0 then xy+yz+zx ≤0.
Solution: We can suppose that x, y > 0, because the inequality is sym-metric and doesn’t change if we replace x, y, z by −x, −y, −z. Wehave
xy + yz + zx = xy + z(x+ y) = xy − (ax+ by)(x+ y)
c.
So we have to prove that cxy ≤ (ax+ by)(x+ y). But
cxy ≤ (a+ b)xy ≤ (ax+ by)(x+ y).
We see that in this case we managed better without using the dualprinciple.
4. Show that in a triangle, the length of a median from a vertex is not lessthan the length of the bisector from the same vertex.
Solution: We use the formula ma =
√2b2 + 2c2 − a2
4and
la =
√bc(b+ c)(b+ c− a)(b+ c+ a)
(b+ c)2. So we have to prove
2b2 + 2c2 − a2
4≤ bc(b+ c− a)(b+ c+ a)
(b+ c)2.
Now by using the dual principle we have
2b2 + 2c2 − a2
4=
2(x+ z)2 + 2(y + z)2 − (x+ y)2
4
=x2 + y2 − 2xy + 4z2 + 4zy + 4zx
4≥ z2 + zy + zx = z(x+ y + z).
102
However
l2a =bc(b+ c− a)(b+ c+ a)
(b+ c)2≤
(b+ c)2
4(b+ c− a)(b+ c+ a)
(b+ c)2
=(b+ c− a)(b+ c+ a)
4= z(x+ y + z).
5. In a triangle, show thatrarbmamb
+rbrcmbmc
+rcramcma
≥ 3.
Solution: We pass by dual principle to x, y, z. Then we must prove∑cyc
x(x+ y + z)
mbmc≥ 3.
We prove more∑cyc
x(x+y+z)m2
b+m2c≥ 3
2. Which transforms to
∑cyc
x(x+ y + z)
(y + z)(x+ y + z) + (x− z)2 + (y − z)2≥ 3
2
or ∑cyc
x2
y + z + [(x− z)2 + (x− y)2]x
x+ y + z
≥ 3
2.
Now applying (Cauchy) this reduces to
(x+ y + z)2 ≥ 3(xy + yz + zx) +∑cyc
x+ y
x+ y + z(x− y)2
which is just∑cyc
z
x+ y + z(x− y)2 ≥ 0 true.
6. Prove that: In the triangle ABC:
ma · cosA
2+mb · cos
B
2+mc · cos
C
2≥ 3
4(a+ b+ c).
Solution: We pass to dual principle. It’s easy to see that
m2a = x(x+ y + z) +
(y − z)2
4, cos
A
2=
√x(x+ y + z)
(x+ y)(x+ z).
Now let’s prove the following: ma cosA
2≥ p
(x
x+ y+
x
x+ z
)- where
p = x + y + z = a+b+c2 . The inequality would follow then by summing
the desired terms.
First, we consider the expression
(1) ma−√x(x+ y + z) =
m2a −
√x(x+ y + z)
2
ma +√x(x+ y + z)
=(y − z)2
4[ma +
√x(x+ y + z)
]103
Denote this expression by E.
Then we must have
(2)[√
x(x+ y + z) + E] √x(x+ y + z)√
(x+ y)(x+ z)− p
(x
x+ y+
x
x+ z
)=
=E√x(x+ y + z)√
(x+ y)(x+ z)− 1
2p
(√x
x+ y−√
x
x+ z
)2
.
We are to prove this expression is greater than zero.
However√x
x+ y−√
x
x+ z=
√x(x+ z)−
√x(x+ y)√
(x+ y)(x+ z)
=x(x+ z)− x(x+ y)[√
x(x+ z) +√x(x+ y)
]√(x+ y)(x+ z)
.
Squaring we get
(3)x2(y − z)2[√
x(x+ z) +√x(x+ y)
]2(x+ y)(x+ z)
.
Substituting 1) and 3) into 2) we produce√x(x+ y + z)(y − z)2
4[ma +
√x(x+ y + z)
] ≥ x2(x+ y + z)(y − z)2
2[√
x(x+ z) +√x(x+ y)
]2(x+ y)(x+ z)
.
Now cancelling common terms and grouping leads to
2[ma +
√x(x+ y + z)
]√x(x+ y + z) ≤
√(x+ y)(x+ z)
(√x+ y +
√x+ z
)2.
However√
(x+ y)(x+ z) ≥√x(x+ y + z) thus is suffices to prove that
2[ma +
√x(x+ y + z)
]≤(√x+ y +
√x+ z
)2= 2x+y+z+2
√(x+ y)(x+ z).
This breaks into the inequality 2ma < 2x+ y + z and√x(x+ y + z) ≤√
(x+ y)(x+ z) which are true (the first one is the well known ma ≤b+ c
2).
1.14 Substitutions
The Dual Principle actually consists of substituting x =b+ c− a
2, y =
a+ c− b2
, z =a+ b− c
2. There are many other substitutions, which solve
many inequalities impossible to solve in other way. Basically every inequalityis solved by substituting some variables into a known inequality, possibly morethan one time.
104
When we have some numbers with product 1, say xyz = 1, it’s convenient to
set x =a
b, y =
b
cthen z =
c
a.
For example the inequality
x
y+y
z+x
z≥ x+ y + z, xyz = 1
reduces toac
b2+ab
c2+bc
a2≥ ab2 + bc2 + ca2
abc,
ora3c3 + a3b3 + b3c3 ≥ a2b3c+ ab2c3 + a3bc2,
oru3 + v3 + w3 ≥ u2v + v2w + w2u,
where u = ab, v = ac, w = bc, and this is a known inequality - we’ve done itbefore.For the inequality
x11 + x1x2
+x2
1 + x2x3+ . . .+
xn1 + xnx1
> 1,
where x1x2 . . . xn = 1, we can substitute xi =aiai+1
, xn =ana1
, but this turns
xi1 + xixi+1
into a messyaiai+2
ai+1(ai + ai+2). The substitution xi =
ai+1
ai, however,
turns it intoai+1
ai + ai+2>
ai+1
a1 + a2 + . . .+ anand it’s done.
Another popular substitution is to work with the inverses of the numbersinstead of them.Look at this example:
If abc = 1 then1
a3(b+ c)+
1
b3(a+ c)+
1
c3(a+ c)≥ 3
2.
This time it’s convenient to set x =1
a= bc, y =
1
b= ac, z =
1
c= ab, the
inequality becoming thenx2
y + z+
y2
x+ z+
z2
x+ y≥ 3
2. However we know that
x2
y + z+
y2
x+ z+
z2
x+ y≥ x+ y + z
2≥ 3
2.
There are also trigonometric substitutions, based on the following relations:
cosA =b2 + c2 − a2
2bc
sinA
2=
√(p− b)(p− c)
bc=
√yz
(x+ y)(x+ z)
cosA
2=
√p(p− a)
bc=
√x(x+ y + z)
(x+ y)(x+ z)
and other analogous ones. (The notations are usual: a, b, c are the sides of the
triangle, A,B,C its angles, p =a+ b+ c
2its semiperimeter, x =
b+ c− a2
,
y =a+ c− b
2, z =
a+ b− c2
).
105
Via this substitutions, the inequality sinx+sin y+sin z ≤ 3
2, for x+y+z =
π
2,
for example, transforms to a purely algebraic one: if we consider the trianglewith angles 2x, 2y, 2z then
sinx+ sin y + sin z =
√y
x+ y
z
x+ z+
√x
x+ y
z
y + z+
√x
x+ z
y
y + z
≤ 1
2
(y
x+ y+
z
x+ z+
x
x+ y+
z
y + z+
x
x+ z+
y
y + z
)=
3
2.
The list of substitutions can be extended further, for example if xyz = x +
y + z + 2 then1
1 + x+
1
1 + y+
1
1 + z= 1 so
1
1 + x=
a
a+ b+ c,
1
1 + y=
b
a+ b+ c,
1
1 + z=
c
a+ b+ c
thus x =b+ c
a, y =
a+ c
b, z =
a+ b
c.
Thus, for example, we can prove that if x, y, z > 0 and xyz = x + y + z + 2
then xy + yz + zx ≥ 2(x + y + z). Indeed, after the substitution x =b+ c
a,
y =a+ c
b, z =
a+ b
c, we have to prove that
(a+ b)(a+ c)
bc+
(b+ a)(b+ c)
ac+
(c+ a)(c+ b)
ab≥ 2(
b
a+a
b+a
c+c
a+b
c+c
b),
ora2
bc+b2
ac+c2
ab+ 3 ≥ b
a+a
b+a
c+c
a+b
c+c
b.
After clearing denominators it becomes
a3 + b3 + c3 + 3abc ≥ ab(a+ b) + bc(b+ c) + ca(c+ a),
or Schur’s Inequality.
Exercises:
1. If a, b ∈ R\{−1, 1} then
∣∣∣∣(1− a2)(1− b2)(1 + a2)(1 + b2)
(1 +
4ab
(1− a2)(1− b2)
)∣∣∣∣ ≤ 1.
Solution: If a = tanx, b = tan y then∣∣∣∣(1− a2)(1− b2)(1 + a2)(1 + b2)
[1 +
4ab
(1− a2)(1− b2)
]∣∣∣∣ =
=
∣∣∣∣(1− tan2 x)(1− tan2 y)
(1 + tan2 x)(1 + tan2 y)
[1 +
4 tanx tan y
(1− tan2 x)(1− tan2 y)
]∣∣∣∣ = | cos(x− y)| ≤ 1.
2. If a, b, c > 0 thena2
(a+ b)(a+ c)+
b2
(b+ a)(b+ c)+
c2
(c+ a)(c+ b)≥ 3
4.
Solution: If we let x = b+ c, y = a+ c, z = a+ b the inequality turns to
(y + z − x)2
yz+
(x+ z − y)2
xz+
(x+ y − z)2
xy≥ 3
106
or after cancelling common terms to
x2
yz+y2
xz+z2
xy+ 3 ≥ x
y+y
x+x
z+z
x+y
z+z
y,
which multiplying by xyz comes to Schur.
3. Let a0 =1
2, and ak+1 = ak +
a2kn
. Show that 1− 1
n< an < 1.
Solution: Let bk =1
ak. Then bk+1 =
11
bk+
1
nb2k
=nb2k
nbk + 1= bk −
bknbk + 1
. Now if bk > 1 thenbk
nbk + 1∈(
1
n+ 1,
1
n
). As b0 = 2, we
can prove by induction on 1 ≤ k ≤ n that bk ∈(
2− k
n, 2− k
n+ 1
).
Particularly for k = n, bk ∈(
1,n+ 2
n+ 1
), which implies the problem.
4. If x, y, z > 0 with xy+yz+zx+2xyz = 1, then1
x+
1
y+
1
z≥ 4(x+y+z).
Solution: As1
x,
1
y,
1
zsatisfy
1
x+
1
y+
1
z+2
1
x· 1y· 1z
= 1, we have numbers
a, b, c with1
x=b+ c
a,
1
y=a+ c
b,
1
z=a+ b
c. The inequality to prove
now becomes
b+ c
a+a+ b
c+a+ c
b≥ 4
(a
b+ c+
c
a+ b+
b
a+ c
)which follows from 4
a
b+ c≤ a
b+a
cand the analogous relations.
5. Show that
√y + z
x+
√z + x
y+
√x+ y
z≥ 4 (x+ y + z)√
(y + z) (z + x) (x+ y)for
every three positive reals x, y, z.
Solution: For simplifying the solution set
x→√y + z, y →
√z + x, z →
√x+ y.
Then x2, y2, z2 are sides of a triangle and hence so do x, y, z that forman acute-angled triangle. The inequality transforms then to∑
cyc
x
y2 + z2 − x2≥ x2 + y2 + z2
xyz.
We can solve this inequality in two ways:
a) Applying cosine theorem to the acute-angled triangle formed by x, y, z
we transform the inequality to∑cyc
x
yz cosA≥ 2
x2 + y2 + z2
xyzand by mul-
tiplying it by xyz we have to prove∑cyc
x2
cosA≥ 2(x2 + y2 + z2).
107
However (x2, y2, z2) and
(1
cosA,
1
cosB,
1
cosC
)are clearly ordered the
same way hence by Chebyshev (see the appropriate chapter on ordering)we have ∑
cyc
x2
cosA≥∑cyc
x2
1
cosA+
1
cosB+
1
cosC3
and the desired inequality follows now from
1
cosA+
1
cosB+
1
cosC≥ 9
cosA+ cosB + cosC
(by CBS) and from the well-known inequality cosA+ cosB+ cos c ≤ 3
2.
b) The inequality is
∑cyc
(x
y2 + z2 − x2− x
yz
)≥ 0 or
∑cyc
x(x2 + yz − y2 − z2)yz(y2 + z2 − x2)
≥ 0.
Now using the fact that x, y, z form a triangle it’s quite clear that x(x2+
yz − y2 − z2), 1
yz(y2 + z2 − x2)are ordered the same way with x, y, z
and hence by Chebyshev inequality it suffices to prove that∑cyc
x(x2 +
yz − xy − xz) ≥ 0 which is just the renowned Schur’s Inequality.
6. If x1, x2, . . . , xn > 0 with1
x1+
1
x2+ . . .+
1
xn= n. Show that
x1 . . . xn − 1 ≥(n− 1
n
)n−1(x1 + . . .+ xn − n).
Solution: Let yi =1
xi. Then
n∑i=1
yi = n. Now multiplying the inequality
by y1 . . . yn we get a new inequality where we are allowed even to haveyi = 0.
Transform our inequality to f(y1, y2, . . . , yn) ≥ 0 where
f(y1, y2, . . . , yn) =
= nn−1(1− y1y2 . . . yn)+
+(n−1)n−1(ny1y2 . . . yn−y2y3 . . . yn−y1y3 . . . yn−. . .−y1y2 . . . yn−1).Consider y1, y2, . . . , yn with minimal value of f and of all such n-uples,one with minimal y21 + y22 + . . .+ y2n.
Now let’s try to decrease f . Pick up arbitrary non-zero yi, yj and fix theothers. Thus yi + yj is constant. Then f will have form ayiyj + b linearin yiyj and takes its minimum at the extremes, hence when yi = yj orone of them is 0. Thus we may replace yi, yj by their mean (type 1) ormake one of them 0 (type 2).
108
We keep doing transformations. We see that we can perform only finitelymany operations of type 2 because they increase the number of zeros inthe sequence. Thus at some time we shall have l zeros and some othernon-zero number every two of them we can replace by their arithmeticmean non-increasing f .
Now we can suppose the non-zero numbers are also equal(otherwise re-placing them by their mean decreases the sum of squares) hence they
aren
n− l. For this very particular case we can perform a trivial manual
checking.
7. Prove that for any seven numbers we may find two numbers x, y for
which 0 ≤ x− y1 + xy
≤ 1√3
.
Solution: Let the numbers be a1, a2, . . . , a7 and let ai = tan bi, where
0 ≤ bi ≤ π. We may partition [0, π] into six intervals of lengthπ
6. Then
some two numbers bi, bj will fit into the same interval then if bi < bj
we will have 0 ≤ bj − bi ≤π
6so 0 ≤ tan(bj − bi) ≤ tan
π
6=
1√3
, thus
0 ≤ aj − ai1 + ajai
and we may take x = aj , y = ai.
1.15 Homogenization and dehomogenization
If we have to solve a non-homogeneous inequality,we may want to make ithomogeneous to deal with more familiar homogeneous inequalities (this isbecause we already have a luggage of known inequalities to which we canreduce it). Look, for example, at the inequality(a2 +
1
b
)(b2 +
1
c
)(c2 +
1
a
)≥ 8
9(a+b+c)
(1
a+
1
b+
1
c
), for a, b, c > 0, abc = 1.
The inequality is not homogeneous, but we can use the condition abc = 1 to
make it homogeneous, by replacing1
a,
1
b,
1
cby bc, ac, ab. It becomes then
(a2 + ac)(b2 + ab)(c2 + bc) ≥ 8
9(a+ b+ c)(ab+ bc+ ca).
We can the remove abc = 1 from the left-hand side and then we get the alreadyknown homogeneous inequality
(a+ b)(b+ c)(c+ a) ≥ 8
9(a+ b+ c)(ab+ bc+ ca).
In other cases is better to dehomogenize an inequality to reduce the number ofvariables or add a good condition on them. For example, if the inequality is injust two variables, we may set one of them be 1 (for example the inequality x3+
y3 ≥ x2y+xy2 reduces to x3+1 ≥ x2+x be dividing by y3 and letting x→ x
y).
109
The inequality then may be reduced to a polynomial in one variable and solvedby algebraic methods applicable to polynomials (i.e. finding the roots). Otherexample of dehomogenization is the AM-GM inequality x1 + x2 + . . .+ xn ≥n n√x1x2 . . . xn. By replacing x1, x2, . . . , xn by
x1n√x1 . . . xn
, . . . ,xn
n√x1 . . . xn
we
can assume x1x2 . . . xn = 1 and then write x1 =y1y2, x2 =
y2y3, . . . , xn =
yny1
.
The inequality then becomes
y1y2
+y2y3
+ . . .+yny1≥ n,
and it can be solved by unimonotonic sequences (refer to the chapter on or-dering).
Exercises
1. If ai, bi, ci > 0 then
(a1b1c1 + . . .+ anbncn)3 ≤ (a31 + . . .+ a3n)(b31 + . . .+ b3n)(c31 + . . .+ c3n).
Solution: The quantity
(a1b1c1 + . . .+ anbncn)3
(a31 + . . .+ a3n)(b31 + . . .+ b3n)(c31 + . . .+ c3n)
doesn’t change if we replace each ai by tai. Therefore we may assumethat a31 + . . .+a3n = 1 and analogously b31 + . . .+b3n = 1, c31 + . . .+c3n = 1.
Then aibic≤a3i + b3i + c3i
3.
By summing this inequalities we get a1b1c1 + . . . + anbncn ≤ 1 and theconclusion follows.
Remark: This also easily follows from Holder’s inequality. The solutionwe gave above is essentially the same as the inequality aibici ≤ 1
3(pa3i +
qb3i+rc3i ) where p =
a31+...+a3n
3√
(a31+...+a3n)(b
31+...+b
3n)(c
31+...+c
3n), q =
b31+...+b3n
3√
(a31+...+a3n)(b
31+...+b
3n)(c
31+...+c
3n), r =
c31+...+c3n
3√
(a31+...+a3n)(b
31+...+b
3n)(c
31+...+c
3n)
- except this method makes it easier to
see. Of course, this is also how we prove Holder’s inequality.
2. If x, y, z > 0 then x3y + y3z + z3x ≥ xyz(x+ y + z).
Solution: Assume that x is the smallest of x, y, z. Then as the inequalityis homogeneous we can assume x = 1, y = 1 + u, z = 1 + v.
The inequality now after cancelling common terms becomes
(u− v)2 + (u3 + v3 − uv2) + u2 + v2 + 2u2v + u3v ≥ 0
and is clearly true.
3. For every positive a, b and c such that a2 + b2 + c2 = 1 prove the
inequality:∑cyc
a
a3 + bc> 3 (the inequality is strict).
110
Solution: By letting x = a2, y = b2, z = c2 and homogenizing we have
to prove the following∑cyc
x(x+ y + z)
x2 + t≥ 3 where t =
√xyz(x+ y + z).
Which is in turn equivalent to
(x+ y + z)∑cyc
x(y2 + t)(z2 + t)− 3(x2 + t)(y2 + t)(z2 + t) ≥ 0.
Now let u = x+ y + z, v = xy + yz + zx, w = xyz. We transform ourinequality to
g(x, y, z) = u(vw + (uv − 3w)√uw + u3w)− 3w2 −
−3(v2 − 2uw)√uw − 3(u2 − 2v)uw − 3uw
√uw ≥ 0.
Now let’s assume that this inequality fails for some x, y, z thus g(x, y, z) <0.
Now keep u,w fixed and let’s try to vary v in order to minimize g.It can run on a set intervals of reals that ensures that the polynomialp(x) = x3− ux2 + vx−w = 0 has solutions or equivalently that the line
y = −v intersects the graph of function x2 − ux − w
xin three points.
Now g(x, y, z) transforms to a quadratic expression in v with leadingcoefficient negative thus it takes minimum at some interval. Howeverthe extremum for v realizes when the line x = −v touches the graph ofthe function somewhere which implies that the equation has a doubleroot. This means that two of x, y, z are equal. So we need to check therelation only for x = y, z = 1. This is quite simple to check: after clearingdenominators this expression transforms to (x2+4x−2)
√2x+ 1 ≥ 4x2+
x− 2.
This inequality can be easily proved as follows: If 4x2+x−2 ≥ 0 then we
shall have x ≥√
33− 1
8> 0.6 and necessarily x2 + 4x− 2 ≥ 0 and thus
by squaring we deduce that we have to prove 2x4+x3+24x2−5x−4 > 0which is clearly true for x > 0.6. If 4x2 + x − 2 < 0 we must look onlyat the case x2 + 4x− 2 < 0 which is x <
√6− 2.
By squaring we must prove the inequality 2x4 + x3 + 24x2 − 5x− 4 < 0which again easily follows from x <
√6− 2.
1.16 Unimonotonic sequences
Suppose that x1 > x2 and y1 > y2. Then we have
(x1y1 + x2y2)− (x1y2 + x2y1) = (x1 − x2)(y1 − y2) ≥ 0.
Now take numbers x1, x2, . . . , xn, y1, y2, . . . , yn and consider the quantity
Q(y1, y2, . . . , yn) = x1y1 + . . .+ xnyn.
111
If xi > xj but yi < yj then by exchanging yi with yj we increase Q by
(xi − xj)(yj − yi) > 0.
If yi > yj then by exchanging yi with yj we decrease Q by (xi − xj)(yi − yj).Now suppose x1 > x2 > . . . > xn.
Let z1, z2, . . . , zn be the permutation that maximizes Q and let z′1, z′2, . . . , z
′n
be the permutation that minimizes Q. If zi > zj for some i < j then we couldincrease Q which contradicts the maximality of Q(z1, z2, . . . , zn). That’s whyz1 ≥ z2 ≥ . . . ≥ zn. Analogously we deduce z′1 ≤ z′2 ≤ . . . ≤ z′n (and soz′1 = zn, z
′2 = zn−1, . . . , z
′n = z1).
So, we proved the following theorem:
Theorem (Rearrangement Inequality). If x1 ≥ x2 ≥ . . . ≥ xn and y1, y2, . . . , ynare real numbers then
x1z1 + . . .+ xnzn ≥ x1y1 + x2y2 + . . .+ xnyn ≥ x1zn + . . .+ xnz1,
where z1 ≥ z2 ≥ . . . ≥ zn is a decreasing permutation of y1, y2, . . . , yn.
The sequences x1, x2, . . . , xn and y1, y2, . . . , yn are called unimonotonic if theyare ordered in the same way (i.e (xi − xj)(yi − yj) ≥ 0 for any i, j). If theyare inversely ordered (i.e. (xi − xj)(yi − yj) ≤ 0) we call them antimono-tonic. By the above theorem, the quantity x1y1 + . . .+ xnyn is maximized byunimonotonic sequences and minimized by antimonotonic sequences.
For example, if yi =1
xiwhere xi > 0 then x1, x2, . . . , xn and y1, y2, . . . , yn are
antimonotonic sequences so
Q(y1, y2, . . . , yn) ≤ Q(y2, . . . , yn, y1).
Thusx1x2
+ . . .+xnx1≥ n
which proves the AM-GM Inequality as it is equivalent to it (see the lastexample from the previous chapter - or the first exercise from this one).
If y1 ≥ y2 ≥ . . . ≥ yn and x1 ≥ x2 ≥ . . . ≥ xn then
x1y1 + x2y2 + . . . xnyn ≥ x1y1+i + x2y2+i + . . .+ xnyn+i
(we consider yn+k = yk). Summing this i = 0, 1, . . . , n− 1 we obtain
Theorem (Chebyshev’s Inequality)
n(x1y1 + . . .+ xnyn) ≥ (x1 + . . .+ xn)(y1 + y2 + . . .+ yn)
If y1 < y2 < . . . < yn, it changes sign.
A consequence of Chebyshev’s inequality is the following inequality (which canalso be proved by induction):
If x1, x2, . . . , xn and y1, y2, . . . , yn are antimonotonic then
x1y1
+ . . .+xnyn≥ nx1 + . . .+ xn
y1 + . . .+ yn.
112
Proof : By Chebyshev’s inequality
x1y1
+ . . .+xnyn≥ 1
n(x1 + x2 + . . .+ xn)
(1
y1+ . . .+
1
yn
).
Since1
y1+ . . .+
1
yn≥ n2
y1 + . . .+ yn,
the conclusion follows.
Exercises
1. Show that if x1, x2, . . . , xn > 0 then x1 + x2 + . . .+ xn ≥ n n√x1x2 . . . xn.
Solution: As the inequality is homogeneous, we can assume x1x2 . . . xn =
1 then we can write x1 =a1a2, x2 =
a2a3, . . . , xn =
ana1
.
And the inequality becomesa1a2
+ . . . +ana1≥ n, so it follows from the
antimonocity of the sequences (a1, a2, . . . , an) and
(1
a1,
1
a2, . . . ,
1
an
).
2. Show that if a, b, c > 0 thenb2 − a2
c+ a+c2 − b2
a+ b+a2 − c2
b+ c≥ 0.
Solution: We can rewrite the inequality as
b2
c+ a+
c2
a+ b+
a2
b+ c≥ a2
c+ a+
b2
a+ b+
c2
b+ c
which follows from the unimonotonicity of (b2, a2, c2) and
(1
a+ c,
1
b+ c,
1
a+ b
).
3.a3
b2 − bc+ c2+
b3
a2 − ac+ c2+
c3
a2 − ab+ b2≥ a+ b+ c.
Solution: We can suppose that a ≥ b ≥ c. Now note that we have
(b2 − bc+ c2)− (a2 − ac+ c2) = (b− a)(b+ a− c).
Hence we have two cases to consider:
a) a, b, c form a triangle. In this case (a3, b3, c3) and(1
b2 − bc+ c2,
1
a2 − ac+ c2,
1
a2 − ab+ b2
)are ordered the same way, so
2
(a3
b2 − bc+ c2+
b3
a2 − ac+ c2+
c3
a2 − ab+ b2
)≥
≥ b3 + c3
b2 − bc+ c2+
a3 + c3
a2 − ac+ c2+
a3 + b3
a2 − ab+ b2= a+ b+ c.
113
b) a > b + c. Let’s denote x =1
b2 − bc+ c2, y =
1
a2 − ac+ c2, z =
1
a2 − ab+ b2. Then x ≥ z ≥ y. Hence a3x+b3y+c3z ≥ a3y+b3x+c3z ≥
a3y+b3z+c3x and also a3x+b3y+c3z ≥ a3z+b3y+c3x ≥ a3z+b3x+c3y.
By summing these two relation we get again
2
(a3
b2 − bc+ c2+
b3
a2 − ac+ c2+
c3
a2 − ab+ b2
)≥
≥ b3 + c3
b2 − bc+ c2+
a3 + c3
a2 − ac+ c2+
a3 + b3
a2 − ab+ b2= a+ b+ c.
4. If α > 0 and x, y, z > 0 then
xα+2
(x+ y)(x+ z)+
yα+2
(y + x)(y + z)+
zα+2
(z + y)(z + x)≥ 1
4(xα + yα + zα).
Solution: We can see that the sequences
(xα, yα, zα) and
(x2
(x+ y)(x+ z),
y2
(y + x)(y + z),
z2
(z + x)(z + y)
)are ordered the same way, hence
xα+2
(x+ y)(x+ z)+
yα+2
(y + x)(y + z)+
zα+2
(z + y)(z + x)≥
≥ 1
3
[x2
(x+ y)(x+ z)+
y2
(y + x)(y + z)+
z2
(z + x)(z + y)
](xα + yα + zα).
Moreover, we havex2
(x+ y)(x+ z)+
y2
(y + x)(y + z)+
z2
(z + x)(z + y)≥ 3
4from CBS Lemma.
5. Let x1 ≤ x2 ≤ . . . ≤ xn and y1 ≤ y2 ≤ . . . ≤ yn be real numbers. Forany permutations (z1, z2, . . . , zn) of (y1, y2, . . . , yn), prove that
(x1−y1)2+(x2−y2)2+. . .+(xn−yn)2 ≤ (x1−z1)2+(x2−y2)2+. . .+(xn−nn)2.
Solution: Notice that y21+y22+. . .+y2n = z21+z22+. . .+z2n. After expansionand simplification, the desired inequality is equivalent to x1y1 + x2y2 +. . .+ xnyn ≥ x1z1 + x2z2 + . . .+ xnzn, which is just the Rearrangementinequality.
6. (Nesbitt’s Inequality) Let a, b, c be positive real numbers. Prove that
a
b+ c+
b
c+ a+
c
a+ b≥ 3
2.
Solution: Without loss of generality, we can suppose that a ≥ b ≥ c,then
1
b+ c≥ 1
c+ a≥ 1
a+ b.
114
Therefore, applying the Rearrangement inequality, we obtain
a
b+ c+
b
c+ a+
c
a+ b= a · 1
b+ c+ b · 1
c+ a+ c · 1
a+ b
≥ a · 1
a+ b+ b · 1
b+ c+ c · 1
c+ a
=a
a+ b+
b
b+ c+
c
c+ a,
and
a
b+ c+
b
c+ a+
c
a+ b= a · 1
b+ c+ b · 1
c+ a+ c · 1
a+ b
≥ a · 1
c+ a+ b · 1
a+ b+ c · 1
b+ c
=a
c+ a+
b
a+ b+
c
a+ b,
Adding up these two inequalities, we obtain
2
(a
b+ c+
b
c+ a+
c
a+ b
)≥ a+ b
a+ b+b+ c
b+ c+c+ a
c+ a= 3.
This completes the proof.
7. Let a1, a2, . . . , an be distinct positive integers. Prove that
a112
+a222
+ . . .+ann2≥ 1 +
1
2+ . . .+
1
n.
Solution: Let (a′1, a′2, . . . , a
′n) be a cyclic permutation of (a1, a2, . . . , an)
such that a′1 < a′2 < . . . < a′n. Then we have a′i ≥ i for all i = 1, 2, . . . , n.Now, we have that
a′1 < a′2 < . . . < a′n and1
12>
1
22> . . . >
1
n2.
Therefore, according to the Rearrangement inequality, we obtain
a112
+a222
+ . . .+ann2
= a1 ·1
12+ a2 ·
1
22+ . . .+ an ·
1
n2
≥ a′1 ·1
12+ a′2 ·
1
22+ . . .+ a′n ·
1
n2
≥ 1 · 1
12+ 2 · 1
22+ . . .+ n · 1
n2
= 1 +1
2+ . . .+
1
n.
This completes our proof. Equality holds if and only if ai = i for alli = 1, 2, . . . , n.
8. Let a, b, c be the side lengths of a triangle. Prove that
a2(b+ c− a) + b2(c+ a− b) + c2(a+ b− c) ≤ 3abc.
115
Solution: We will firstly show that if a ≥ b, then
a(b+ c− a) ≤ b(c+ a− b).
Indeed, we have
b(c+ a− b)− a(b+ c− a) = (a− b)(a+ b− c) ≥ 0.
Now, we see that the inequality is symmetric, hence we can assume thata ≥ b ≥ c. Then, from the above inequality, we have
a ≥ b ≥ c and c(a+ b− c) ≥ b(c+ a− b) ≥ a(b+ c− a).
Thus, according to the Rearrangement inequality, we obtain
a · a(b+ c− a) + b · b(c+ a− b) + c · c(a+ b− c) ≤≤ b · a(b+ c− a) + c · b(c+ a− b) + a · c(a+ b− c),
and
a · a(b+ c− a) + b · b(c+ a− b) + c · c(a+ b− c) ≤≤ c · a(b+ c− a) + a · b(c+ a− b) + b · c(a+ b− c).
Adding both these inequalities, we have
2[a2(b+ c− a) + b2(c+ a− b) + c2(a+ b− c)
]≤ 6abc,
which is
a2(b+ c− a) + b2(c+ a− b) + c2(a+ b− c) ≤ 3abc.
9. Let a, b, c be the side lengths of a triangle. Prove that
a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.
Solution: Without loss of generality, we can suppose that a = max {a, b, c}and consider 2 cases
Case 1. If a ≥ b ≥ c, then from the above problem, we have
1
a≤ 1
b≤ 1
cand a(b+ c− a) ≤ b(c+ a− b) ≤ c(a+ b− c).
And the Rearrangement inequality yields that
a+ b+ c =1
a· a(b+ c− a) +
1
b· b(c+ a− b) +
1
c· c(a+ b− c)
≥ 1
c· a(b+ c− a) +
1
a· b(c+ a− b) +
1
b· c(a+ b− c)
= a+ b+ c− a2b(a− b) + b2c(b− c) + c2a(c− a)
abc.
This implies that
a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.
116
Case 2. If a ≥ c ≥ b, then from the above problem, we have
1
a≤ 1
c≤ 1
band a(b+ c− a) ≤ c(a+ b− c) ≤ b(c+ a− b).
Thus, by Rearrangement inequality, we see that
a+ b+ c =1
a· a(b+ c− a) +
1
c· c(a+ b− c) +
1
b· b(c+ a− b)
≥ 1
c· a(b+ c− a) +
1
b· c(a+ b− c) +
1
a· b(c+ a− b)
= a+ b+ c− a2b(a− b) + b2c(b− c) + c2a(c− a)
abc.
Thena2b(a− b) + b2c(b− c) + c2a(c− a ≥ 0,
and our proof is complete.
10. Let a, b, c be positive real numbers. Prove that
a
b+b
c+c
a≥ a+ c
b+ c+b+ a
c+ a+c+ b
a+ b.
Solution: We rewrite our inequality as(a
b− a
b+ c
)+
(b
c− b
c+ a
)+
(c
a− c
a+ b
)≥ b
a+ b+
c
b+ c+
a
c+ a,
ca
b(b+ c)+
ab
c(c+ a)+
bc
a(a+ b)≥ b
a+ b+
c
b+ c+
a
c+ a.
Applying Cauchy Schwarz inequality,(b
a+ b+
c
b+ c+
a
c+ a
)2
≤
≤[
ab
c(a+ b)+
bc
a(b+ c)+
ca
b(c+ a)
] [bc
a(a+ b)+
ca
b(b+ c)+
ab
c(c+ a)
].
It remains to show that
ab
c(a+ b)+
bc
a(b+ c)+
ca
b(c+ a)≤ ca
b(b+ c)+
ab
c(c+ a)+
bc
a(a+ b).
Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then
xy
z≥ zx
y≥ yz
xand
1
x+ y≤ 1
z + x≤ 1
y + z.
Therefore, according to the Rearrangement inequality, we have
bc
a· 1
a+ b+ab
c· 1
c+ a+ca
b· 1
b+ c≥ xy
z· 1
x+ y+zx
y· 1
z + x+yz
x· 1
y + z
=xy
z(x+ y)+
yz
x(y + z)+
zx
y(z + x)
=ab
c(a+ b)+
bc
a(b+ c)+
ca
b(c+ a).
Our proof is complete. Equality holds if and only if a = b = c.
117
11. Let a, b, c be positive real numbers. Prove that
a+ b
b+ c+b+ c
c+ a+c+ a
a+ b≤ (a+ b+ c)2
ab+ bc+ ca.
Solution: Rewrite our inequality as
(a+ b) [a(b+ c) + bc]
b+ c+
(b+ c) [b(c+ a) + ca]
c+ a+
(c+ a) [c(a+ b) + ab]
a+ b≤ (a+b+c)2,
bc(a+ b)
b+ c+ca(b+ c)
c+ a+ab(c+ a)
a+ b≤ ab+ bc+ ca.
Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then
x+ y ≥ z + x ≥ y + z andxy
x+ y≥ zx
z + x≥ yz
y + z.
Therefore, according to the Rearrangement inequality, we obtain
ab+ bc+ ca = xy + yz + zx
= xy · xy
x+ y+ zx · zx
z + x+ yz · yz
y + z
≥ (c+ a) · ab
a+ b+ (b+ c) · ca
c+ a+ (a+ b) · bc
b+ c
=bc(a+ b)
b+ c+ca(b+ c)
c+ a+ab(c+ a)
a+ b.
Our proof is complete.
12. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 4.Prove that
a2bc+ b2cd+ c2da+ d2ab ≤ 4.
Solution: Let (x, y, z, t) is a cyclic permutation of (a, b, c, d) such thatx ≥ y ≥ z ≥ t, then xyz ≥ xyt ≥ xzt ≥ yzt. Therefore, according to theRearrangement inequality, we obtain
x · xyz + y · xyt+ z · xzt+ t · yzt ≥ a · abc+ b · bcd+ c · cda+ d · dab= a2bc+ b2cd+ c2da+ d2ab.
That is
a2bc+ b2cd+ c2da+ d2ab ≤ x · xyz + y · xyt+ z · xzt+ t · yzt
= (xy + tz)(xz + ty) ≤(xy + tz + xz + ty
2
)2
=
[(x+ t)(y + z)
2
]2≤[
(x+ y + z + t)2
8
]2= 4.
Remark. We can also prove this inequality by AM-GM inequality asfollows:
118
We have
a2bc+ b2cd+ c2da+ d2ab = ac(ab+ cd) + bd(bc+ da).
If ab+ cd ≥ bc+ da, then
a2bc+ b2cd+ c2da+ d2ab ≤ ac(ab+ cd) + bd(ab+ cd)
= (ac+ bd)(ab+ cd) ≤(ac+ bd+ ab+ cd
2
)2
=
[(a+ d)(b+ c)
2
]2≤[
(a+ b+ c+ d)2
8
]2= 4.
If bc+ da ≥ ab+ cd, then
a2bc+ b2cd+ c2da+ d2ab ≤ ac(bc+ da) + bd(bc+ da)
= (ac+ bd)(bc+ da) ≤(ac+ bd+ bc+ da
2
)2
=
[(a+ b)(c+ d)
2
]2≤[
(a+ b+ c+ d)2
8
]2= 4.
This completes the proof.
13. Let a, b, c, d be positive real numbers. Prove that(a
a+ b+ c
)2
+
(b
b+ c+ d
)2
+
(c
c+ d+ a
)2
+
(d
d+ a+ b
)2
≥ 4
9.
Solution: Let (x, y, z, t) is a cyclic permutation of (a, b, c, d) such thatx ≥ y ≥ z ≥ t, then
1
(x+ y + z)2≥ 1
(x+ y + t)2≥ 1
(x+ z + t)2≥ 1
(y + z + t)2.
Therefore, according to the Rearrangement inequality, we obtain
x2 · 1
(x+ y + z)2+ y2 · 1
(x+ y + t)2+ z2 · 1
(x+ z + t)2+ t2 · 1
(y + z + t)2≤
≤(
a
a+ b+ c
)2
+
(b
b+ c+ d
)2
+
(c
c+ d+ a
)2
+
(d
d+ a+ b
)2
.
It remains to show that
x2
(x+ y + z)2+
y2
(x+ y + t)2+
z2
(x+ z + t)2+
t2
(y + z + t)2≥ 4
9.
The inequality being homogeneous, hence we can assume that x + y +z + t = 1. The above inequality becomes[
x2
(1− t)2+
t2
(1− x)2
]+
[y2
(1− z)2+
z2
(1− y)2
]≥ 4
9.
119
Setting u = x+ t (0 < u < 1), and v = xt, we have
x2
(1− t)2+
t2
(1− x)2=
2v2 − 2(u− 1)(2u− 1)v + u2(1− u)2
(1− u+ v)2= f(v),
and
f ′(v) =2(1− u)
[(3− 2u)v − (1− u)3
](1− u+ v)3
,
f ′(v) = 0⇔ v =(1− u)3
3− 2u.
From now, by making variation board, we obtain
f(v) ≥ f(
(1− u)3
3− 2u
)=−2u2 + 4u− 1
(2− u)2∀v > 0.
Similarly, if we put w = y + z = 1− u, then we also have
y2
(1− z)2+
z2
(1− y)2≥ −2w2 + 4w − 1
(2− w)2=−2(1− u)2 + 4(1− u)− 1
(1 + u)2=
1− 2u2
(1 + u)2.
It remains to show that
−2u2 + 4u− 1
(2− u)2+
1− 2u2
(1 + u)2≥ 4
9,
which can be easily simplified to
(1− 2u)2(11 + 10u− 10u2)
9(1 + u)2(2− u)2≥ 0,
which is clearly nonnegative and our proof is complete.
Remark. We can also prove this inequality using Holder inequality. SeeOld And New Inequalities 2, Vo Quoc Ba Can - Cosmin Pohoata, GILpublishing house, 2008.
14. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that
a3b2 + b3c2 + c3a2 ≤ 3.
Solution: Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥y ≥ z, then x2y2 ≥ x2z2 ≥ y2z2. Hence, the Rearrangement inequalityyields that
a3b2 + b3c2 + c3a2 = a · a2b2 + b · b2c2 + c · c2a2
≤ x · x2y2 + y · x2z2 + z · y2z2
= y(x3y + yz3 + x2z2
)≤ y
(x2 · x
2 + y2
2+ z2 · y
2 + z2
2+ x2z2
)=
1
2y(x2 + z2)(x2 + y2 + z2) =
3
2y(x2 + z2)
≤ 3
√√√√(y2 + x2+z2
2 + x2+z2
2
3
)3
= 3.
120
This completes our proof. Equality holds if and only if a = b = c = 1.
Remark. We can also prove this inequality using Cauchy Schwartz in-equality. See Old And New Inequalities 2.
15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Provethat
a2b+ b2c+ c2a ≤ 4
27.
)
Solution: Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥y ≥ z, then xy ≥ xz ≥ yz. Hence, the Rearrangement inequality yields
a2b+ b2c+ c2a = a · ab+ b · bc+ c · ca= x · xy + y · xz + z · yz= y(x2 + z2 + xz) ≤ y(x+ z)2
≤ 1
2
(2y + x+ z + x+ z
3
)3
=4
27.
16. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that
a2b+ b2c+ c2a ≤ 2 + abc.
Solution: Similar to the above problem, let (x, y, z) be a cyclic permuta-tion of (a, b, c) such that x ≥ y ≥ z, then by Rearrangement inequality,we also have that
a2b+ b2c+ c2a ≤ y(x2 + z2 + xz).
And we can deduce our inequality to
y(x2 + z2 + xz) ≤ 2 + xyz,
y(x2 + z2) ≤ 2,
which is true because
y(x2 + z2) = y(3− y2) = 2− (y + 2)(y − 1)2 ≤ 2.
17. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat
a2 + b2 + c2
ab+ bc+ ca+
4abc
a2b+ b2c+ c2a+ abc≥ 2.
Solution: Similar to the above problem, let (x, y, z) be a cyclic permuta-tion of (a, b, c) such that x ≥ y ≥ z, then by Rearrangement unequality,we also have that
a2b+ b2c+ c2a ≤ y(x2 + z2 + xz).
121
And we can deduce our inequality to
x2 + y2 + z2
xy + yz + zx+
4xyz
y(x2 + z2 + xz) + xyz≥ 2,
x2 + y2 + z2
xy + yz + zx≥ 2(x2 + z2)
(x+ z)2,
which can be easily simplified to
(xy + yz − x2 − z2)2 ≥ 0,
Equality holds if and only if a = b = c or a = b, c = 0 and its cyclicpermutations.
Remark. By the same manner, we can prove that
4(a2 + b2 + c2)
ab+ bc+ ca+
9abc
a2b+ b2c+ c2a≥ 7,
2(a2 + b2 + c2)
(a+ b+ c)2+
abc
a2b+ b2c+ c2a≥ 1.
18. Let a, b, c be positive real numbers. Prove that
a√a+ b
+b√b+ c
+c√c+ a
≥√a+√b+√c√
2.
Solution: Setting x =√a, y =
√b, z =
√c and squaring both sides, we
can rewrite our inequality as
2∑cyc
x4
x2 + y2+ 4∑cyc
x2y2√(x2 + y2)(y2 + z2)
≥ (x+ y + z)2.
Let (m,n, p) be a cyclic permutation of (x, y, z) such that m ≥ n ≥ p,then
m2n2√m2 + n2
≥ p2m2√p2 +m2
≥ n2p2√n2 + p2
and1√
m2 + n2≤ 1√
p2 +m2≤ 1√
n2 + p2.
Therefore, using Rearrangement inequality, we obtain∑cyc
x2y2
x2 + y2=
=m2n2
m2 + n2+
p2m2
p2 +m2+
n2p2
n2 + p2
=m2n2√m2 + n2
· 1√m2 + n2
+p2m2√p2 +m2
· 1√p2 +m2
+n2p2√n2 + p2
· 1√n2 + p2
≤ x2y2√x2 + y2
· 1√y2 + z2
+z2x2√z2 + x2
· 1√x2 + y2
+y2z2√y2 + z2
· 1√y2 + z2
=∑cyc
x2y2√(x2 + y2)(y2 + z2)
.
122
It remains to show that
2∑cyc
x4
x2 + y2+ 4∑cyc
x2y2
x2 + y2≥ (x+ y + z)2,
which is obviously true because
2∑cyc
x4
x2 + y2=
∑cyc
x4 + y4
x2 + y2+∑cyc
x4 − y4
x2 + y2
=∑cyc
x4 + y4
x2 + y2+∑cyc
(x2 − y2)
=∑cyc
x4 + y4
x2 + y2,
and ∑cyc
x4 + y4
x2 + y2+ 4∑cyc
x2y2
x2 + y2=
=∑cyc
(x4 + y4 + 4x2y2
x2 + y2− x2 + y2 + 4xy
2
)+∑cyc
x2 + y2 + 4xy
2
=∑cyc
(x− y)4
2(x2 + y2)+
(∑cyc
x
)2
≥
(∑cyc
x
)2
.
Remark. In the same manner, we can prove that√a3
a2 + ab+ b2+
√b3
b2 + bc+ c2+
√c3
c2 + ca+ a2≥√a+√b+√c√
3.
19. Let x, y, z be positive real numbers. Prove that
x√x+ y
+y√y + z
+z√z + x
≤ 3√
3
4·
√(x+ y)(y + z)(z + x)
xy + yz + zx.
Solution: We cam rewrite our inequality as
∑cyc
x√(x+ y)(x+ z)
· 1√(x+ y)(y + z)
≤ 3√
3
4√xy + yz + zx
.
Let (a, b, c) be a cyclic permutation of (x, y, z) such that a ≥ b ≥ c, then
a√(a+ b)(a+ c)
≥ b√(b+ c)(b+ a)
≥ c√(c+ a)(c+ b)
and
1√(a+ b)(a+ c)
≤ 1√(b+ c)(b+ a)
≤ 1√(c+ a)(c+ b)
.
123
Hence, by Rearrangement inequality, we obtain∑cyc
x√(x+ y)(x+ z)
· 1√(x+ y)(y + z)
≤
≤ a√(a+ b)(a+ c)
· 1√(c+ a)(c+ b)
+b√
(b+ c)(b+ a)· 1√
(b+ c)(b+ a)+
+c√
(c+ a)(c+ b)· 1√
(a+ b)(a+ c)
=1√
(b+ c)(b+ a)
[1 +
b√(b+ c)(b+ a)
].
It remains to show that√ab+ bc+ ca
(b+ c)(b+ a)
[1 +
b√(b+ c)(b+ a)
]≤ 3√
3
4.
Indeed, setting u = b√(b+c)(b+a)
≤ 1, then√
xy+yz+zx(y+z)(y+x) =
√1− u2. Ap-
plying AM-GM inequality, we have√ab+ bc+ ca
(b+ c)(b+ a)
[1 +
b√(b+ c)(b+ a)
]= (1 + u)
√1− u2 =
√(1 + u)3(1− u)
=
√(1 + u) · (1 + u) · (1 + u) · 3(1− u)
3
≤
√(3/2)4
3=
3√
3
4.
Our inequality is proved. Equality holds if and only if x = y = z.
20. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat
a√a+ b
+b√b+ c
+c√c+ a
≤ 5
4
√a+ b+ c.
Solution: After squaring, we can rewrite our inequality as∑cyc
a2
a+ b+ 2∑cyc
ab√(a+ b)(b+ c)
≤ 25
16
∑cyc
a,
2∑cyc
ab√(a+ b)(b+ c)
≤ 9
16
∑cyc
a+∑cyc
ab
a+ b.
Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥ y ≥ z, then
xy√x+ y
≥ zx√z + x
≥ yz√y + z
,
and1√y + z
≥ 1√z + x
≥ 1√x+ y
.
124
Therefore, according to the Rearrangement inequality, we obtain
ab√(a+ b)(b+ c)
+bc√
(b+ c)(c+ a)+
ca√(c+ a)(a+ b)
=
=ab√a+ b
· 1√b+ c
+bc√b+ c
· 1√c+ a
+ca√c+ a
· 1√a+ b
≤ xy√x+ y
· 1√y + z
+zx√z + x
· 1√z + x
+yz√y + z
· 1√x+ y
=y(x+ z)√
(x+ y)(y + z)+
xz
x+ z.
From now, we can deduce our inequality to
2y(x+ z)√(x+ y)(y + z)
+2xz
x+ z≤ 9
16
∑cyc
x+∑cyc
xy
x+ y,
2y(x+ z)√(x+ y)(y + z)
≤ 9
16(x+ y + z) +
xy
x+ y+
yz
y + z− zx
z + x.
Now, using AM-GM inequality, we have that
2√(x+ y)(y + z)
≤ x+ y + z
2(x+ y)(y + z)+
2
x+ y + z.
and it remains to show that
y(x+ z)(x+ y + z)
2(x+ y)(y + z)+
2y(x+ z)
x+ y + z≤ 9
16(x+y+z)+
xy
x+ y+
yz
y + z− zx
z + x,
y(x+ z)(x+ y + z)
2(x+ y)(y + z)+
2y(x+ z)
x+ y + z≤ 9
16(x+y+z)+
y(xy + yz + 2xz)
(x+ y)(y + z)− zx
z + x,
Since this inequality is homogeneous, we may assume that x + z = 1,put t = xz, then
t− y(1− y) = xz − y(x+ z − y) = (y − x)(y − z) ≤ 0.
Hence 0 ≤ t ≤ y(1− y). The above inequality becomes
y(y + 1)
2(y2 + y + t)+
2y
y + 1≤ 9
16(y + 1) +
y(y + 2t)
y2 + y + t− t,
f(t) =9
16(y + 1)− 2y
y + 1+y(4t+ y − 1)
2(t+ y + y2)− t ≥ 0.
After expanding, we can see that our inequality has the form g(t) =At2+Bt+C ≥ 0, with A < 0. Therefore, in order to prove this inequality,we just need to prove that it holds when t = 0 and t = y(1 − y), or inthe other words, we must prove that f(0) ≥ 0 and f(y(1− y)) ≥ 0. Butit is true because
f(0) =9
16(y + 1)− 2y
y + 1+
y(y − 1)
2(y + y2)=
(1− 3y)2
16(1 + y)≥ 0,
f(y(1− y)) =9
16(y + 1)− 2y
y + 1+y [4y(1− y) + y − 1]
2 [y(1− y) + y + y2]
=5 + 2y + 13y2 − 16y3
16(y + 1)> 0 (since 1 ≥ y ≥ 0).
Thus f(t) ≥ 0 and our proof is complete.
125
21. Let a, b, c be positive real numbers such that a+ b+ c = 3. Prove that√a
b2 + 3+
√b
c2 + 3+
√c
a2 + 3≤ 3
2.
Solution: Let (x, y, z) be a cyclic permutation of (a, b, c) such that x ≥y ≥ z, then
√x ≥ √y ≥
√z and
1√z2 + 3
≥ 1√y2 + 3
≥ 1√x2 + 3
.
Therefore, by Rearrangement inequality, we obtain√a
b2 + 3+
√b
c2 + 3+
√c
a2 + 3=
=√a · 1√
b2 + 3+√b · 1√
c2 + 3+√c · 1√
a2 + 3
≤√x · 1√
z2 + 3+√y · 1√
y2 + 3+√z · 1√
x2 + 3
=
√x
z2 + 3+
√z
x2 + 3+
√y
y2 + 3.
And we can deduce our inequality to√x
z2 + 3+
√z
x2 + 3+
√y
y2 + 3≤ 3
2.
Setting t = xz, then 0 ≤ t ≤ (x+z)2
4 = (3−y)24 . By Cauchy-Schwartz
inequality, we have√x
z2 + 3+
√z
x2 + 3≤
√(x+ z)
(1
z2 + 3+
1
x2 + 3
)
=
√(3− y)(x2 + z2 + 6)
x2z2 + 3(x2 + z2) + 9
=
√(3− y)(y2 − 6y + 15− 2t)
t2 − 6t+ 3y2 − 18y + 36.
Consider the function
f(t) =y2 − 6y + 15− 2t
t2 − 6t+ 3y2 − 18y + 36
We have
f ′(t) =2[t2 − (y2 − 6y + 15)t+ 9
](t2 − 6t+ 3y − 18y + 3)2
,
f ′(t) = 0⇒
t = t1 =y2−6y+15+(3−y)
√y2−6y+21
2
t = t2 =y2−6y+15−(3−y)
√y2−6y+21
2
.
126
If y ≥ 1 then we have t1 ≥ t2 ≥ (3−y)24 , therefore since 0 ≤ t ≤ (3−y)2
4 ,
we have f ′(t) ≥ 0, and f(t) is increasing for all 0 ≤ t ≤ (3−y)24 . Thus
f(t) ≤ f(
(3− y)2
4
)=
8
y2 − 6y + 21.
Hence√x
z2 + 3+
√z
x2 + 3+
√y
y2 + 3≤
√8(3− y)
y2 − 6y + 21+
√y
y2 + 3≤ 3
2.
Since√8(3− y)
y2 − 6y + 21+
√y
y2 + 3≤
√(2 + 1)
[4(3− y)
y2 − 6y + 21+
y
y2 + 3
]
=3
2
√1− (y2 + 15)(y − 1)2
(y2 + 3)(y2 − 6y + 21)≤ 3
2.
If 1 ≥ y ≥ 0, then we have t1 ≥ (3−y)24 ≥ t2 > 0, therefore we obtain
f(t) ≤ f(t2) =2
(3− y)(√
y2 − 6y + 21 + y − 3) .
Hence√x
z2 + 3+
√z
x2 + 3+
√y
y2 + 3≤√
2√y2 − 6y + 21 + y − 3
+
√y
y2 + 3.
We need to prove√2√
y2 − 6y + 21 + y − 3+
√y
y2 + 3≤ 3
2.
We have4(y2 − 6y + 21)− (9− y)2 = 3(y − 1)2 ≥ 0,
It follows that√y2 − 6y + 21 + y − 3 ≥ 9− y
2+ y − 3 =
y + 3
2.
and it suffices to prove that
2√y + 3
+
√y
y2 + 3≤ 3
2,
which is true because
2√y + 3
+
√y
y2 + 3≤
√(2 + 1)
(2
y + 3+
y
y2 + 3
)
=3
2
√1− (y − 1)2(y + 1)
(y + 3)(y2 + 3)≤ 3
2.
Equality holds if and only if a = b = c = 1.
127
1.17 Working backwards
Consider the following hard problem:Non-negative reals x, y, z satisfy x2 + y2 + z2 + xyz = 4. Then xy + yz +zx− xyz ≤ 2.The condition is hard to use, however the conclusion is quite simple. For thisreason let us reverse the problem:Assume that xy + yz + zx− xyz = 2, we shall then prove that
x2 + y2 + z2 + xyz ≥ 4.
This new question implies the original one: if we would find a, b, c with
a2 + b2 + c2 + abc = 4
and ab+bc+ca−abc > 2, we could take a 0 < t < 1 with t2(ab+bc+ca)−t3abc =2, then setting x = ta, y = tb, z = tc we would get
x2 + y2 + z2 + xyz < a2 + b2 + c2 + abc = 4,
contradiction.So, let’s prove the new problem.
Assume that x ≤ y ≤ z. We have z =2− xy
x+ y − xy.
Now if x+ y < xy so1
x+
1
y< 1 we deduce y + z < yz, x+ z < xz so
x(y + z − yz) + y(x+ z − xz) + z(x+ y − xy) ≥ 0
so2(xy + yz + zx− xyz)− xyz ≤ 0 so xyz ≥ 4
and then
x2 + y2 + z2 + xyz ≥ xy + yz + zx− xyz + 2xyz ≥ 10 > 4.
So assume x + y − xy > 0. Set xy = a, x + y − xy = b. Then we must provethat if a+ bz = 2 then (a+ b)2 − 2a+ z2 + az ≥ 4.
By replacing z =2− ab
the relation to prove becomes(b2 − b+ 1
)a2 + 2a(b3 − b2 + b− 2) + (b2 − 2)2.
Regarded as an equation in a, its discriminant is
−b2(b− 1)(b2 − 5b+ 5)
so it can be positive for b < 1 or b ∈
(5−√
5
2,5 +√
5
2
).
If b < 1. Notice that we must have
(x+ y)2 ≥ 4xy or 4a ≤ (a+ b)2 or a2 + 2a(b− 2) + b2 ≥ 0,
128
which yields a2 ≥ 2a(2− b)− b2. And we can deduce our inequality to(b2 − b+ 1
) [2a(2− b)− b2
]+ 2a(b3 − b2 + b− 2) + (b2 − 2)2 ≥ 0,
or(1− b)(4 + 4b− b2 − 4ab) ≥ 0.
Moreover, as z ≥ y ≥ z, we have 2z ≥ x + y = a + b, therefore 2 = a + bz ≥
a+(a+ b)b
2, or 0 ≤ a ≤ 2− b. It follows that 4 + 4b− b2− 4ab ≥ 4 + 4b− b2−
4(2− b)b = −4b+ 3b2 + 4 = 2b2 + (b− 2)2 > 0.
If b ∈
(5−√
5
2,5 +√
5
2
), our inequality is trivial because b3− b2 + b−2 > 0.
This completes our proof.
Exercise:
1. If x2 + y2 + z2 = 2 then x+ y + z − xyz ≤ 2.
Solution: If one of x, y, z is negative, for example z < 0, then we have
z − xyz = z(1 − xy) ≤ 0 as from AM-GM 1 − xy ≥ 1 − x2 + y2
2=z2
2.
Therefore x+y+z−xyz = x+y+z(1−xy) ≤ x+y ≤√
2(x2 + y2) ≤ 2from Cauchy Schwartz.
If x, y, z are all non-negative. As above, we prove that if x+y+z−xyz = 2
then x2 + y2 + z2 ≥ 2. z =2− x− y
1− xy. Let a = x + y, b = xy. Then
z =2− a1− b
, and we need to prove that a2 − 2b +(2− a)2
(1− b)2≥ 2 or that
a2(b− 1)2 + (a− 2)2 − 2b(b− 1)2 ≥ 2(b− 1)2, or (a− 2)2 + a2(b− 1)2 −2(1− b)2(1 + b) ≥ 0,
If b ≥ 1 then a2 ≥ 4b ≥ 2(1 + b) and the expression in clearly non-negative.
If b < 1 then the discriminant, looked as an equation in a, is 2b2(b− 1)3,so negative, and in this case the conclusion also follows.
1.18 Mixing variables
In the previous chapter we took a quantity
Q(y1, y2, . . . , yn) = x1y1 + x2y2 + . . .+ xnyn
and increased or decreased it by exchanging some of the yj . In this chapterwe will take a similar quantity as a function of some variables (usually it’ssymmetric in these variables) and change the variables thus increasing or de-creasing the function. We distinguish two sorts of change: one is replacing avariable by a constant or a value that equals other variable (i.e. set xi = xj),and the other by replacing two (or more variables) by a new common value
129
(usually some sort of mean) - this way is particularly useful when we havesome conditions on the variables. This method is called the method of mixingvariables and it is used to reduce the number of variables until we can provethe inequality directly.
Let’s illustrate this method. Take Schur’s inequality:
a3 + b3 + c3 + 3abc− a2b− ab2 − a2c− ac2 − b2c− bc2 ≥ 0.
Take now
f(a, b, c) = a3 + b3 + c3 + 3abc− a2b− ab2 − a2c− ac2 − b2c− bc2.
Thenf(a, a, c)− f(a, b, c) = (a− b)(b2 + 2ac− a2 − bc− c2).
Since the inequality is symmetric in a, b, c, we can suppose that c < a < b.Then a− b < 0 and
b2 + 2ac− a2 − bc− c2 = (b2 − a2) + (ac− bc) + (ac− c2)= (b− a)(a+ b− c) + (ac− c2) ≥ 0
so f(a, a, c) < f(a, b, c) and so the inequality is sufficient to prove for the casewhen the two variables are equal. But in this case it reduces to c(a− c)2 ≥ 0,true!
Take now a harder inequality:
If x2 + y2 + z2 + xyz = 4, x, y, z > 0 then x+ y + z ≤ 3.The condition is quite uncomfortable to tackle. On the contrary the conclu-sion sounds simple. We have seen such inequalities in the section on workingbackwardsThat’s why let’s reverse them: prove that if x+ y + z = 3 then
x2 + y2 + z2 + xyz ≥ 4.
This would provide a solution for the original inequality: if we would have
x+y+z > 3 then dividing x, y, z by a constant a > 1 such thatx
a+y
a+z
a= 3
we would have(xa
)2+(ya
)2+(za
)2+x
a
y
a
z
a< x2 + y2 + z2 + xyz = 4
contradiction.So, let’s prove the new inequality.Let f(x, y, z) = x2 + y2 + z2 + xyz.
Since x+ y + z is fixed, consider f
(x+ y
2,x+ y
2, z
)we have
f(x, y, z)− f(x+ y
2,x+ y
2, z
)=
(x− y)2
2− z(x− y)2
4> 0
if we suppose z < 2 which we can do since x+ y + z = 3 and the inequality is
symmetric in x, y, z. So f(x, y, z) ≥ f(x+ y
2,x+ y
2, z). The inequality is just
enough to prove when two of the variables are equal, so y = x, z = 1 − 2x.The condition now becomes (x− 1)2(5− 2x) ≥ 0 and is true.
130
Finally, one can not make two variables equal but even make their differencebigger. Consider the inequality
a2b+ b2c+ c2a ≤ 4
27if a+ b+ c = 1, a, b, c ≥ 0.
Let f(a, b, c) = a2b+ b2c+ c2a.Then
f(0, a+ b, c)− f(a, b, c) = (ca2 + 2abc− c2a− a2b)= a(ac+ 2bc− c2 − ab) = a [bc− (c− a)(c− b)] .
So if c is the between b and a then f(0, a + b, c) ≥ f(a, b, c). But we canclearly assume this since the inequality is cyclic. The inequality is so sufficientto prove when one variable is 0. So if a = 0, b = x, c = 1− x it’s equivalent to
x2(1− x) ≤ 4
27which follows from the AM-GM inequality for
x
2,x
2, 1− x.
Exercises
1. If a, b, c > 0 then 2(a2 + b2 + c2) + 3(abc)23 ≥ (a+ b+ c)2.
Solution: Let’s try to mix a, b into their geometric mean, trying to
invariate the more complicated (abc)23 .
The difference 2(a2 + b2 + c2) + 2(abc)23 − (a+ b+ c)2 will then change
by
−2(a− b)2 − (2√ab+ c)2 + (a+ b+ c)2 = −(a− b)2 + 2
(a+ b− 2
√ab)c
= −(a− b)2 +2(a− b)2c
a+ b+ 2√ab≤ 0
when c is the smallest of a, b, c.
So we may assume that a = b. Take a = b = 1, c = x3 ≤ 1. Theinequality becomes 4 + 2x6 + 3x2 − (2 + x3)2 ≥ 0 or x6 − 4x3 + 3x2 ≥ 0or x2(x− 1)2(x2 + 2x+ 3) ≥ 0, which is true.
2. If a, b, c > 0 with abc = 1 then a2 + b2 + c2 +3 ≥ a+ b+ c+ab+ bc+ ca.
Solution: Let f(a, b, c) = a2+b2+c2+3−ab−bc−ca−a−b−c. To preservethe condition abc = 1, let’s mix b, c into their geometric mean. We com-
pute f(a, b, c) − f(a,√bc,√bc)
=(√
b−√c)2 [(√
b+√c)2− 1− a
].
Now if a = min{a, b, c} ≤ 1 then(√
b+√c)2≥ 4√bc ≥ 4 ≥ 1 + a, so f
decreases. So assume b = c =1
x, a = x2.
We then compute f to be
x4 + 3 +1
x2− 2
x− x2 − 2x =
x6 − x4 − 2x3 + 3x2 − 2x+ 1
x2
=(x− 1)2(x4 + 2x3 + 3x2 + 1)
x2≥ 0.
131
3. Let a, b, c, d ≥ 0 with a+ b+ c+ d = 1. Show that
abc+ bcd+ cda+ dab ≤ 1
27+
176
27abcd.
Solution: This is a classic example of mixing variables. Since the in-equality is symmetric, we remember the principle saying ”in almost allcases, the extreme values of a symmetric function are achieved whenalmost all variables are equal”. As we have that the sum of a, b, c, d, itis natural to try to mix two numbers into their arithmetic mean.
So, let f(a, b, c, d) = abc+ bcd+ cda+ dab− 1
27− 176
27abcd. Then
f
(a+ b
2,a+ b
2, c, d
)− f(a, b, c, d) =
=
[(a+ b)2
4− ab
](c+ d)− 176
27cd
[(a+ b)2
4− ab
].
Since(a+ b)2
4− ab is non-negative, f
(a+ b
2,a+ b
2, c, d
)≥ f(a, b, c, d)
when c+d ≥ 176
27cd. As cd ≤ (c+ d)2
4, f(
a+ b
2,a+ b
2, c, d) ≥ f(a, b, c, d)
when c+ d ≥ 176
27
(c+ d)2
4or
27
44≥ c+ d. This certainly happens when
a ≥ b ≥ c ≥ d, so we have f(a, b, c, d) ≤ f(m,m, c, d) where m =a+ b
2.
Then m+d ≤ 1
2<
27
44, so we can replace (m, c) with their mean n, to get
f(a, b, c, d) ≤ f(m,m, c, d) ≤ f(m,n, n, d). Analogously, we can replaceany two of the three greater numbers (except d) by their arithmeticmean.
We can now conclude that f(a, b, c, d) ≤ f(a+ b+ c
3,a+ b+ c
3,a+ b+ c
3, d
)by passing to the limit (remember the topic ”limits in inequalities”). So
we must prove that f(x, x, x, 1− 3x) ≤ 0 where1
4≤ x =
a+ b+ c
3≤ 1
3.
This is easy to do, as this equals
3x2(1− 3x) + x3 − 1
27− 176
27x3(1− 3x) =
1
27(528x4 − 392x3 + 81x2 − 1)
=1
27(3x− 1)(4x− 1)2(11x+ 1).
4. Let a, b, c, d ≥ 0. Show that
a4 + b4 + c4 + d4 + 2abcd ≥ a2b2 + b2c2 + c2d2 + d2a2 + b2b2 + a2c2.
Solution: We shall present two solutions by mixing here. The first is themix by product, which invariates the term 2abcd. Indeed, let
f(a, b, c, d) = a4+b4+c4+d4+2abcd−(a2b2+b2c2+c2d2+d2a2+b2b2+a2c2).
132
Then
f(a, b, c, d)− f(√
ab,√ab, c, d
)=
= a4 + b4 − 2a2b2 − (c2 + d2)(a2 + b2 − 2ab)
= (a− b)2[(a+ b)2 − c2 − d2
].
This is nonnegative when (a+b)2 ≥ c2+d2 so if we suppose a ≥ b ≥ c ≥ d,then f(a, b, c, d) ≥ f(m,m, c, d) where m =
√ab.
As in the problem above, we can mix any two of the larger three numberinto their geometric product, hence by passing to the limit it suffices toprove the inequality for a = b = c ≥ d. As the inequality is homogeneous,we can set a = b = c = 1, d = x ≤ 1. We have to prove that x4+3+2x ≥3 + 3x2, or that x3 + 2 ≥ 3x, which is obvious by AM-GM.
The second solution is the unusual sort of mix, in which we replace onevariable by another. Define f as above.
Let’s look at f(a, b, c, d)− f(a, a, c, d). It equals
b4 − a4 + 2acd(b− a)− (b2 − a2)(a2 + c2 + d2) =
= (b− a)[(b+ a)(b2 + a2) + 2bcd− (a2 + c2 + d2)(b+ a)
].
So it’s positive when b ≥ a and
(b2 + a2)(b+ a) + 2acd ≥ (a2 + c2 + d2)(a+ b),
or when b ≤ a and (b2 + a2)(b+ a) + 2acd ≤ (a2 + c2 + d2)(a+ b). As
(a2 + b2)(a+ b) + 2acd− (a2 + c2 + d2)(a+ b) =
= b2(a+ b) + 2acd− (c2 + d2)(a+ b)
= (b2 + ab− c2 − d2)b− (c− d)2a,
we can see that this is positive when b > a are the largest among a, b, c, dand is negative when a > b are the smallest among a, b, c, d.
Then, if we assume a ≥ b ≥ c ≥ d, by our conclusion we deduce
f(a, b, c, d) ≥ f(b, b, c, d) ≥ f(b, b, c, c).
Finally, f(b, b, c, c) = (b2 − c2)2 ≥ 0.
1.19 Limits in inequalities
Limits is a subject that apparently has nothing to do with inequalities. Itdoes, as we shall list above three applications for inequalities. All of theseapplications are based on the following crucial fact:
(F): If xi is a sequence converging to a constant a, f, g are continuous functionsand f(xi) ≥ g(xi), then f(a) ≥ g(a).
133
The proof of this fact is very simple: if we suppose that g(a) − f(a) = ε > 0then we may choose some n with
|f(xi)− a| <ε
2, |g(xi)− a| <
ε
2
and thus
(f(xi)− g(xi))− (f(a)− g(a)) ≤ |f(xi)− f(a)|+ |g(xi)− g(a)| < ε.
But this is as f(xi) − g(xi) is a nonnegative number, but f(a) − g(a) = −ε.This fact can easily be extended to functions of more variables. An importantnote here is that the inequalities f(xi) ≥ g(xi) can be strict, but it doesn’tfollow from here that the inequality f(a) ≥ g(a) is strict. Look for example
at the inequalities1
n> 0, but lim
n→∞
1
n= 0.
• Sometimes we have to prove an inequality like Young’s one:ap
p+bq
q≥ ab,
1
p+
1
q= 1, p, q > 0. We can prove them for p =
m+ n
n, q =
m+ n
nbeing
rationals, as multiplying to m + n this inequality is a direct consequence ofAM-GM. Now, as every real can be approximated by rationals, we can find
rationals pi, qi with1
pi+
1
qi= 1, |pi − p| <
1
i, |qi − q| <
1
i. As pi tend to p,
qi tend to q, from (F) we deduce that the inequality is true for p, q.
• Sometimes we have to prove an inequalityf(x1, x2, . . . , xn) ≥
≥ f(x1 + x2 + . . .+ xn
n,x1 + x2 + . . .+ xn
n, . . . ,
x1 + x2 + . . .+ xnn
)and we can prove that
f(x1, x2, . . . , xn) ≥ f(x1, x2, . . . , xi−1,
xi + xj2
, xi+1, . . . , xj−1,xi + xj
2, xj+1, . . . , xn
)for any 1 ≤ i < j ≤ n. By such operations we can actually make our sequenceof numbers tend to
a1 + a2 + . . .+ ann
, . . . ,a1 + a2 + . . .+ an
n.
Indeed, consider the invariant
E(x1, x2, . . . , xn) =n−1∑i=1
∑j=i+1n
(xi − xj)2.
Using the inequality
(x− a)2 + (x− b)2 ≥ 2
(x− a+ b
2
)2
,
we can prove that replacing xi, xj byxi + xj
2, we decrease E by at least
(xi − xj)2. Thus, taking |xi − xj | be the greatest of all possible differences
134
between two numbers, we can decrease E to at most
[1− 2
n(n− 1)
]E. So
we can make E as close to zero as possible, particularly making any of thepossible differences between two number as small as possible. Since our trans-formations preserve the sum, it’s now clear that al the numbers get as close
toa1 + a2 + . . .+ an
nas desired. Since f is continuous, we now deduce the
claim.Like this, for example, we can prove AM-GM:
a1 + a2 + . . .+ an ≥ n√a1a2 . . . an,
and even in two ways. The first way is almost identical to what we’ve told
above. However, we can do it another way: take f(x1, x2, . . . , xn) =x1x2 . . . xnn√x1x2 . . . xn
.
Then
f(x1, x2, . . . , xn) ≥ f(x1, . . . , xi−1,√xixj , xi+1, . . . , xj−1,
√xixj , xj+1, . . . , xn)
(the product of the number is the same, but the sum decreases from AM-GMfor two numbers). Next, we can proceed analogously as above (for example,work with lnxi instead of xi).
• Limits are also applicable with integral inequalities, as the integral is in factitself a limit. Like this, many of the classic inequalities translate into integralones.For example, CBS (Cauchy-Schwartz). Inequality rewrites as(∫ b
af2(x)dx
)(∫ b
ag2(x)dx
)≥(∫ b
af(x)g(x)dx
)2
.
Indeed, as
∫ b
ah(x)dx = lim
n→∞
f(x1) + f(x2) + . . .+ f(xn)
n, where xi =
(n− i)a+ ib
n,
(∫ b
af2(x)dx
)(∫ b
ag2(x)dx
)−(∫ b
af(x)g(x)dx
)2
is the limit of
f2(x1) + . . .+ f2(xn)
n
g2(x1) + . . .+ g2(xn)
n−(f(x1)g(x1) + . . .+ f(xn)g(xn)
n
)2
,
which is positive from CBS. Using (F), we deduce our claim.
1.20 Derivatives in Inequalities
When we have to prove a inequality of the form f(a) ≥ f(b), a > b, we canprove that f is increasing on [b; a]. But to prove that f is increasing on [b, a]it suffices to prove that f ′(x) ≥ 0.Consider for example the already discussed inequality
an1 + an2 + . . .+ ann ≥ na1a2 . . . an, an > 0 (AM-GM).
135
We can suppose that a1 ≥ a2 ≥ . . . ≥ an. Now let
f(a1, a2, . . . , an) = an1 + an2 + . . .+ ann − na1a2 . . . an.
Regarded as a function in a1, it is increasing, as
f ′(a1) = nan−11 − (n− 1)a2a3 . . . an ≥ 0.
So
f(a1, a2, . . . , an) ≥ f(a2, a2, a3, . . . , an).
Now we can regard it as a function in a2. It’s derivative is again positive, asit is 2an−12 − 2a2a3 . . . an. Now, the method of proof is clear: by induction onk we prove that
f(a1, a2, . . . , an) ≥ f(ak, ak, . . . , ak, ak+1, ak+2, . . . , an).
Indeed, the basis is proven, and for the induction step we note that
f(ak, ak, . . . , ak, ak+1, ak+2, . . . , an) ≥ f(ak+1, ak+1, . . . , ak+1, ak+1, ak+2, . . . , an)
because the function
g(x) = g(x, x, . . . , x, ak+1, ak+2, . . . , an)
is increasing as its derivative is
nkxn−1 − nkxk−1ak+1ak+2 . . . an ≥ 0, g(ak) ≥ g(ak+1).
Finally, because f(an, an, . . . , an) ≥ 0, we deduce our claim.
Another use of derivatives in inequalities comes from the combination of Weier-strass Theorem with Fermat’s Theorem.
Recall that Weierstrass’ theorem states that every continuous function on acompact set attains its extremes, and Fermat’s Theorem says that if x ∈ (a, b),is an extremal point of a continuous derivable function f then f ′(x) = 0.As proving an inequality f(x1, x2, . . . , xn) ≥ 0 is equivalent to min{f} ≥0, it suffices to consider x1, . . . , xn that realize this minimum (Weierstrass’stheorem ensures that it is achieved). Using Fermat’s theorem, we obtaininformation about these points, and often we can explicitly find then andverify the inequality directly.
Let’s solve the already familiar inequality
x2y + y2z + z2x ≤ 4
27, x+ y + z = 1
If we suppose x ≥ y ≥ z and substitute z = 1− x− y then we must prove
f(x, y) = x2y + y2(1− x− y) + (1− x− y)2x ≤ 4
27for x, y
in the compact set bounded by the conditions x ≥ y ≥ 0, y ≥ 1 − x − y andx+ y ≤ 1.
136
Now let (x0, y0) maximize f (this is true by Weierstrass Theorem). Therefore,from Fermat’s Theorem, either some of these conditions is an equality, eitherf ′(x) = f ′(y) = 0. In the first case, we either have one of the variables (x, y, z)zero, or two variables equal. If one is zero, we have discussed this case before, ifx = y then z = 1−2x and the inequality becomes x3+x2(1−x)+(1−2x)2x ≤4
27or f(x) = 4x3−2x2+x ≤ 4
27, for x ∈
[0,
1
2
]. As f ′(x) = 12x2−4x+1 ≥ 0,
we must prove this for x =1
2, which is true. Finally, let’s get to the cream of
the solution: the last case, when f ′x(x, y) = f ′y(x, y) = 0.
First, write out f explicitly:
x3 − y3 + 3x2y − 2x2 − 2xy + y2 + x
.
Then
f ′x(x, y) = (3x− 1) (x+ 2y − 1)
f ′y(x, y) = (3x+ 3y − 2)(x− y)
So, f ′x(x, y) = 0 when x =1
3or x+2y = 1 and f ′y(x, y) = 0 when 3x+3y−2 = 0
or x = y. If x = y, then since f ′x(x, y) = 0 we must have x = y =1
3which
gives us x = y = z =1
3and if 3x + 3y = 2 = 2(x + y + z), we will have that
x + y = 2z, which also gives us x = y = z since x ≥ y ≥ z. Therefore, in
this case, we can see that extreme attains when x = y = z =1
3but with this
value, it is trivial that f(x, y) <4
27. This ends our proof.
Exercises
1. Show that for a, b ≥ 0 we have 3a3 + 7b3 ≥ 9ab2.
Solution: If b = 0 it’s trivial so assume b > 0. By putting b = 1, a = x,we need to prove f(x) = 3x3 + 7 − 9x ≥ 0. As f ′(x) = 9(x2 − 1), itsuffices to look at the case x = 1, and in this case we get a positive value.
2. Let a, b, c > 0. Then 4
√a+ 3
√b+√c ≥ 40
√abc.
Solution: Put it as(a+ 3
√b+√c)10≥ abc. Let f(x) =
(x+ 3
√b+√c)10−
bcx. As f ′′(x) ≥ 0 we conclude f ′(x) ≥ f ′(0). Now we prove f ′(0) ≥ 0.
As f ′(x) = 10(x+ 3
√b+√c)9−bc, we have to prove that 10 (b+
√c)
3 ≥bc.
Now take g(x) = 10(x+√c)3−cx, then g′′(x) ≥ 0 and so g′(x) ≥ g′(0) =
29c ≥ 0 so g′ is positive and g is increasing so g(b) ≥ g(0) ≥ 0. Hencef ′(0) ≥ 0 so f ′ is increasing and positive and then f(a) ≥ f(0) > 0, asdesired.
137
3. Show that1
n+ 1+ . . .+
1
3n< ln 3.
Solution: We prove that ln(x+ 1)− lnx ≥ 1
x+ 1, for x ≥ 1. Indeed, by
Lagrange Theorem there is y ∈ [x, x + 1] with ln(x + 1) − ln(x) =1
y≥
1
x+ 1.
Thus
1
n+ 1+ . . .+
1
3n< ln(n+ 1)− ln(n) + . . .+ ln(3n)− ln(3n− 1)
= ln(3n)− ln(n) = ln 3.
4. Let a, b, c > 0 with a+ b+ c = 6. Find the minimum value of
P =
(a+
1
a
)(b+
1
b
)(c+
1
c
).
Solution: The expressions P is quite complicated to link with the relationa+b+c = 6, so we must to try to simplify P , looking at just two variables.
Thus, let’s compute the minimal value of
(a+
1
a
)(b+
1
b
)when a+b =
t is constant. This is a function in a:
(a+
1
a
)(t− a+
1
t− a
). When
a is close to the extreme values 0 and t, the expression tends to infinity,so the minimum is attained somewhere in the interior of the interval,thus at a = x where the derivative vanishes. The derivative is, as easilyto compute,
g(x) = t− 2x+1
t− x− 1
x− 1
x2
(t− x+
1
t− x
)+
1
(t− x)2
(x+
1
x
).
By symmetry, we see that when a = b i.e. 2x = t the derivative vanishes,so let’s extract t− 2x out of the expression:
g(x) = (t− 2x) +2x− tx(t− x)
+1
x2
(x+
1
x+ x− t− 1
t− x
)+
+
(x+
1
x
)[1
(t− x)2− 1
x2
]= (t− 2x) +
2x− tx(t− x)
+(t− 2x)(x2 − tx+ 1)
x3(t− x)− t(x2 + 1)(t− 2x)
x3(t− x)2
=(t− 2x)
[x2(x− t)2 − t2 − 1
]x2 (t− x)2
.
So, the derivative vanishes when x =t
2or when x2(t − x)2 = t2 + 1.
Thus we have established that for fixed a + b,
(a+
1
a
)(b+
1
b
)takes
its minimal value when a = b or a2b2 = (a + b)2 + 1. It’s quite hard to
138
compute a that satisfies the second relation and compare this value for
the value int
2. So we can’t actually compute the minimal value for all
t.
However, AM-GM helps us: we know that x(t−x) ≤ t2
4, so x2(t−x)2 ≤
t4
16. So when t2+1 >
t4
16, the minimum value occurs exactly when a = b.
We note that all t < 4 satisfy t2 + 1 >t4
16because t2 ≥ t4
16. Since
a+ b+ c = 6, some two of a+ b+ c sum to at most 4, that’s why we cansuppose they are equal.
Indeed, let a ≤ b ≤ c realize the minimum of P , then as a + b ≤ 4 wemust have a = b otherwise we could change a, b to keep a + b (and c)
fixed but decrease
(a+
1
a
)(b+
1
b
). Set a = b = x. If c = x then
x = 2. If not, by the same reasoning we deduce that a2c2 = (a+ c)2 + 1so x2(6 − 2x)2 = (6 − x)2 + 1, or 4x4 − 24x3 + 5x2 + 12x = 37. This
cannot actually happen for x ≤ 2, because if x <3
2then 4x4 < 24x3 and
5x2 + 12x < 5 · 1.52 + 12 · 1.5 = 29.25 < 37,
and if3
2≤ x ≤ 2 then 4x4 − 24x3 ≤ 8x3 − 24x3 = −16x3, and clearly
−16x3 + 5x2 + 12x < 0 < 37.
So, x = 2 and the minimum value realizes for a = b = c = 2 and equals125
8.
1.21 Convexity
Convexity, despite being slightly more complicated to define than monotonicityor continuity, is a very important property of a function. Here is one of themost used definitions for convexity:
Definition: A function defined on a interval I is called convex if for any 0 < α,β < 1, α+ β = 1, x, y ∈ I
f(αx+ βy) ≤ αf(x) + βf(y).
If the inequality is strict or α, β ∈ {0; 1}, the function is called strictly convex.(Note that when one of α, β is negative the inequality changes sign).When −f is convex, f is called concave.
Convexity is much more used with continuous and twice derivable functions,and here’s why:
• For a continuous twice derivable function f , f is convex if f ′′(x) ≥ 0. Iff ′′(x) > 0 for all x, the function is called strictly convex.
139
The proof of this fact goes as follows: clearly f ′(x) is increasing hence∫ a+t
af ′(x)dx ≤
∫ a+p+kt
a+pf ′(x)dx,
if a < b, p, t > 0. This means
k(f(a+ t)− f(a)) ≤ f(a+ p+ kt)− f(a+ p).
Now if we let p = t we get
kf(a+ t)− kf(a) ≥ f(a+ (k + 1)t)− f(a+ t)
or
f(a+ t) ≤ kf(a) + f(a+ kt)
k + 1,
and this is exactly the definition of convexity.
• A continuous function f is convex if
f(x) + f(y) ≥ 2f
(x+ y
2
). One implication is obvious. For the other, we firstly prove by induction that
for anyk
2n∈ [0, 1], k, n ∈ N, we have
k
2nf(x) +
2n − k2n
f(y) ≥ f(kx+ (2n − k)y
2n
).
The basis of the induction is k = 1 and follows form the condition. Now, if wehave proven the statement for n− 1, let’s prove it for n.
If k is even, thenk
2n=
k
22n−1
and we apply the induction hypothesis.
If k is odd, then we k − 1, k + 1 are even and so
k − 1
2nf(x) +
2n − k + 1
2nf(y) ≥ f
(k − 1
2nx+
2n − k + 1
2ny
)and
k + 1
2nf(x) +
2n − k − 1
2nf(y) ≥ f
(k + 1
2nx+
2n − k − 1
2ny
).
Summing these two identities and using the fact that
f
(k − 1
2nx+
2n − k + 1
2ny
)+f
(k + 1
2nx+
2n − k − 1
2ny
)≥ 2f
(k
2nx+
2n − k2n
y
),
we get the induction step done.
Therefore, we have proven
αf(x) + βf(y) ≥ f(αx+ βy), α+ β = 1
140
for α of the formk
2n. Since every real can be represented as the limit of
numbers of formk
2n, using the continuity of the function we deduce the formula
(see ”Limits in Inequalities”).
A useful reformulation of the definition goes as follows: if z is between x andy, then
f(z) ≤ z − yx− y
f(x) +z − xy − x
f(y)
(you may check thatz − yx− y
+z − xy − x
= 1 and that xz − yx− y
+ yz − xy − z
= z).
In this setting one can define convexity for functions that are not necessarilydefined on an interval (even though in practice we don’t really deal with suchfunctions).
A very important property of the convex function is that they realize themaximal value on a interval at the extremities of the interval (and henceconcave functions realize their minimal value at the extremities of the interval).This follows directly from the definition and this is sometimes used in problemslike this one:
Let a, b, c ∈ (0, 1). Then
a
b+ c+ 1+
b
c+ a+ 1+
c
a+ b+ 1+ (1− a)(1− b)(1− c) ≤ 1.
We may consider this as a function in
a : f(x) =x
b+ c+ 1+
c
x+ b+ 1+
b
x+ c+ 1+ (1− x)(1− b)(1− c).
We can easily see by taking the second derivative that it is convex, so takes itsmaximal values at 0 or 1. Analogously for b and c we conclude it’s enough toinvestigate the cases when all a, b, c are either 0 or 1, which are easy to handle.
But the most important property of convex functions is Jensen’s inequality,which is presented next.
1.22 Jensen’s Inequality
The notion of convexity gets most of its importance in lights of the famousJensen’s Inequality.
Suppose that f is continuous. We have seen from the definition of the inequal-ity that if z is a number between x and y then f(z) is smaller from that alinear combination of f(x) and f(y), and moreover this combination appliedto x and y gives z.Let’s look at this situation from a geometrical point of view:Draw the graph of f(x). Consider the pointsX(x, f(x)), Y (y, f(y)) and Z(z, f(z)).If z is between x and y, then the condition of convexity tells us that
f(z) ≤ z − yx− y
f(x) +z − xy − x
f(y)
141
which actually means that Z is below the line segment XY .From here we can deduce that every point in (XY ) is above the graph off . This means that the area above the graph of f is a convex subset (hencewhy this definition). It’s clear that the reverse implication holds: if the areaabove the graph is convex, then function if convex (the statements ”f(z) ≤z − yx− y
f(x) +z − xy − x
f(y)” and ”Z is below XY ” are equivalent). A good geo-
metric image of convex functions is therefore a kind of ”trough” representingthe graph of the function, like a parabola with positive leading coefficient.Now, take n pointsA1(x1, f(x1)), A2(x2, f(x2)), . . . , An(xn, f(xn)) on the graphof f , and assign to them positive weights λ1, . . . , λn with λ1+λ2+. . .+λn = 1.The point G = (λ1x1 + . . . + λnxn, λ1f(x1) + . . . + λnxn) is then the centerof gravity of the system of points. But it’s known that the center of grav-ity of some points (with positive weights) lies inside the convex hull of thesepoints. Indeed, if it wasn’t true, there would be a line l such that it separatesthe points A1, . . . , An and G. But then taking a system of coordinates withl being the ordinate, we would see that G has positive abscise but Ai havenegative abscises. This contradicts the obvious condition that the abscise ofG is a linear combination with positive coefficients λ1, . . . , λn of abscises ofA1, A2, . . . , An and thus must be negative. Therefore G must lie inside theconvex hull of A1, . . . , An. As the area above the graph of f is convex, G mustlie above the graph of f . This means that
λ1f(x1) + . . .+ λnf(xn) ≥ f(λ1x1 + . . .+ λnxn).
Thus, we have proved the following inequality:Theorem(Jensen’s Inequality) If f is a convex function on a interval I andx1, x2, . . . , xn ∈ I, then for any λ1, . . . , λn ≥ 0 with sum 1 we have
λ1f(x1) + . . .+ λnf(xn) ≥ f(λ1x1 + . . .+ λnxn).
Jensen’s inequality has a lot of applications in inequalities. It clearly getsreversed when f is concave.
For example, the function ex is convex, as its second derivative is ex > 0.
Taking now λi =1
nand applying Jensen we deduce
n∑i=1
exi
n≥ e
∑ni=1 xin .
Now taking ai = exi , we obtain exactly AM-GM inequality. Alternatively, onecan prove the AM-GM inequality using the concave function lnx.Many functions are either convex or concave. Here is a list of most common-used concave and convex functions:
− ax is convex for a > 1 and concave for 0 < a < 1.− lnx is concave. So is loga x when a > 1. When a < 1, loga x is convex.− xt is convex when t ∈ (−∞, 0)
⋃(1,∞) and concave when t ∈ (0, 1)
− sinx is convex on (π, 2π) and concave on (0, π).
142
− cosx is convex on
(π
2,3π
2
)and concave on
(−π
2,π
2
).
− tanx or cotx is convex on(π
2, π)⋃(3π
2, 2π
)and concave on
(0,π
2
)⋃(π,
3π
2
).
− |x− a| is convex, although not differentiable at a.
Exercises
1. In a triangle show that cosA+ cosB + cosC ≤ 3
2.
Solution: Suppose that A ≤ B ≤ C, then 0 < A ≤ B < π2 . As the func-
tion cosx is concave on[0,π
2
], we have cosA+ cosB ≤ 2 cos
A+B
2=
2 sinC
2.It suffices to prove that 2 sin
C
2+ cosC ≤ 3
2, which can be trans-
formed into −1
2
(2 sin
C
2− 1
)2
≤ 0, true!
2. If a, b, c ∈ (0, 1) then√a− a2 +
√2b− b2 +
√3c− c2 ≤
√6(a+ b+ c)− (a+ b+ c)2.
Solution: Set a = x, b = 2z, c = 3z. The inequality becomes
√x− x2 + 2
√y − y2 + 3
√z − z2 ≤ 6
√x+ 2y + z
6−(x+ 2y + 3z
6
)2
.
It follows from Jensen’s inequality applied to the concave function√x− x2.
3. If a, b, x, y, z > 0 and x+ y + z = 1 then(a+
b
x
)(a+
b
y
)(a+
b
z
)≥ (a+ 3b)3.
Solution: The inequality follows from Jensen applied to the convex func-
tion ln
(a+
b
x
).
4. Show that for a convex function f , we have
f(x) + f(y) + f(z) + f
(2x+ y
3
)+ f
(2y + z
3
)+ f
(2z + x
3
)≥
≥ 2
(f
(x+ 2y
3
)+ f
(y + 2z
3
)+ f
(z + 2x
3
)).
Solution: We have f(x) + f
(2z + x
3
)≥ 2f
(z + 2x
3
)by Jensen. By
summing with the analogously deduced relations we get the result.
143
5. Let a0, a1, . . . , an ∈(π
4,π
2
)with
n∑i=0
tan ai −π
4. Show that
tan a0 tan a1 . . . tan an ≥ nn+1.
Solution: Set bi = tan ai −π
4, then 0 < bi < 1 and b0 + . . .+ bn ≥
n− 1
n+ 1.
We have to show thatn∏i=0
1 + bi1− bi
≥ nn+1. This is straightforward by
Jensen, because the increasing function f(x) = ln
(1 + x
1− x
)is convex (it
has increasing derivative2
1− x2).
We can also show this by using just AM-GM: if we set xi = 1− bi thenas
n∑i=0
bi ≥ n− 1,
we get 1 + bk ≥n∑
i=0,i 6=kbi. Hence by AM-GM 1 + bi ≥ n
√√√√ n∏i=0,i 6=k
xi. Mul-
tiplying this by all i and dividing by x0x1 . . . xn, we get the conclusion.
1.23 The shrinking principle and Karamata’s In-equality
Take a convex function f and two numbers x, y. For any α, β ∈ [0, 1] withα+ β = 1 we have
αf(x) + βf(y) ≥ f(αx+ βy).
Also
βf(x) + αf(y) ≥ f(βx+ αy).
Summing we get
f(x) + f(y) ≥ f(αx+ βy) + f(βx+ αy).
Noticing the fact that (αx+ βy) + (βx+αy) = x+ y, we can reformulate thisresult as follows:
If z, t are between x, y and z + t = x+ y then
f(x) + f(y) ≥ f(z) + f(t),
for any convex f .
We see that z, t are actually obtained by taking x, y and ”pulling” them closerone to another, like ”shrinking” the interval (x; y). That’s why we can callthis result ”the shrinking principle”.
144
If we take now a sum f(x1)+ . . .+f(xn) and ”shrink” some pair of them xi, xjto a pair x′i, x
′j between xi and xj we decrease the sum f(x1)+ . . .+f(xn). We
can then perform more shrinkings. Thus, if y1, y2, . . . , yn are obtained fromx1, x2, . . . , xn by some shrinkings, we have
f(x1) + f(x2) + . . .+ f(xn) ≥ f(y1) + f(y2) + . . .+ f(yn).
It is now natural to ask how to characterize all the sequences y1, . . . , yn thatare obtained from x1, x2, . . . , xn by several applications of shrinking. A firstobservation is that the sum is constant. Another observation is, as y1, . . . , ynare ”closer” than x1, . . . , xn, the greatest of xi is bigger than the greatest ofyi, and the least of xi is smaller than the least of yi. We now come to thefollowing definition:
Definition. A sequence x1 ≥ x2 ≥ . . . ≥ xn majorizes another sequencey1 ≥ y2 ≥ . . . ≥ yn if and only if the following conditions hold:
(i) x1 + x2 + . . .+ xn = y1 + y2 + . . .+ yn(ii) x1 + x2 + . . .+ xm ≥ y1 + y2 + . . .+ ym for any 1 ≤ m ≤ n.
We claim that a sequence y1 ≥ y2 ≥ . . . ≥ yn can be obtained from
x1 ≥ x2 ≥ . . . ≥ xn
by shrinking if and only if it is majorized by x1, · · · , xn.
Let’s prove the first implication: that any sequence obtained by is majorizedby the original sequence. Since the relation of majorization is transitive, it’senough to prove that shrinking once yields a sequence majorized by the originalone.
Indeed, suppose we replace xi > xj by xi > yi > yj > xj , xi + xj = yi + yj .
Then the condition (i) clearly holds, so let’s prove (ii):
x1 + x2 + . . . + xm ≥ S, where S is the sum of m largest numbers from thenew sequence. If m < i then actually x1 + . . . + xm = S. If j > m ≥ i butthe number yj is not between the m th largest numbers of the new sequence,a strict inequality holds. Finally if m < j again equality holds.
Now let’s prove the second implication: if (yi) is majorized by (xi), it can beobtained from (xi) by some consecutive applications of shrinking.
Pick up the least k s.t. xk 6= yk and the greatest l s.t. xl 6= yl. Clearlyxk > yk, xl < yl. If xk+xl ≥ yk+yl, then we can shrink xk, xl to yk, xk+xl−ykotherwise we shrink xk, xl to xk + xl − yl, yl. It’s easy to prove that the newsequence is majorized by (xi) and majorizes (yi), but it has more commonterms with (yi). Hence, performing some such operations we can get al termsof the sequence equal to the terms of (yi), as we desired.
An immediate corollary of this reasoning is the Karamata’s inequality:
Theorem (Karamata’s Inequality) If (xi) majorizes (yi) and f is a convexfunction then
n∑i=1
f(xi) ≥n∑i=1
f(yi).
Another inequality, proved just like Karamata with the help of the shrinkingprinciple, is Muirhead’s inequality:
145
If (pi) majorizes (qi) and f is a convex function, then∑σ∈Sn
f
(n∑i=1
pixσ(i)
)≥∑σ∈Sn
f
(n∑i=1
qixσ(i)
).
The proof is left to the reader as an exercise. The suggestion is that theshrinkages that turn (pi) into (qi) induce shrinkages turn that turn (
∑pixσ(i))
into (∑qixσ(i)).
A particular case of this inequality, obtained for the function ex, is very com-monly used in solving symmetric homogeneous olympiad-style inequalities,and has the following form:
If x1, . . . , xn are positive numbers and (pi) majorizes (qi), then∑σ∈Sn
∏xpiσ(i) ≥
∑σ∈Sn
∏xqiσ(i).
For example, taking the sequence (4, 1, 0) that majorizes (3, 1, 1), we get that
x4y + xy4 + x4z + z4x+ y4z + z4y ≥ 2(x3yz + y3xz + z3xy).
One of Karamata’s applications is Popoviciu’s Inequality:
f(x)+f(y)+f(z)+3f
(x+ y + z
3
)≥ 2
(f
(x+ y
2
)+ f
(y + z
2
)+ f
(z + x
2
)).
Exercises
1. Assume that f is a convex increasing positive-valued function, and
x1 ≥ x2 ≥ . . . ≥ xn, y1 ≥ y2 . . . ≥ yn
satisfy x1 ≥ y1, x1+x2 ≥ y1+y2, . . . , x1+x2+. . .+xn ≥ y1+y2+. . .+yn.Then show that f(x1)+f(x2)+ . . .+f(xn) ≥ f(y1)+f(y2)+ . . .+f(yn).
Solution: We use induction on n. Let
ci = x1 + x2 + . . .+ xi − y1 − y2 − . . .− yi.
If one of ci is zero, then the inequality follows from Karamata applied to(x1, x2, . . . , xi) and (y1, y2, . . . , yi) and induction hypothesis applied to
(xi+1, . . . , xn) and (yi+1, . . . , yn).
If all of the ci are greater than zero, let ck be the smallest.
We can see that
f(y1)+. . .+f(yn) ≤ f(y1 +
ckk
)+. . .+f
(yk +
ckk
)+f (yk+1)+. . .+f(yn),
and(y1 +
ckn, . . . , yk +
ckn, yk+1, . . . , yn
)still satisfies the condition with
respect to (x1, x2, . . . , xn). Now the inequality follows from Karamata for
(x1, x2, . . . , xk) and(y1 +
ckn, . . . , yk +
ckn
)and induction hypothesis for
(xk+1, . . . , xn) and(yk+1, . . . , yn).
146
2. Show that 2a2
+ 2b2
+ 2c2 ≥ 2ab + 2ac + 2bc.
Solution: It follows from the previous exercise applied to the increasingconvex function 2x. Another solution is possible by applying Jensen tothe function ee
x.
3. Show that for a, b, c > 0 we have
1
a+
1
b+
1
c+
9
a+ b+ c≥ 4
(1
a+ b+
1
b+ c+
1
c+ a
).
Solution: This is just Popoviciu’s Inequality applied to the convex func-
tion1
x. Compare this solution to the one we given in the SOS section.
4. Prove that for any positive real numbers a1, . . . , an we have the inequal-ity:
n∑i=1
a21a22 + . . .+ a2n
≥n∑i=1
a1a2 + . . .+ an
.
Solution:
Lemma Consider positive reals a1 ≥ . . . ≥ an and b1 ≥ . . . ≥ bn with
a1 + . . . + an = b1 + . . . + bn andaiaj≥ bibj
for i < j. Then a1, . . . , an
majorizes b1, . . . , bn.
Proof : The proof is by induction. We see thataia1≤ bib1
for i = 1, . . . , n.
Summing we getS
a1≤ S
b1hence a1 ≥ b1 (S = a1+. . .+an = b1+. . .+bn)
Hence a1 ≥ b1. Further, we look at the new sequences a2, a3, . . . , anand b2, . . . , bn. By induction hypothesis, we get a2, a3, . . . , an majorizesb2, . . . , bn. Combined with a1 ≥ b1 the lemma is proven.
Now let xi =a2i
a21 + · · · a2n, yi =
aia1 + · · ·+ an
. We see that
a2ia21 + . . .+ a2i−1 + a2i+1 + . . .+ a2n
=xi
1− xi
andai
a1 + . . .+ ai−1 + ai+1 + . . .+ an=
yi1− yi
.
If we suppose a1 ≥ a2 ≥ . . . ≥ an then x1 ≥ . . . ≥ xn and y1 ≥ y2 ≥
. . . ≥ yn. Furtherxixj
=a2ia2j≥ ai
aj=
yiyj
for i < j. Hence by Lemma
x1, . . . , xn majorizes y1, . . . , yn and by noticing that f(x) =x
1− xis
convex, we apply Karamata QED.
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1.24 Schur’s Inequality
A lot of contest inequalities, after homogenization, clearing denominators andother algebraic manipulations, reduce to symmetric homogeneous inequalitiesin some variables. One useful tool in handling these inequalities is, as we haveseen above, Muirhead’s inequality. Another tool, also very helpful, is Schur’sinequality:
a3 + b3 + c3 + 3abc− a2b− b2c− c2a− ab2 − bc2 − ca2 ≥ 0.
It can be rewritten as
a(a− b)(a− c) + b(b− a)(b− c) + c(c− a)(c− b) ≥ 0.
Now, a generalization follows:If f is a convex or monotonic positive valued function, then for any a, b, c fromthe value domain the following inequality holds:
f(a)(a− b)(a− c) + f(b)(b− c)(b− a) + f(c)(c− a)(c− b) ≥ 0.
Proof. Let’s suppose that a ≥ b ≥ c. Then the inequality to prove is
f(a)(a− b)(a− c) + f(c)(a− c)(b− c) ≥ f(b)(a− b)(b− c).
If f is increasing, then
f(a) ≥ f(b), a− c ≥ b− c
sof(a)(a− b)(a− c) ≥ f(b)(a− b)(b− c).
As f(c)(a− c)(b− c) ≥ 0, the inequality follows.If f is decreasing, analogously we break the inequality into
f(c)(a− c)(b− c) ≥ f(b)(a− b)(b− c)
and f(a)(a− b)(a− c) ≥ 0. Finally, when f is convex then Jensen’s inequalityfor b between a and c gives
b− ca− c
f(a) +a− ba− c
f(c) ≥ f(b),
which multiplied by (a− b)(b− c) gives
f(a)(b− c)2(a− b)
a− c+ f(c)
(a− b)2(b− c)a− c
≥ f(b)(b− a)(b− c).
To conclude the proof, it suffices to notice that
(b− c)2(a− b)a− c
≤ (a− c)(a− b)
and(a− b)2(b− c)
a− c≤ (a− c)(b− c).
This inequality is particularly used for function f of the form at, for examplefor f(x) = x2 we get a4+b4+c4+abc(a+b+c) ≥ a3(b+c)+b3(a+c)+c3(a+b).In this form, the inequality has already been encountered and used severaltimes in the book, mostly for t = 1 but also for some other values.
148
1.25 The generalized Means
We have met the Arithmetic Mean, the Quadratic Mean and the HarmonicMean. All these means have the form
M(t) =
n∑i=1
ati
n
1t
Moreover, the inequalities between them for the case of two variables are
M(−1) ≤M(1) ≤M(2).
So, we might propose that M(x) is an increasing function in x for any fixedsystem of numbers a1, a2, . . . , an.The aim of this paragraph is to prove this inequality. The function is definedfor all x 6= 0, so to prove our assertion we need to prove that M ′(x) > 0 forx 6= 0 and to prove that M(x) ≥ M(y) for x > 0 > y. The last inequalityfollows from AM-GM, as AM-GM for ati gives M(t) ≥ G for t > 0 and M(t) ≤G for t < 0, where G is the geometric mean of ai.
Now, let’s compute M ′(x). Note F =
n∑i=1
ati
n, for simplicity. Then
M ′(x) =
(elnFx
)′= M(x)
F ′x− F lnF
x2F.
So, we must prove thatF ′x− F lnF ≥ 0.
It rewrites as
1
n
n∑i=1
axi ln aix− ln
n∑i=1
axi
n
n∑i=1
axi
,and is clearly equivalent to
1
n
n∑i=1
axi ln
naxin∑i=1
axi
≥ 0.
Multiplying this byn2∑ni=1 a
xi
we have to prove
n∑i=1
ti ln ti ≥ 0, where
ti =naxi
ax1 + ax2 + . . .+ axn
149
thusn∑i=1
ti = n.
This follows from the convexity of the function x lnx, whose second derivative
is1
x> 0. Moreover, this function is strictly convex hence if not all ai are equal
then M(x) is strictly increasing, as we wish.We have remarked that f is increasing and continuous in any point exceptx = 0, where it is not defined. However we know that for x > 0 f(x) > Gand for x < 0 f(x) < G. So, let’s prove that lim
x→0M(x) = G and then we
can put M(0) = G. Dividing by G, we have to prove that limx→0
(n∑i=1
axi
) 1x
=
1, if a1a2 . . . an = 1. Now we use Taylor expansion for the function axk as
1 + ln akx+∞∑i=2
lni akxi
i!.
If we denote by A the maximum number among | ln ai|, then we see that
|axk − (1− ln akx)| < nx2A2∞∑i=0
(Ax)i
(i+ 2)!< x2A2 < xε
for any ε if x is sufficiently small.
Thereforen∑i=1
axi −
(n∑i=1
1 + ln akx
)< nxε.
Dividing by n and using the fact thatn∑i=1
ln ai = 0 we get 1 ≤n∑i=1
axi ≤ 1 + εx.
Finally using the fact that (1 + εx)1x < eε we deduce that 1 < M(x) < eε for
x sufficiently small and this means that M(x) tends to 0 as desired.Finally, it’s easy to prove that
M(+∞) = limx→+∞
M(x) = max{ai} and M(−∞) = limx→−∞
M(x) = min{ai}.
Concluding, M(x) is a non-decreasing (actually strictly increasing unless allvariables are equal in which case it is constant) function that spans all valuesbetween min{ai} and max{ai} i.e.
min{ai} = M(−∞) < M(0) = n√a1a2 . . . an < M(+∞) = max{ai}.
1.26 Inequalities between the symmetric sums
All polynomial expressions that are symmetric in n variables can be expressedin terms of their symmetric sums - this is a content of a well-known theoremin algebra. Therefore, in solving symmetric inequalities one would like tocompare the symmetric sums among each other.More precisely, take x1, x2, . . . , xn be some real numbers and consider thepolynomial
P (X) = (X − x1)(X − x2) . . . (X − xn).
150
Its coefficient at xn−k is clearly
(−1)kσk = (−1)k∑
1≤i1<...<ik≤nxi1xi2 . . . xik .
The numbers σi are called the symmetric sums of x1, . . . , xn. Clearly σk hasdegree k and has
(nk
)monomials.
Therefore if xi > 0 the numbers mk = k
√σk(nk
) are a kind of mean of the xi
(when all of them equal a, mk also equals a).The polynomial P (X) has all roots real and distinct hence so does P ′(X) byRolle’s Theorem. Repeating this procedure k times we obtain P (k)(X) hasn − k real and distinct roots. Then the polynomial Q(X) = xn−kP (k)(X)
also has n − k real roots. Then, Q(n−k−2)X is a quadratic with two real roots,
hence its discriminant must be positive. Now, let’s compute its coefficients.All coefficients (−1)iσi vanish for i < k. Moreover, all coefficients (−1)iσi fori > k+2 vanish also as after the ”reversal” (the passing from P (k) to Q) thereare not among the three leading coefficients and vanish. Thus, the coefficientsthat are left are (−1)kσk, (−1)k+1σk+1, (−1)k+2σk+2. Now one can computethat the remaining quadratic is actually
k!(n− k)!
2(−1)kσkx
2+(k+1)!(n−k−1)!(−1)k+1σk+1x+(k + 2)!(n− k − 2)!
2σk+2.
This is written simpler by passing to mi:
(−1)kn!
(1
2mkkx
2 −mk+1k+1x+
1
2mk+2k+2
).
It’s discriminant is m2(k+1)k+1 −mk
kmk+2k+2 ≥ 0.
This can be also rewritten as(mk
mk+1
)k≤(mk+1
mk+2
)k+2
.
Raising this to k+ 1th power and multiplying them for k = 0, 1, . . . , i+ 1 and
cancelling common terms we deduce 1 ≤(
mi
mi+1
)i(i+1)
so mi ≥ mi+1.
The inequality mi ≥ mi+1 is a pretty strong inequality. For example, theinequality 3(x2 + y2 + z2) ≥ (x+ y+ z)2 is equivalent to 3(σ21 − 2σ2) ≥ σ21 i.e.σ21 ≥ 3σ2 which follows directly from the inequality m1 ≥ m2 for n = 3.We invite the reader to play with these relations and find more inequalities.Note: other symmetric inequalities, like Schur’s inequality, can also be writtenin terms of symmetric sums. In the following chapter, we will concentrate onthe case n = 3, which gives most applications.
1.27 The pqr technique
This inequality is devoted to a special kind of substitution. It is a particularcase of the symmetric sums from the previous chapter, for three variables.
151
Recall from the previous chapter that if x1, x2, . . . , xn are variables, there arespecial expressions called the symmetric sums: the coefficients of the polyno-mial
(t− x1)(t− x2) . . . (t− xn)
Explicitly, they are
σ1 =
n∑i=1
xi, σ2 =∑
1≤i<j≤nxixj , . . . , σn = x1x2 . . . xn
It is a well-known result in elementary algebra that every polynomial whichis symmetric in x1, . . . , xn can be expressed as an expression in σ1, σ2, . . . , σn(more precisely as a polynomial in the variables σ1, . . . , σn). It is thereforesometimes convenient to rewrite a symmetric inequality in terms of the sym-metric sums.
This idea is most often applied in the case n = 3, when there are three sym-metric sums, that are usually denoted as p = x1 + x2 + x3, q = x1x2 + x2x3 +x3x1, q = x1x2x3, hence the name ”the pqr technique”.
There are two important ideas that facilitate the application of this method.
First, we should think about the relationship between p, q, r when we use theabove substitution. Some simple inequalities, proven before (as inequalities inx1, x2, x3) are
p2 ≥ 3q, q2 ≥ 3pr. (1.1)
These inequalities are just obtained by making the substitution in the wellknown inequalities (x + y + z)2 ≥ 3(xy + yz + zx) and (X + Y + Z)2 ≥3(XY + Y Z + ZX) where X = yz, Y = xz, Z = xy.
Another useful inequality that tells information about p, q, r is Schur’s inequal-ity
ak(a− b)(a− c) + bk(b− c)(b− a) + ck(c− a)(c− b) ≥ 0
for any a, b, c ≥ 0 if k ∈ R and any a, b, c ∈ R if k = 2n (n ∈ N). The proof ofthis inequality has been presented before. Here we just consider its applicationin proving inequalities. We usually use the case k = 1 (for a, b, c ≥ 0) andk = 2 since they are easily transformed into pqr form. When k = 1 anda, b, c ≥ 0, one gets a(a− b)(a− c) + b(b− c)(b− a) + c(c− a)(c− b) ≥ 0.Wehave
a(a− b)(a− c) + b(b− c)(b− a) + c(c− a)(c− b) =
= a(a2 − ab− ac+ bc) + b(b2 − bc− ba+ ca) + c(c2 − ca− cb+ ab)
= a [a(a+ b+ c)− 2(ab+ bc+ ca) + 3bc] + b [b(a+ b+ c)− 2(ab+ bc+ ca) + 3ca]
+c [c(a+ b+ c)− 2(ab+ bc+ ca) + 3ab]
= (a2 + b2 + c2)(a+ b+ c)− 2(a+ b+ c)(ab+ bc+ ca) + 9abc
=[(a+ b+ c)2 − 4(ab+ bc+ ca)
](a+ b+ c) + 9abc
= 9r + p(p2 − 4q).
And thus, we get
r ≥ p(4q − p2)9
. (1.2)
152
When k = 2, we have a2(a− b)(a− c) + b2(b− c)(b− a) + c2(c− a)(c− b) ≥ 0.And after proceeding as above, we will get
6pr ≥ (p2 − q)(4q − p2). (1.3)
These are some relations we usually use in proving inequalities. But some-times, these inequalities are not strong enough to help us, so we will needanother estimate. We will present it at the end of this section.Secondly, we should establish for ourselves some identities so that we can easilychange the inequality into pqr form without having to to waste time to do thisthing every time we try this method. Here are some easy identities.
a2 + b2 + c2 = p2 − 2qa3 + b3 + c3 = p3 − 3pq + 3ra4 + b4 + c4 = p4 − 4p2q + 2q2 + 4pra5 + b5 + c5 = p5 − 5p3q + 5pq2 + 5(p2 − q)ra2b2 + b2c2 + c2a2 = q2 − 2pra3b3 + b3c3 + c3a3 = q3 − 3pqr + 3r2
(a+ b)(b+ c)(c+ a) = pq − rab(a+ b) + bc(b+ c) + ca(c+ a) = pq − 3ra3(b+ c) + b3(c+ a) + c3(a+ b) = p2q − 2q2 − pra4(b+ c) + b4(c+ a) + c4(a+ b) = qp3 − 3pq2 + (5q − p2)ra3(b2 + c2) + b3(c2 + a2) + c3(a2 + b2) = pq2 − (2p2 + q)r(a+ b)(a+ c) + (b+ c)(b+ a) + (c+ a)(c+ b) = p2 + q
Finally, let us apply this technique in practice.
Example 1. Let a, b, c > 0 such that a+ b+ c = 1. Show that4
ab+ bc+ ca+
81abc ≥ 15.
Solution: Our inequality is equivalent to4
q+ 81r ≥ 15 .Now, since a, b, c
are positive real numbers, the Schur’s inequality in degree three (i.e. k = 1)gives us 9r ≥ p(4q − p2) = 4q − 1. Using this inequality, it suffices to showthat 4
q + 9(4q − 1) ≥ 15, or 1q + 9q ≥ 6 which is obviously true by AM-GM
inequality.
Example 2. If a, b, c are positive reals such that abc = 1, thena+ b+ c
3≥
5
√a2 + b2 + c2
3
Solution: Our inequality is equivalent to p5 ≥ 81(p2−2q). Since q2 ≥ 3pr = 3p,
we have 3 ≤ q2
p . Therefore 81(p2 − 2q) ≤ 27q2(p2−2q)p . It suffices to prove that
p6 ≥ 27q2(p2 − 2q), or equivalently p6 + 54q3 ≥ 27p2q2. But this inequality istrue since by AM-GM inequality, we have
p6 + 54q3 = p6 + 27q3 + 27q3 ≥ 3 3√p6 · 27q3 · 27q3 = 27p2q2.
Example 3. Let a, b, c be real numbers such that a2 + b2 + c2 = 2. Prove thata+ b+ c ≤ 2 + abc.
153
Solution: From the given condition, we get q = p2−22 and we may write our
inequality as r + 2 − p ≥ 0. If p ≤ 1, then the inequality is trivial sinceby AM-GM inequality, we have 2 = a2 + b2 + c2 ≥ 3
3√a2b2c2 ≥ 2
3√a2b2c2,
then |r| ≤ 1. If p > 1, then by Schur’s inequality for fourth degree, we get
6pr ≥ (p2 − q)(4q − p2) =
(p2 − p2 − 2
2
)[2(p2 − 2)− p2
]=
(p2 + 2)(p2 − 4)
2,
or r ≥ (p2 + 2)(p2 − 4)
12p. It suffices to show that (p2+2)(p2−4)
12p + 2 − p ≥ 0, or
(4p+ p2 − 2)(p− 2)2 ≥ 0 which is obviously true since p ≥ 1.
Example 4. Let a, b, c > 0 such that abc = 1. Show that (a+ b)(b+ c)(c+ a) ≥4(a+ b+ c− 1).
Solution: We may write our inequality as pq−r ≥ 4(p−1), or p(q−4)+3 ≥ 0.If q ≥ 4, then the inequality is trivial. If q ≤ 4, since q2 ≥ 3pr = 3p. We have
p(q−4)+3 ≥ q2
3·(q−4)+3 =
1
3(q−3)(q2−q−3). On the other hand, we have
q = ab+bc+ca ≥ 33√a2b2c2 = 3. Therefore q2−q−3 ≥ 3q−q−3 = 2q−3 > 0.
Our proof is complete.As we said above, sometimes the above estimations are not always enough, forexample with the following inequality
1
a+
1
b+
1
c+ 48(ab+ bc+ ca) ≥ 25.
for any a, b, c > 0 such that a+ b+ c = 1.We cannot apply the above estimations to prove this inequality since it hasa ”strange” equality case, that is
(12 ,
14 ,
14
), but all above estimations are just
used for the case a = b = c or a = b, c = 0 or b = c = 0. That is the mainreason.Here is a stronger estimate:
Denote a+ b+ c = p, ab+ bc+ ca = p2−t23 , abc = r, where t ≥ 0, then we have
(p+ t)2(p− 2t)
27≤ r ≤ (p− t)2(p+ 2t)
27.
This estimate attains equality when (a− b)(b− c)(c− a) = 0.
Exercises
1. Let a, b, c > 0 such that a+ b+ c = 1. Show that a3 + b3 + c3 + 4abc ≥ 1
4.
Solution: After transforming the inequality into pqr form, we need toprove 1 − 3q + 7r ≥ 1
4 . By Schur’s inequality for third degree, we have
4q ≤ 1 + 9r, hence 1− 3q + 7r ≥ 1− 3
4(1 + 9r) + 7r =
1
4+
1
4r ≥ 1
4.
2. Let a, b, c be real numbers such that a2 + b2 + c2 = 9. Prove that 2(a+b+ c)− abc ≤ 10.
Solution: With this problem, we will get confused since the originalinequality attains equality for (a, b, c) = (2, 2,−1). With this equality
154
point, we cannot apply the Schur’s inequality to prove it since the Schur’sInequality attains equality for (t, t, 0). But there is a key for us to applyit, that is to change the equality case from (2, 2,−1) to (t, t, 0). To dothis, we will use the substitution a = x+ 1, b = y+ 1, c = z+ 1, then theoriginal condition can be rewritten as (x−1)2+(y−1)2+(z−1)2 = 9, orp2 − 2p− 2q = 6.And we need to prove 2 [(x− 1) + (y − 1) + (z − 1)]−
(x−1)(y−1)(z−1) ≤ 10, or p+ q− r ≤ 15, or p+p2 − 2p− 6
2− r ≤ 15,
or 2r ≥ p2 − 36. From (x− 1)2 + (y − 1)2 + (z − 1)2 = 9, we have
|(x− 1)(y − 1)(z − 1)| ≤
√[(x− 1)2 + (y − 1)2 + (z − 1)2
3
]3= 3√
3.
Hence
2 [(x− 1) + (y − 1) + (z − 1)]− (x− 1)(y − 1)(z − 1) ≤≤ 2 [(x− 1) + (y − 1) + (z − 1)] + |(x− 1)(y − 1)(z − 1)|= 2p+ |(x− 1)(y − 1)(z − 1)| − 6 ≤ 2p+ 3
√3− 6.
Thus, if 2p ≤ 16− 3√
3, then our inequality is proved. If 2p ≥ 16− 3√
3,then by Schur’s inequality for fourth degree, we have
6pr ≥ (p2 − q)(4q − p2) =
(p2 − p2 − 2p− 6
2
)(4 · p
2 − 2p− 6
2− p2
)=
1
2(p+ 2)(p− 6)(p2 + 2p+ 6),
or
2r ≥ (p+ 2)(p− 6)(p2 + 2p+ 6)
6p.
It suffices to show that(p+ 2)(p− 6)(p2 + 2p+ 6)
6p≥ p2 − 36, or (p −
6)2(p2 + 4p− 2) ≥ 0 which is obviously true since p ≥ 16−3√3
2 > 1.
3. Let a, b, c ≥ 0 such that a+b+c = 2. Prove that a2b2+b2c2+c2a2+abc ≤1.
Solution: We have p = 2 and our inequality is equivalent to q2− 3r ≤ 1.By Schur’s inequality for third degree, we have 9r ≥ p(4q−p2) = 8(q−1),
or q ≤ 8 + 9r
8. Hence q2−3r ≤
(8 + 9r
8
)2
−3r = 1+3
64r(27r−16) ≤ 1 as
by AM-GM inequality, we have r ≤(a+ b+ c
3
)3
=8
27<
16
27.
4. Let a, b, c be positive real numbers such that abc = 1. Prove that
1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a≤ 1
a+ 2+
1
b+ 2+
1
c+ 2.
Solution: After some computations (with notice that r = 1), we can findthat
1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a=
p2 + 4p+ 3 + q
p2 + 2p+ pq + q,
155
and1
a+ 2+
1
b+ 2+
1
c+ 2=
4p+ 12 + q
4p+ 9 + 2p.
Hence, our inequality is equivalent to4p+ 12 + q
4p+ 9 + 2p≥ p2 + 4p+ 3 + q
p2 + 2p+ pq + q,
or (3q−5)p2 + (q2 + 6q−24)p− q2−3q−27 ≥ 0. By AM-GM inequality,we have p, q ≥ 3, therefore q2 + 6q − 24 > 0. Hence
(3q − 5)p2 + (q2 + 6q − 24)p− q2 − 3q − 27 ≥≥ (3q − 5) · 32 + (q2 + 6q − 24) · 3− q2 − 3q − 27
= 2(q − 3)(q + 24) ≥ 0.
5. If a, b, c > 0 thena
b+ c+
b
c+ a+
c
a+ b+
abc
2(a3 + b3 + c3)≥ 5
3.
Solution: We normalize for p = 1, then we have
a
b+ c+
b
c+ a+
c
a+ b= (a+ b+ c)
(1
b+ c+
1
c+ a+
1
a+ b
)− 3
=q + 1
q − r− 3,
andabc
a3 + b3 + c3=
r
1− 3q + 3r.
Hence, our inequality is equivalent toq + 1
q − r+
r
2(1− 3q + 3r)≥ 14
3. By
Schur’s inequality for third degree, we have r ≥ 4q − 1
9. Hence
q + 1
q − r≥
q + 1
q − 4q − 1
9
=9(q + 1)
5q + 1, and
r
2(1− 3q + 3r)≥
4q−19
2
(1− 3q + 3 · 4q − 1
9
) =
4q − 1
6(2− 5q).We need to prove
9(q + 1)
5q + 1+
4q − 1
6(2− 5q)≥ 14
3, or
(17− 50q)(1− 3q)
2(5q + 1)(2− 5q)≥
0 which is true since17
50>
1
3≥ q.
6. Let a, b, c ≥ 0 such that ab + bc + ca + abc = 4. Prove that a + b + c ≥ab+ bc+ ca.
Solution: From the given condition, we have q + r = 4. And we need toprove p ≥ q. If p ≥ 4, then it is trivial since p ≥ 4 ≥ q. If 4 ≥ p, by AM-GM inequality, we can easily check that p ≥ 3. And the Schur’s inequality
for third degree gives us r ≥ p(4q − p2)9
. Then q +p(4q − p2)
9≤ 4, or
q ≤ p3 + 36
4p+ 9. Hence p− q ≥ p− p3 + 36
4p+ 9=
(4− p)(p2 − 9)
4p+ 9≥ 0.
7. Let a, b, c ≥ 0. Prove thata2 + b2 + c2
ab+ bc+ ca+
8abc
(a+ b)(b+ c)(c+ a)≥ 2.
156
Solution: Also, we normalize that p = 1, then our inequality becomes1− 2q
q+
8r
q − r≥ 2, or
8r
q − r≥ 4q − 1
q.By Schur’s inequality for third
degree and fourth degree, we have r ≥ max
{(4q − 1)(1− q)
6, 4q−19
}.
Then8r
q − r≥ 8r
q − 4q−19
=72r
1 + 5q≥ 12(4q − 1)(1− q)
1 + 5q. It suffices to
show that12(4q − 1)(1− q)
1 + 5q≥ 4q − 1
q, or
(1− 3q)(1− 4q)2
q(1 + 5q)≥ 0 which
is true.
8. Let a, b, c ≥ 0 such that ab+bc+ca+6abc = 9. Prove that a+b+c+3abc ≥6.
Solution: From the given condition, we have q + 6r = 9 and we needto prove p + 3r ≥ 6, or 2p − q − 3 ≥ 0. If p ≥ 6, then it is trivial since2p−q−3 ≥ 2·6−9−3 = 0. If 6 ≥ p, by AM-GM inequality, we get p ≥ 3.
And the Schur’s inequality for third degree gives us r ≥ p(4q − p2)9
.
Hence q +2p(4q − p2)
3≤ 9, or q ≤ 2p3 + 27
8p+ 3. Thus 2p − q − 3 ≥ 2p −
2p3 + 27
8p+ 3− 3 =
2(6− p)(p− 3)(p+ 1)
8p+ 3≥ 0.
9. Let a, b, c ≥ 0 such that a2 + b2 + c2 = 3. Prove that 12 + 9abc ≥7(ab+ bc+ ca).
Solution: The given condition can be rewritten as p2 − 2q = 3, and weneed to prove 12 + 9r ≥ 7q. By Schur’s inequality for third degree, wehave 9r ≥ p(4q − p2) = p
[2(p2 − 3)− p2
]= p(p2 − 6). Hence
12 + 9r − 7q ≥ 12 + p(p2 − 6)− 7q = 12 + p(p2 − 6)− 7 · p2 − 3
2
=1
2(2p+ 5)(p− 3)2 ≥ 0.
10. If a, b, c are positive reals such that abc = 1. Then 1 +3
a+ b+ c≥
6
ab+ bc+ ca.
Solution: The inequality can be rewritten as 1 +3
p− 6
q≥ 0. By AM-GM
inequality, we have q2 ≥ 3pr = 3p, hence3
p≥ 9
q2. Thus 1 +
3
p− 6
q≥
9
q2− 6
q+ 1 =
(3
q− 1
)2
≥ 0.
11. If a, b, c > 0, abc = 1. Then 2(a2+b2+c2)+12 ≥ 3(a+b+c)+3(ab+bc+ca).
Solution: Rewrite the original inequality in the pqr form 2p2− 3p− 7q+12 ≥ 0. By Schur’s inequality for third degree, we have p(4q−p2) ≤ 9r =
157
9, hence q ≤ p3+94p , thus
2p2 − 3p− 7q + 12 ≥ 2p2 − 3p− 7 · p3 + 9
4p+ 12 =
(p− 3)(p2 − 9p+ 21)
4p.
By AM-GM inequality, we have p ≥ 3 and p2−9p+21 ≥(2√
21− 9)p >
0. Our proof is complete.
12. If a, b, c are nonnegative real numbers, no two of which are zero. Deter-
mine the least constant a such that
(x+ y + z
3
)a(xy + yz + zx
3
)3−a2≥
(x+ y)(y + z)(z + x)
8.
Solution: Let a = b = 1, c → 0, then we get a ≥ 3 ln 3− 4 ln 2
2 ln 2− ln 3= a0 ≈
1.81884 . . . We will show that it is the value which we find, that is
(x+ y + z
3
)a0 (xy + yz + zx
3
)3−a02≥ (x+ y)(y + z)(z + x)
8.
Since this inequality is homogeneous, we can normalize that p = 1, then
it becomes r +8q
3−a02
33+a02
− q ≥ 0. If 1 ≥ 4q, then we have8q
3−a02
33+a02
− q =
q3−a02
(8
33+a02
− qa0−12
)≥ q
3−a02
8
33+a02
−(
1
4
)a0−12
= 0. If 13 ≥ q ≥
14 , then by Schur’s inequality for third degree, we have r ≥ 4q − 1
9. It
suffices to show that4q − 1
9+
8q3−a02
33+a02
−q ≥ 0, or f(q) = −5
9q+
8q3−a02
33+a02
−
1
9≥ 0. We have f ′′(q) = −2(3− a0)(a0 − 1)
33+a02 · q
1+a02
< 0. Hence f(q) is concave,
thus f(q) ≥ min
{f
(1
3
), f
(1
4
)}= 0. Our proof is complete.
13. Let a, b, c ≥ 0 such that a+b+c = 1. Prove that1
a+ b+
1
b+ c+
1
c+ a+
2abc ≥ 247
54.
Solution: The inequality is equivalent to 1+qq−r + 2r ≥ 247
54 , or 1+rq−r + 2r −
19354 ≥ 0. By Schur’s inequality for third degree, we have q ≤ 1+9r
4 . Hence
1 + r
q − r+ 2r − 193
54≥ 1 + r
1+9r4 − r
+ 2r − 193
54=
4(r + 1)
5r + 1+ 2r − 193
54
=(23− 20r)(1− 27r)
54(5r + 1)≥ 0
aince by AM-GM inequality, we have 27r ≤ 1.
158
14. If a, b, c ≥ 0, prove that a4(b+ c) + b4(c+a) + c4(a+ b) ≤ 1
12(a+ b+ c)5.
Solution: We normalize that p = 1, then the inequality becomes (5q −1)r+ q−3q2 ≤ 1
12, or 12(5q−1)r ≤ (1−6q)2. If q ≤ 1
5 , then it is trivial.
If1
3≥ q ≥ 1
5, then from 3r ≤ q2, we have
12(5q − 1)r − (1− 6q)2 ≤ 4q2(5q − 1)− (1− 6q)2
= 20q3 − 40q2 + 12q − 1
=
(20q3 − 36q2 + 12q − 28
25
)− 4q2 +
3
25
=4
25(5q − 1)2(5q − 7)− 4q2 +
3
25
≤ −4q2 +3
25≤ 3
25− 4 ·
(1
5
)2
= − 1
25< 0.
15. If a, b, c are nonnegative real numbers, no two of which are zero. Provethat(
a
b+ c
)2
+
(b
c+ a
)2
+
(c
a+ b
)2
+10abc
(a+ b)(b+ c)(c+ a)≥ 2.
Solution: Denote x =2a
b+ c, y =
2b
c+ a, z =
2c
a+ b, then we have 1
x+2 +
1y+2 + 1
z+2 = 1, or xy + yz + zx+ xyz = 4. Now, put p = x+ y + z, q =xy+ yz+ zx and r = xyz, then we have q+ r = 4, and we need to provex2 + y2 + z2 + 5xyz ≥ 8, or p2 − 2q + 5(4− q) ≥ 8, or p2 − 7q + 12 ≥ 0.If p ≥ 4, then it is trivial since p2 − 7q + 12 ≥ 42 − 7 · 4 + 12 = 0.If 4 ≥ p, then we can easily check by AM-GM inequality that p ≥ 3.
Using Schur’s inequality for third degree, we have r ≥ p(4q − p2)9
, hence
q +p(4q − p2)
9≤ 4, or q ≤ p3 + 36
4p+ 9. It follows that p2 − 7q + 12 ≥
p2 − 7 · p3 + 36
4p+ 9+ 12 =
3(p− 3)(16− p2)4p+ 9
≥ 0.
1.28 The tangent line technique and its extensions
The technique of using tangent line is an elegant method that provides niceand simple solutions for many inequalities we solve. But as we see, the originaltechnique can only help us solve the inequalities with only one equality casewhen all variables are equal. This is really a big disadvantage of this technique.In the latter part of this section, we will expand the technique to a morecomplicated version which can sometimes solve inequalities that the originaltangent line technique cannot tackle.
159
Starting with some examples
Example 1. If a, b, c are positive real numbers, then
a
b+ c+
b
c+ a+
c
a+ b≥ 3
2.
Solution: Without loss of generality, we may assume that a+ b+ c = 3, thenour inequality becomes
a
3− a+
b
3− b+
c
3− c≥ 3
2.
For any x ≤ 3, we have4x
3− x− 3x+ 1 =
3(x− 1)2
3− x≥ 0. Hence
4a
3− a+
4b
3− b+
4c
3− c≥ (3a− 1) + (3b− 1) + (3c− 1) = 6.
Example 2. Let a, b, c be positive real numbers, then prove that
(2a+ b+ c)2
2a2 + (b+ c)2+
(2b+ c+ a)2
2b2 + (c+ a)2+
(2c+ a+ b)2
2c2 + (a+ b)2≤ 8.
Solution: Suppose that a+ b+ c = 3, then our inequality becomes
(3 + a)2
2a2 + (3− a)2+
(3 + b)2
2b2 + (3− b)2+
(3 + c)2
2c2 + (3− c)2≤ 8,
or equivalently
a2 + 6a+ 9
a2 − 2a+ 3+b2 + 6b+ 9
b2 − 2b+ 3+c2 + 6c+ 9
c2 − 2c+ 3≤ 24.
For any x > 0, we havex2 + 6x+ 9
x2 − 2x+ 3− (4x + 4) = −(4x+ 3)(x− 1)2
x2 − 2x+ 3≤ 0.
Hence
a2 + 6a+ 9
a2 − 2a+ 3+b2 + 6b+ 9
b2 − 2b+ 3+c2 + 6c+ 9
c2 − 2c+ 3≤ (4a+ 4) + (4b+ 4) + (4c+ 4) = 24.
Example 3. Let a, b, c be positive real numbers such that abc = 1. Prove that√a2 + 1 +
√b2 + 1 +
√c2 + 1 ≤
√2(a+ b+ c).
Solution: Consider the function f(x) =√x2 + 1−
√2x+
1√2
lnx with x > 0,
we have
f ′(x) =x√
x2 + 1+
1√2x−√
2 =x√
x2 + 1+
1− 2x√2x
.
It is easy to check that f ′(x) > 0 if x ≤ 1
2. And in the case x >
1
2, we have
f ′(x) =x√
x2 + 1+
1− 2x√2x
=
x2
x2 + 1− (1− 2x)2
2x2
x√x2 + 1
+2x− 1√
2x
=(1− x)(2x3 − 2x2 + 3x− 1)
2x2(x2 + 1)
(x√
x2 + 1+
2x− 1√2x
) .160
And since 2x3−2x2 +3x−1 = x3 +x(x−1)2 +(2x−1) > 0 (for any x > 1/2),we obtain f ′(x) = 0 if and only if x = 1. By writing the variation board, wecan easily check that f(x) ≤ f(1) = 0 ∀x > 0.Hence
√a2 + 1 +
√b2 + 1 +
√c2 + 1 ≤
√2(a+ b+ c)− 1√
2(ln a+ ln b+ ln c)
=√
2(a+ b+ c).
Establishing the basic technique
While considering the above examples the reader may ask wonder about theorigin of the intermediate inequalities used. For example, with Example 1, the
fact is4x
3− x≥ 3x−1, and with Example 3, the fact is
√x2 + 1 ≤
√2x− 1√
2lnx.
The method applies when we have to prove an inequality of the form f(x1) +f(x2)+ . . .+f(xn) ≥ 0 but working directly with f(x) is troublesome . In thatcase we try to find a function g(x) such that f(x) ≥ g(x) and it is easier towork with g(x), then we will only need to prove the inequality g(x1) + g(x2) +. . .+g(xn) ≥ 0. As a matter of fact, g(x) is usually created from what are givenin the hypothesis. For example, if the hypothesis is x1+x2+ . . .+xn = n, thenwe may predict g(x) = k(x− 1). If the hypothesis is x1x2 . . . xn = 1, then wemay predict g(x) = k lnx. If the hypothesis is xk1 + xk2 + . . .+ xkn = n, then wemay predict g(x) = k(xk − 1). In general, we will consider inequalities whereequality is attained when all variables are equal. Notice that these predictionscannot be mechanical, they must depend on the case of equality (for example,the above predictions of g(x) count on the case x1 = x2 = . . . = xn = 1). Howdo we find the valid value of k? In the case when we want to compare f withx− 1, and f(x) is differentiable, then we must choose k = f ′(1) since we needto choose k such that the sign of h(x) = f(x)− g(x) does not change when xgets through the point x = 1. Similarly, if we are subject to x1 . . . xn = 1 sowe compare f to lnx we must have k = −f ′(1) because lnx has derivative −1at 1.
In the above examples, the estimate 4x3−x ≥ 3x− 1 is obtained in exactly this
way: we want to prove∑ 4x
3−x ≥ 6 i.e.∑
( 4x3−x − 2) ≥ 0. As 4x
3−x − 2 has
derivative equal to 3 when evaluated at 1, we propose the inequality 4x3−x−2 ≥
3(x− 1) i.e. 4x3−x ≥ 3x− 1, which is true.
This way is very neat and powerful with simple inequalities. But with sharperones, it is more difficult to apply useless since the inequality f(x) ≥ g(x) doesnot always holds. For example, consider with the following
Example 4. Let a, b, c be real numbers such that a+ b+ c = 1. Prove that
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 9
10.
(Poland 1997)
Solution: Using the above way, we can establish the following inequalityx
x2 + 1≤ 36x+ 3
50. But unfortunately, this inequality can only holds when
161
x ≥ −3
4since
36x+ 3
50− x
x2 + 1=
(4x+ 3)(3x− 1)2
50(x2 + 1)while the original prob-
lem is asking us to prove it in the case a, b, c are arbitrary real numbers. Hence,we need to examine the small cases in order to prove it.
Case 1. If min {a, b, c} ≥ −3
4, then we have
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 36
50(a+ b+ c) +
9
50=
9
10.
Case 2. Assume that one of a, b, c is less than −3
4, for example c < −3
4. Then,
by AM-GM inequality, we haveb
b2 + 1≤ 1
2.Hence, our inequality is proved if
a
a2 + 1≤ 2
5, or a ≤ 1
2∨a ≥ 2. It suffices for us to consider the case 2 ≥ a ≥ 1
2.
And by the same manner, we only need to prove the inequality in the case
2 ≥ a, b ≥ 12 , then we have −3
4> c = 1 − a − b ≥ −3, hence
c
c2 + 1≤ − 3
10.
Thusa
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 1
2+
1
2− 3
10=
7
10<
9
10.
In this example, the tangent line technique still works, but we need somecasework for when it fails. This is one way to extend the method, but itmakes the solutions longer and less elegant. In the next subsection, we willpresent another extension of the technique.
An extension of tangent line technique
Example 5. Let a, b, c be nonnegative real numbers such that a + b + c = 3.Prove that
1
2a2 + 3+
1
2b2 + 3+
1
2c2 + 3≥ 3
5.
Solution: The original technique cannot be applied directly since this inequal-ity attains equality not only when a = b = c but also for a = b, c = 0.Without loss of generality, assume that a ≥ b ≥ c. We will consider 2 cases
Case 1. If c ≥ 1
4, then we have
1
2x2 + 3+
4
25x− 9
25=
2(4x− 1)(x− 1)2
25(2x2 + 3)≥ 0
∀x ≥ 1
4, hence
1
2a2 + 3+
1
2b2 + 3+
1
2c2 + 3≥
(9
25− 4
25a
)+
(9
25− 4
25b
)+
(9
25− 4
25c
)=
3
5.
Case 2. If c ≤ 1
4, then
1
2x2 + 3+
8
75x− 22
75=
(4x+ 1)(2x− 3)2
75(2x2 + 3)≥ 0 ∀x ≥ 0,
hence
1
2a2 + 3+
1
2b2 + 3≥ 44
75− 8
75(a+ b) =
44
75− 8
75(3− c)
=8
75c+
4
15.
162
It suffices to show that1
2c2 + 3+
8
75c ≥ 1
3which is true since
1
2c2 + 3+
8
75c− 1
3−
2c[8c2 + 23c+ 12(1− 4c)
]75(2c2 + 3)
≥ 0.
The idea of this solution is quite similar to the original technique, but is onelevel higher in complexity. In order to prove the inequality f(x1)+f(x2)+. . .+f(xn) ≥ 0, we try establish the inequality f(x) ≥ g(x) and find when it holdsfor all x1, x2, . . . , xn. In the other case, there exist xk, xk+1, . . . , xn such thatf(xj) ≤ g(xj) for all j = k, . . . , n then will find another function h(x) suchthat f(x) ≥ h(x) for any x1, x2, . . . , xk−1, then we may reduce our inequalityinto h(x1) + h(x2) + . . .+ h(xk−1) + f(xk) + . . .+ f(xn) ≥ 0. And from now,we will prove this inequality based on the relationship between x1, x2, . . . , xn.There is many ways to establish the function h(x). In the case we considered,after consider the first case (which yields the first equality case a = b = c),in the second case, we have established the function h(x) based on the second
equality case a = b =3
2, c = 0, since it is better to choose h(x) as a linear
function: h(x) = kx + m and we need to have f(x) ≥ h(x) for all x ∈ [0, 3] ,
and it has equality when x =3
2, we can choose k = f ′
(3
2
), then we can
easily choose m (notice that this way is quite similar to the first case).
Example 6. Let a, b, c, d be nonnegative real numbers such that a+b+c+d = 2.Prove that
1
3a2 + 1+
1
3b2 + 1+
1
3c2 + 1+
1
3d2 + 1≥ 16
7.
Solution: Notice that this inequality attains equality when a = b = c = d = 1
and a = b = c =2
3, d = 0, we will use the idea above to prove it. Suppose
that a ≥ b ≥ c ≥ d, then we have 2 cases
Case 1. If d ≥ 1
12, then we have
1
3x2 + 1+
48
49x−52
49=
3(12x− 1)(2x− 1)2
39(3x2 + 1)≥ 0
∀x ≥ 1
12, hence
1
3a2 + 1+
1
3b2 + 1+
1
3c2 + 1+
1
3d2 + 1≥
≥(
52
49− 48
49a
)+
(52
49− 48
49b
)+
(52
49− 48
49c
)+
(52
49− 48
49d
)= 2.
Case 2. If d ≤ 1
12, then we have
1
3x2 + 1+
36
49x− 45
49=
(12x+ 1)(3x− 2)2
49(3x2 + 1)≥ 0
∀x ≥ 0, hence
1
3a2 + 1+
1
3b2 + 1+
1
3c2 + 1≥ 135
49− 36
49(a+ b+ c)
=135
49− 36
49(2− d) =
36
49d+
9
7.
163
It suffices to prove that1
3d2 + 1+
36
49d ≥ 1, which is true since
1
3d2 + 1+
36
49d− 1 =
3d[36d2 + 95d+ 12(1− 12d)]
49(3d2 + 1)≥ 0.
Example 7. Let a, b, c be nonnegative real numbers, not all are zero. Showthat
a2
2a2 + (b+ c)2+
b2
2b2 + (c+ a)2+
c2
2c2 + (a+ b)2≤ 2
3.
Solution: This inequality attains equality only for a = b, c = 0. But we stillcan use the above idea to prove it. Indeed, without loss of generality, assumethat a+ b+ c = 1 and a ≥ b ≥ c, then our inequality becomes
a2
3a2 − 2a+ 1+
b2
3b2 − 2b+ 1+
c2
3c2 − 2c+ 1≤ 2
3.
There are 2 cases
Case 1. If b ≥ 1
6, then we have
x2
3x2 − 2x+ 1+ 1
9−8
9x = −(6x− 1)(2x− 1)2
9(3x2 − 2x+ 1)≤
0 ∀x ≥ 1
6, hence
a2
3a2 − 2a+ 1+
b2
3b2 − 2b+ 1≤ 8
9(a+ b)− 2
9=
8
9(1− c)− 2
9=
2
3− 8
9c.
Butc2
3c2 − 2c+ 1− 8
9c = −c[(1− 3c)(17− 24c) + 7]
27(3c2 − 2c+ 1)≤ 0. Therefore, our in-
equality is proved.
Case 2. If1
6≥ b ≥ c, then a = 1− b− c ≥ 2
3, we have
x2
3x2 − 2x+ 1− 2
9x =
x(6x− 1)(2− x)
9(3x2 − 2x+ 1)≤ 0 ∀x ≤ 1
6,
Henceb2
3b2 − 2b+ 1+
c2
3c2 − 2c+ 1≤ 2
9(b+ c) =
2
9− 2
9a.
It suffices to show thata2
3a2 − 2a+ 1− 2
9a− 4
9≤ 0, which is true since
a2
3a2 − 2a+ 1− 2
9a− 4
9= −(3a− 2)(6a2 + 3a− 4) + 4
27(3a2 − 2a+ 1)≤ 0.
Another extension
Example 8. Let a, b, c be nonnegative real numbers such that a + b + c = 3.Prove that
1
9− ab+
1
9− bc+
1
9− ca≤ 3
8.
Solution: This inequality has the form of f(x1) + f(x2) + f(x3) ≥ 0 which issimilar to the above inequalities, that we have considered but the difference is
164
that x1, x2, x3 are equal to ab, bc, ca and not the original variables. hypothesisis related to a, b, c not ab, bc, ca. Even more, when we try to establish theinequality f(x) ≤ g(x) where g(x) has the form k(x − 1) then we actuallyobtain a reversed inequality. Indeed, by this way, we would want to establish
the inequality1
9− x≤ x+ 7
64. But
1
9− x− x+ 7
64=
(x− 1)2
64(9− x)≥ 0 which is
not what we want.This can still be remedied. Taking notice that ab, bc, ca < 3, it suffices to
consider x < 3, then we have1
9− x=x+ 7
64+
(x− 1)2
64(9− x)≤ x+ 7
64+
(x− 1)2
64(9− 3)=
x2 + 4x+ 43
384. Then we get
1
9− ab+
1
9− bc+
1
9− ca≤ 1
384(a2b2 + b2c2 + c2a2 + 4ab+ 4bc+ 4ca+ 43).
It suffices to show that a2b2 + b2c2 + c2a2 + 4(ab + bc + ca) ≤ 15. Put x =ab + bc + ca, then x ≤ 3 and by Schur’s inequality degree a(a − b)(a − c) +b(b− c)(b− a) + c(c− a)(c− b) ≥ 0, we have abc ≥ max
{0, 4x−93
}.
If 4x ≤ 9, then
a2b2 + b2c2 + c2a2 + 4(ab+ bc+ ca) = x2 + 4x− 6abc
≤ x2 + 4x ≤ 225
16< 15.
If 4x ≥ 9, then
a2b2 + b2c2 + c2a2 + 4(ab+ bc+ ca) = x2 + 4x− 6abc
≤ x2 + 4x− 2(4x− 9)
= (x− 1)(x− 3) + 15 ≤ 15.
Example 9. Let a, b, c be positive real number such that a4+b4+c4 = 3. Provethat
1
4− ab+
1
4− bc+
1
4− ca≤ 1.
Solution: Notice that ab, bc, ca <3
2, and for all x <
3
2, we have
1
4− x=x+ 2
9+
(x− 1)2
9(4− x)≤ x+ 2
9+
(x− 1)2
9
(4− 3
2
) =2x2 + x+ 12
45.
We have
1
4− ab+
1
4− bc+
1
4− ca≤ 1
45
[2(a2b2 + b2c2 + c2a2) + ab+ bc+ ca+ 12
].
On the other hand, by AM-GM inequality, we have a2b2 + b2c2 + c2a2 ≤
a4 + b4 + c4 = 3 and ab+ bc+ ca ≤ a2b2 + b2c2 + c2a2 + 3
2≤ 3. From this, we
get the result.
165
Section for self study
As we saw above, the simple and intuitive tangent line technique can be ex-panded in many complicated, creative or lucrative ways to solve harder in-equalities. Unfortunately, not all these expansions can be categorized. Inpractice, one needs to start with the original idea and modify it accordinglyto suit the needs of every particular problem. In this section, we present someexamples that can be solved in this way.
Example 10. Let a, b, c be positive real numbers such that abc = 1. Show that
a
a2 + 3+
b
b2 + 3+
c
c2 + 3≤ 3
4.
Solution: Notice thatx
x2 + 3− 3x(x+ 1)
8(x2 + x+ 1)= − x(3x+ 1)(x− 1)2
8(x2 + 3)(x2 + x+ 1)≤ 0
∀x ≥ 0. Hence
a
a2 + 3+
b
b2 + 3+
c
c2 + 3≤ 3
8
(a2 + a
a2 + a+ 1+
b2 + b
b2 + b+ 1+
c2 + c
c2 + c+ 1
)=
3
8
(3− 1
a2 + a+ 1− 1
b2 + b+ 1− 1
c2 + c+ 1
).
It suffices to prove that
1
a2 + a+ 1+
1
b2 + b+ 1+
1
c2 + c+ 1≥ 1.
Put a =yz
x2, b =
zx
y2, c =
xy
z2where x, y, z > 0, our inequality becomes
x4
x4 + x2yz + y2z2+
y4
y4 + y2zx+ z2x2+
z4
z4 + z2xy + x2y2≥ 1.
which is true since by Cauchy Schwartz inequality and AM-GM inequality, wehave
x4
x4 + x2yz + y2z2+
y4
y4 + y2zx+ z2x2+
z4
z4 + z2xy + x2y2≥
≥ (x2 + y2 + z2)2
x4 + y4 + z4 + x2y2 + y2z2 + z2x2 + xyz(x+ y + z)
≥ (x2 + y2 + z2)2
x4 + y4 + z4 + 2(x2y2 + y2z2 + z2x2)= 1.
Example 11. Let a, b, c be positive real number such that a+ b+ c = 2. Provethat
bc
a2 + 1+
ca
b2 + 1+
ab
c2 + 1≤ 1.
Solution: Our inequality is equivalent to
abc
(a
a2 + 1+
b
b2 + 1+
c
c2 + 1
)+ 1− (ab+ bc+ ca) ≥ 0.
166
Note that for all x ≥ 0, we have 1x2+1
− 1 +1
2x =
x(x− 1)2
x2 + 1≥ 0. Hence
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≥ a
(1− 1
2a
)+ b
(1− 1
2b
)+ c
(1− 1
2c
)= 2− 1
2(a2 + b2 + c2) = ab+ bc+ ca.
It suffices to show that abc(ab + bc + ca) + 1 − (ab + bc + ca) ≥ 0. Put x =
ab+ bc+ ca ≤ 1
3(a+ b+ c)2 =
4
3, then by Schur’s inequality for fourth degree
a2(a−b)(a−c)+b2(b−c)(b−a)+c2(c−a)(c−b) ≥ 0, we get r ≥ (x− 1)(4− x)
3.
Therefore
abc(ab+ bc+ ca) + 1− (ab+ bc+ ca) = xabc+ 1− x
≥ x(x− 1)(4− x)
3+ 1− x
=(3− x)(x− 1)2
3≥ 0.
Example 12. Let a, b, c be nonnegative real numbers such that a + b + c = 1.Prove that
a2 + bc
a2 + 1+b2 + ca
b2 + 1+c2 + ab
c2 + 1≤ 13
20.
Solution: Notice that for all 0 ≤ x ≤ 1, we have
x
x2 + 1− 12
25x− 4
25= −(3x+ 4)(2x− 1)2
25(x2 + 1)≤ 0,
and1
x2 + 1− 1 +
1
2x2 =
x2(x2 − 1)
2(x2 + 1)≤ 0.
We have
a2
a2 + 1+
b2
b2 + 1+
c2
c2 + 1≤ a
(12
25a+
4
25
)+ b
(12
25b+
4
25
)+ c
(12
25c+
4
25
)=
12
25(a2 + b2 + c2) +
4
25,
and
bc
a2 + 1+
ca
b2 + 1+
ab
c2 + 1≤ bc
(1− 1
2a2)
+ ca
(1− 1
2b2)
+ ab
(1− 1
2c2)
= ab+ bc+ ca− 1
2abc.
We need to prove
12
25(a2 + b2 + c2) +
4
25+ ab+ bc+ ca− 1
2abc ≤ 13
20,
Put q = ab + bc + ca, r = abc then the inequality becomes 1225(1 − 2q) + 4
25 +q − 1
2r ≤1320 , or (1 − 4q + 9r) + 41r ≥ 0 which is obviously true by Schur’s
inequality for third degree.
Exercises
167
1. Show that if a+ b+ c = 6 then a4 + b4 + c4 ≥ 2(a3 + b3 + c3).
Solution: We have a4 − 2a3 − 8(a − 2) =[(a+ 1)2 + 3
](a − 2)2 ≥ 0,
hence
a4 + b4 + c4 − 2(a3 + b3 + c3) ≥ 8(a− 2) + 8(b− 2) + 8(c− 2) = 0.
2. Let a, b, c > 0, a+ b+ c = 1. Prove thata
1 + bc+
b
1 + ca+
c
1 + ab≥ 9
10.
Solution: We have1
1 + x+
81
100x− 99
100=
(9x− 1)2
100(x+ 1)≥ 0 ∀x ≥ 0.Hence
a
1 + bc+
b
1 + ca+
c
1 + ab≥
≥ a
(99
100− 81
100bc
)+ b
(99
100− 81
100ca
)+ c
(99
100− 81
100ab
)=
99
100− 243
100abc ≥ 99
100− 243
100·(a+ b+ c
3
)3
=9
10.
3. Show that if a, b, c > 0 and abc = 1 then a2 + b2 + c2 + 9(ab+ bc+ ca) ≥10(a+ b+ c).
Solution: The inequality is equivalent to
a2 + b2 + c2 + 9
(1
a+
1
b+
1
c
)− 10(a+ b+ c) ≥ 0.
Consider the function f(x) = x2 +9
x− 10x+ 17 lnx with x > 0, we have
f ′(x) = 2x− 9
x2− 10 +
17
x=
(x− 1)[2(x− 2)2 + 1
]x2
,
Hence f ′(x) = 0 if and only if x = 1. Now, by making variation board,we get f(x) ≥ f(1) = 0 ∀x > 0. Therefore
a2 +b2 +c2 +9
(1
a+
1
b+
1
c
)−10(a+b+c) ≥ −17(ln a+ln b+ln c) = 0.
4. Show that if a, b, c > 0 then
(b+ c− a)2
a2 + (b+ c)2+
(c+ a− b)2
b2 + (c+ a)2+
(a+ b− c)2
c2 + (a+ b)2≥ 3
5.
Solution: Assume that a+ b+ c = 1, then our inequality becomes
(1− 2a)2
2a2 − 2a+ 1+
(1− 2b)2
2b2 − 2b+ 1+
(1− 2c)2
2c2 − 2c+ 1≥ 3
5.
We have(1− 2a)2
2a2 − 2a+ 1− 23
25+
54
25a =
2(6a+ 1)(1− 3a)2
25(2a2 − 2a+ 1)≥ 0. Hence
(1− 2a)2
2a2 − 2a+ 1+
(1− 2b)2
2b2 − 2b+ 1+
(1− 2c)2
2c2 − 2c+ 1≥ 69
25− 54
25(a+ b+ c) =
3
5.
168
5. Let a, b, c > 0, a2 + b2 + c2 = 3. Prove that1
a+
1
b+
1
c+
4
3(a+ b+ c) ≥ 7.
Solution: We have1
a+
4
3a− a2 + 13
6=
(6− a)(a− 1)2
6a≥ 0. Hence
1
a+
1
b+
1
c+
4
3(a+ b+ c) ≥ a2 + 13
6+b2 + 13
6+c2 + 13
6= 7.
6. For any positive real numbers a, b, c such that a+b+c = 1, then 10(a3 +b3 + c3)− 9(a5 + b5 + c5) ≥ 1.
Solution: We have 10a3−9a5+16
27−25
9a =
(16 + 21a− 18a2 − 27a3)(1− 3a)2
27.
Hence, assume that a ≥ b ≥ c, and we have 2 cases
Case 1. If 16 + 21a− 18a2 − 27a3 ≥ 0, then we observe that
16 + 21b− 18b2 − 27b3 ≥ 0, 16 + 21c− 18c2 − 27c3 ≥ 0.
We have
10(a3 + b3 + c3)− 9(a5 + b5 + c5) ≥ 25
9(a+ b+ c)− 16
9= 1.
Case 2. If 16 + 21a − 18a2 − 27a3 ≤ 0, then we have 27a3 + 18a2 ≥
21a + 16 ≥ 37a, or a ≥ 2√
30− 3
9>
2
3, it follows that b, c <
1
3. In this
case, we have
10b3 − 9b5 ≥ 9b3, 10c3 − 9c5 ≥ 9c3.
Hence
10b3 − 9b5 + 10c3 − 9c5 ≥ 9(b3 + c3) ≥ 9
4(b+ c)3 =
9
4(1− a)3.
It suffices to show that 10a3 − 9a5 +9
4(1− a)3 ≥ 1 which is true since
10a3 − 9a5 +9
4(1− a)3 − 1 =
(1− a)(3a− 1)(12a3 + 16a2 + 7a− 5)
4
≥ 0.
(since a >
2
3
)This completes our proof.
7. Let a, b, c be positive real numbers, then
a√b+ c
+b√c+ a
+c√a+ b
≥√
3
2(a+ b+ c).
Solution: Assume that a+ b+ c = 1, then our inequality becomes
a√1− a
+b√
1− b+
c√1− c
≥√
3
2.
169
We will show thata√
1− a≥ 5√
6
8
(a− 1
15
). Indeed, put 1 − a = 6t2
with t ≥ 0, then we have a = 1− 6t2, therefore
a√1− a
− 5√
6
8
(a− 1
15
)=
1− 6t2√6t− 5√
6
8
(1− 6t2 − 1
15
)=
√6(5t+ 2)(3t− 1)2
12t≥ 0.
Hence
a√1− a
+b√
1− b+
c√1− c
≥ 5√
6
8
(a+ b+ c− 1
5
)=
√3
2.
8. Let a, b, c be real numbers such that a2 + b2 + c2 = 1. Prove that
1
1− ab+
1
1− bc+
1
1− ca≤ 9
2.
Solution: Since
1
1− ab+
1
1− bc+
1
1− ca≤ 1
1− |ab|+
1
1− |bc|+
1
1− |ca|,
and|a|2 + |b|2 + |c|2 = a2 + b2 + c2 = 1.
It suffices to consider the case a, b, c ≥ 0. Now, for all x ≤ 1
2, we have
1
1− x=
9x+ 3
4+
(1− 3x)2
4(1− x)≤ 9x+ 3
4+
(1− 3x)2
4
(1− 1
2
) =18x2 − 3x+ 5
4.
Now, since ab, bc, ca ≤ 1
2, we get
1
1− ab+
1
1− bc+
1
1− ca≤ 3
4
[6(a2b2 + b2c2 + c2a2)− (ab+ bc+ ca) + 5
].
It suffices to show that
6(a2b2 + b2c2 + c2a2)− (ab+ bc+ ca) ≤ 1,
or
6(a2b2 + b2c2 + c2a2)− (ab+ bc+ ca)(a2 + b2 + c2) ≤ (a2 + b2 + c2)2,
or ∑cyc
a2(a− b)(a− c) + 2ab(a− b)2 + 2bc(b− c)2 + 2ca(c− a)2 ≥ 0.
which is obviously true by Schur’s inequality for fourth degree.
170
9. Let a, b, c be real numbers such that a2 + b2 + c2 + d2 = 1. Prove that
1
1− ab+
1
1− bc+
1
1− cd+
1
1− da≤ 16
3.
Solution: For all x ≤ 1
2, we have
1
1− x=
16x+ 8
9+
(1− 4x)2
9(1− x)≤ 16x+ 8
9+
(1− 4x)2
9
(1− 1
2
)=
32x2 + 10
9.
Now, since ab, bc, cd, da ≤ 1
2, we have
1
1− ab+
1
1− bc+
1
1− cd+
1
1− da≤ 32(a2b2 + b2c2 + c2d2 + d2a2) + 40
9
=32(a2 + c2)(b2 + d2) + 40
9
≤ 8(a2 + b2 + c2 + d2)2 + 40
9
=16
3.
10. Suppose a, b, c are real numbers such that a2 + b2 + c2 = 1. Show that
1
3 + a2 − 2bc+
1
3 + b2 − 2ca+
1
3 + c2 − 2ab≤ 9
8.
Solution: Since
1
3 + a2 − 2bc+
1
3 + b2 − 2ca+
1
3 + c2 − 2ab
≤ 1
3 + |a|2 − 2 |bc|+
1
3 + |b|2 − 2 |ca|+
1
3 + |c|2 − 2 |ab|,
and
|a|2 + |b|2 + |c|2 = a2 + b2 + c2 = 1.
It suffices for us to consider the case a, b, c ≥ 0. Now, for all x ∈ [−1, 1] ,we have
1
3 + x=
21− 9x
64+
(1 + 3x)2
64(3 + x)≤ 21− 9x
64+
(1 + 3x)2
64(3− 1)
=9x2 − 12x+ 43
128.
171
Since a2 − 2bc, b2 − 2ca, c2 − 2ab ∈ [−1, 1] (it is easy to check), we have
1
3 + a2 − 2bc+
1
3 + b2 − 2ca+
1
3 + c2 − 2ab
≤ 3
128
[3∑cyc
(a2 − 2bc)2 − 4∑cyc
(a2 − 2bc) + 43
]
=3
128
[3∑cyc
(a2 − 2bc)2 + 8∑cyc
ab+ 39
].
It suffices to show that
3∑cyc
(a2 − 2bc)2 + 8∑cyc
ab ≤ 9,
or
3∑cyc
(a2 − 2bc)2 + 8(ab+ bc+ ca)(a2 + b2 + c2) ≤ 9(a2 + b2 + c2)2,
or6∑cyc
a4 + 6∑cyc
a2b2 + 4abc∑cyc
a ≥ 8∑cyc
ab(a2 + b2).
By Schur’s inequality for fourth degree, we have
4∑cyc
a4 + 4abc∑cyc
a ≥ 4∑cyc
ab(a2 + b2).
We have to prove
2∑cyc
a4 + 6∑cyc
a2b2 ≥ 4∑cyc
ab(a2 + b2).
which is true since
a4 + b4 + 6a2b2 − 4ab(a2 + b2) = (a− b)4 ≥ 0.
11. Let a, b, c be positive real numbers such that abc = 1. Prove that
1
3a2 + (a− 1)2+
1
3b2 + (b− 1)2+
1
3c2 + (c− 1)2≥ 1.
Solution: Assume that a ≥ b ≥ c, then if c ≤ 1
2, we have
1
3c2 + (c− 1)2=
1
4c2 − 2c+ 1=
1
2c(2c− 1) + 1≥ 1.
In the case a ≥ b ≥ c ≥ 1
2, then consider the function f(x) =
1
3x2 + (x− 1)2−
1
3+
2
3lnx with x ≥ 1
2, we have f ′(x) =
2(x− 1)(16x3 − 1)
3x(4x2 − 2x+ 1)2, hence
172
f ′(x) = 0 if and only if x = 1 (since x ≥ 1
2). Now, by writing the
variation board, we have f(x) ≥ f(1) = 0 ∀x ≥ 1
2. Therefore
1
3a2 + (a− 1)2+
1
3b2 + (b− 1)2+
1
3c2 + (c− 1)2≥ 1−2
3(ln a+ln b+ln c) = 1.
Remark. There is another approach for this inequality: we have
1
3a2 + (a− 1)2− 1
a4 + a2 + 1=
a(a+ 2)(a− 1)2
(4a2 − 2a+ 1)(a4 + a2 + 1)≥ 0.
Hence, it suffices to show that
1
a4 + a2 + 1+
1
b4 + b2 + 1+
1
c4 + c2 + 1≥ 1.
This inequality has been proved in Example 10.
12. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat
a2
5a2 + (b+ c)2+
b2
5b2 + (c+ a)2+
c2
5c2 + (a+ b)2≤ 1
3.
Solution: Normalize by declaring a + b + c = 1 and assume a ≥ b ≥ c,then our inequality becomes
a2
6a2 − 2a+ 1+
b2
6b2 − 2b+ 1+
c2
6c2 − 2c+ 1≤ 1
3.
Consider 2 cases
Case 1. If c ≥ 1
8, then
x2
6x2 − 2x+ 1− 12x− 1
27= −(8x− 1)(3x− 1)2
27(6x2 − 2x+ 1)≤
0 ∀x ≥ 1
8. Hence
a2
6a2 − 2a+ 1+
b2
6b2 − 2b+ 1+
c2
6c2 − 2c+ 1≤ 12a− 1
27+
12b− 1
27+
12c− 1
27
=1
3.
Case 2. If c ≤ 1
8, then
x2
6x2 − 2x+ 1− 4x+ 1
18= −(6x+ 1)(2x− 1)2
18(6x2 − 2x+ 1)≤
0 ∀x ≥ 0. Hence
a2
6a2 − 2a+ 1+
b2
6b2 − 2b+ 1≤ 4(a+ b) + 2
18=
4(1− c) + 2
18=
1
3− 2
9c.
It suffices to show that c2
6c2−2c+1− 2
9c ≤ 0 which is true since
c2
6c2 − 2c+ 1− 2
9c = −c[12c2 + 3c+ 2(1− 8c)]
9(6c2 − 2c+ 1)≤ 0.
173
13. Let a, b, c, d be positive real numbers such that a+ b+ c+ d = 4. Provethat
(a2 + 1)(b2 + 1)(c2 + 1)(d2 + 1) ≥ (a+ 1)(b+ 1)(c+ 1)(d+ 1).
Solution: Assume that a ≥ b ≥ c ≥ d. Then our inequality is equivalentto ∑
cyc
[ln(a2 + 1)− ln(a+ 1)] ≥ 0.
We have 2 cases
Case 1. If a ≤ 2, consider the function f(x) = ln(x2 + 1)− ln(x+ 1)−1
2(x− 1) with x ≤ 2, we have
f ′(x) =(x− 1)(3− x2)2(x2 + 1)(x+ 1)
From this, we get f ′(x) = 0 if and only if x = 1 or x =√
3 and by writingthe variation board, we have f(x) ≥ min {f(1), f(2)} = 0 ∀x ≤ 2. Hence∑
cyc
[ln(a2 + 1)− ln(a+ 1)] ≥ 1
2
∑cyc
(a− 1) = 0.
Case 2. If a ≥ 2, then 2 ≥ b ≥ c ≥ d, consider the function g(x) =
ln(x2 + 1)− ln(x+ 1)− 7
65(3x− 2)− ln
13
15with x ≤ 2, we have
g′(x) =(3x− 2)(43 + 10x− 7x2)
65(x2 + 1)(x+ 1).
Since x ≤ 2, we have g′(x) = 0 if and only if x =2
3and by writing the
variation board, we get
g(x) ≥ g(
2
3
)= 0 ∀x ≤ 2.
Hence
[ln(b2 + 1)− ln(b+ 1)] + [ln(c2 + 1)− ln(c+ 1)] + [ln(d2 + 1)− ln(d+ 1)]
≥ 7
65(3b+ 3c+ 3d− 6) + 3 ln
13
15=
21
65(2− a) + 3 ln
13
15.
We need to prove
h(a) = ln(a2 + 1)− ln(a+ 1) +21
65(2− a) + 3 ln
13
15≥ 0.
We have
h′(a) =(3a− 2)(43 + 10a− 7a2)
65(a2 + 1)(a+ 1),
and since a ≥ 2, we get h′(a) = 0 if and only if a = 5+√326
7 . Writing thevariation board again, we have h(a) ≥ min {h(2), h(4)} > 0.
174
14. Let a, b, c be nonnegative real numbers, no two of which are zero. Provethat √
1 +48a
b+ c+
√1 +
48b
c+ a+
√1 +
48c
a+ b≥ 15.
Solution: Suppose that a+ b+ c = 1 and a ≥ b ≥ c, then our inequalitybecomes √
1 + 47a
1− a+
√1 + 47b
1− b+
√1 + 47c
1− c≥ 15.
We have 2 cases
Case 1. If c ≥ 2
27, then
1 + 47x
1− x−(
54
5x+
7
5
)2
=12(27x− 2)(3x− 1)2
25(1− x)≥ 0 ∀1 ≥ x ≥ 2
27.
Hence√1 + 47a
1− a+
√1 + 47b
1− b+
√1 + 47c
1− c≥ 54
5(a+ b+ c) +
21
5= 15.
Case 2. If c ≤ 2
27, then
1 + 47x
1− x−(
96
7x+
1
7
)2
=48(48x+ 1)(2x− 1)2
49(1− x)≥ 0 ∀1 ≥ x ≥ 0.
Hence √1 + 47a
1− a+
√1 + 47b
1− b≥ 96
7(a+ b) +
2
7= 14− 96
7c.
We need to prove√
1+47c1−c ≥ 1 + 96
7 c. Put 1+47c1−c = t2 (t ≥ 0) , then we
have c = t2−1t2+47
and since 0 ≤ c ≤ 227 , we get 11
5 ≥ t ≥ 1, our inequality
becomes t ≥ 1 + 96(t2−1)7(t2+47)
, or (t − 1)(7t2 − 96t + 233) ≥ 0 which is true
since 115 ≥ t ≥ 1.
15. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Provethat
1− 4a2
1 + 3a− 3a2+
1− 4b2
1 + 3b− 3b2+
1− 4c2
1 + 3c− 3c2≤ 1.
Solution: Without loss of generality, assume that a ≥ b ≥ c. We have 2cases
Case 1. If c ≥ 1
9, then
1− 4x2
1 + 3x− 3x2+
27x− 14
15=
(1− 9x)(1− 3x)2
15(1 + 3x− 3x2)≤ 0 ∀1 ≥ x ≥ 1
9.
Hence
1− 4a2
1 + 3a− 3a2+
1− 4b2
1 + 3b− 3b2+
1− 4c2
1 + 3c− 3c2≤ 42− 27(a+ b+ c)
15= 1.
175
Case 2. If c ≤ 1
9, then
1− 4x2
1 + 3x− 3x2− 8(1− 2x)
7= −(12x+ 1)(2x− 1)2
7(1 + 3x− 3x2)≤ 0 ∀1 ≥ x ≥ 0.
Hence
1− 4a2
1 + 3a− 3a2+
1− 4b2
1 + 3b− 3b2≤ 8
7(2− 2a− 2b) =
16
7c.
We need to prove 167 c+ 1−4c2
1+3c−3c2 ≤ 1 which is true since
16
7c+
1− 4c2
1 + 3c− 3c2− 1 = −c(48c2 − 41c+ 5)
7(1 + 3c− 3c2)≤ 0
(since c ≤ 1
9
).
16. Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 2.Prove that
a2
(a2 + 1)2+
b2
(b2 + 1)2+
c2
(c2 + 1)2+
d2
(d2 + 1)2≤ 16
25.
Solution: Without loss of generality, assume that a ≥ b ≥ c ≥ d. Wehave 2 cases
Case 1. If 12d3 + 11d2 + 32d ≥ 4, then
x2
(x2 + 1)2−48x− 4
125= −(12x3 + 11x2 + 32x− 4)(2x− 1)2
125(x2 + 1)2≤ 0
for all x such that 12x3 + 11x2 + 32x ≥ 4. Hence
a2
(a2 + 1)2+
b2
(b2 + 1)2+
c2
(c2 + 1)2+
d2
(d2 + 1)2≤
≤ 48a− 4
125+
48b− 4
125+
48c− 4
125+
48d− 4
125=
16
25.
Case 2. If 4 ≥ 12d3 + 11d2 + 32d ≥ 32d, it follows that d ≤ 1
8, we have
x2
(x2 + 1)2−108(5x+ 1)
2197= −(60x3 + 92x2 + 216x+ 27)(3x− 2)2
2197(x2 + 1)2≤ 0 ∀x ≥ 0.
Hence
a2
(a2 + 1)2+
b2
(b2 + 1)2+
c2
(c2 + 1)2≤ 108 [5(a+ b+ c) + 3]
2197
=108 (5(2− d) + 3)
2197
=108
169− 540
2197d.
It suffices to prove that
d2
(d2 + 1)2− 540
2197a+
108
169≤ 16
25,
176
or169d2
(d2 + 1)2− 540
13d ≤ 4
25.
We have
169d2
(d2 + 1)2− 540
13d− 4
25≤ 169d2
(d2 + 1)2− 540
15a− 4
25
=169d2
(d2 + 1)2− 36a− 4
25
= −4 + 900d− 4217a2 + 1800d3 + 4d4 + 900d5
25(d2 + 1)2
≤ −4 + 8 · 900d2 − 4217d2 + 1800d3 + 4d4 + 900d5
25(d2 + 1)2
= −4 + 2983d2 + 1800d3 + 4d4 + 900d5
25(d2 + 1)2< 0.
17. Let a1, a2, . . . , an (n ≥ 2) be nonnegative real numbers such that a1 +a2 + . . .+ an = n. Prove that
(n− 1)(a31 + a32 + . . .+ a3n) + n2 ≥ (2n− 1)(a21 + a22 + . . .+ a2n).
Solution: If n = 2, then the inequality becomes equality. If n = 3,put q = a1a2 + a2a3 + a3a1, r = a1a2a3, then the inequality becomes2(27 − 9q + 3r) + 9 ≥ 5(9 − 2q), or 3r + 9 − 4q ≥ 0 which is justSchur’s inequality for third degree. Consider the case n ≥ 4, assumethat a1 ≥ a2 ≥ . . . ≥ an. We have 2 cases
Case 1. If an ≥1
n− 1, then we have
(n−1)x3−(2n−1)x2+(n+1)x−1 = (x−1)2[(n−1)x−1] ≥ 0 ∀x ≥ 1
n− 1.
Hence
(n− 1)
n∑i=1
a3i ≥n∑i=1
[(2n− 1)a2i − (n+ 1)ai + 1],
or
(n− 1)n∑i=1
a3i + n2 ≥ (2n− 1)n∑i=1
a2i .
Case 2. If an ≤1
n− 1, then we have
(n− 1)x3 − (2n− 1)x2 +n(n− 1)(n− 2)x+ n2
(n− 1)2=
=[(n− 1)x− n]2[(n− 1)x+ 1]
(n− 1)2≥ 0 ∀x ≥ 0.
Hence
n−1∑i=1
[(n− 1)a3i − (2n− 1)a2i +
n(n− 1)(n− 2)ai + n2
(n− 1)2
]≥ 0,
177
or
(n− 1)n−1∑i=1
a3i − (2n− 1)n−1∑i=1
a2i + n2 ≥ n(n− 2)
n− 1an.
We need to prove
(n− 1)a3n − (2n− 1)a2n +n(n− 2)
n− 1an ≥ 0,
oran[(n− 1)(2− an)[1− (n− 1)an] + n2 − 4n+ 2]
n− 1≥ 0
which is true since n ≥ 4 and an ≤ 1n−1 .
1.29 Using identities to prove inequalities
This is another collection of loosely related methods which provide short so-lutions to hard problems. The idea is to find special identities which help ussolve the problem. Let us consider some examples:
Example 1. Let x, y be real numbers such that x 6= −y, show that
x2 + y2 +
(1 + xy
x+ y
)2
≥ 2.
Solution: Put z = −1+xyx+y then we have xy + yz + zx = −1 and our original
inequality is equivalent to x2 +y2 +z2 ≥ 2, or x2 +y2 +z2 ≥ −2(xy+yz+zx),which is just simply to (x+ y + z)2 ≥ 0.
Example 2. If a, b, c are distinct real numbers, then
1 + a2b2
(a− b)2+
1 + b2c2
(b− c)2+
1 + c2a2
(c− a)2≥ 3
2.
Solution: Notice that(1− aba− b
− 1
)(1− bcb− c
− 1
)(1− cac− a
− 1
)=
=(1− a)(1 + b)(1− b)(1 + c)(1− c)(1 + a)
(a− b)(b− c)(c− a)
=(1 + a)(1− b)
a− b· (1 + b)(1− c)
b− c· (1 + c)(1− a)
c− a
=
(1− aba− b
+ 1
)(1− bcb− c
+ 1
)(1− cac− a
+ 1
),
and (1 + ab
a− b− i)(
1 + bc
b− c− i)(
1 + ca
c− a− i)
=
=(i+ a)(b− i)(i+ b)(c− i)(i+ c)(a− i)
(a− b)(b− c)(c− a)
=(a− i)(i+ b)
a− b· (b− i)(i+ c)
b− c· (c− i)(i+ a)
c− a
=
(1 + ab
a− b+ i
)(1 + bc
b− c+ i
)(1 + ca
c− a+ i
),
178
Hence
1− aba− b
· 1− bcb− c
+1− bcb− c
· 1− cac− a
+1− cac− a
· 1− aba− b
= −1,
1 + ab
a− b· 1 + bc
b− c+
1 + bc
b− c· 1 + ca
c− a+
1 + ca
c− a· 1 + ab
a− b= 1.
Furthermore, for any x, y, z ∈ R, we always have (x−y)2+(y−z)2+(z−x)2 ≥ 0,and (x+ y + z)2 ≥ 0. Hence x2 + y2 + z2 ≥ xy + yz + zx, and x2 + y2 + z2 ≥−2(xy + yz + zx). Therefore(
1 + ab
a− b
)2
+
(1 + bc
b− c
)2
+
(1 + ca
c− a
)2
≥∑cyc
1 + ab
a− b· 1 + bc
b− c= 1.
and (1− aba− b
)2
+
(1− bcb− c
)2
+
(1− cac− a
)2
≥ −2∑cyc
1− aba− b
· 1− bcb− c
= 2.
Thus
2(1 + a2b2)
(a− b)2+
2(1 + b2c2)
(b− c)2+
2(1 + c2a2)
(c− a)2=
∑cyc
(1 + ab)2 + (1− ab)2
(a− b)2
=∑cyc
(1 + ab
a− b
)2
+∑cyc
(1− aba− b
)2
≥ 3.
Example 3. Let a, b, c be positive real numbers such that abc = 1. Prove that
4
(a
(a+ 1)2+
b
(b+ 1)2+
c
(c+ 1)2
)≤ 1 +
16
(a+ 1)(b+ 1)(c+ 1).
Solution: Put x =1− a1 + a
, y =1− b1 + b
, z =1− c1 + c
, then we have
−1 ≤ x, y, z ≤ 1, a =1− x1 + x
, b =1− y1 + y
, c =1− z1 + z
.
It is easy to verify that (1 − x)(1 − y)(1 − z) = (1 + x)(1 + y)(1 + z), hencex+ y + z + xyz = 0. But
4a
(a+ 1)2=
4 · 1−x1+x(1 +
1− x1 + x
)2 = 1− x2, 2
a+ 1=
2
1 +1− x1 + x
= 1 + x.
Therefore, our inequality is equivalent to
1− x2 + 1− y2 + 1− z2 ≤ 1 + 2(1 + x)(1 + y)(1 + z),
orx2 + y2 + z2 + 2(xy + yz + zx) + 2(x+ y + z + xyz) ≥ 0.
179
or
(x+ y + z)2 ≥ 0.
Example 4. Let a, b, c be positive real numbers, then show that
a
b+ c+
b
c+ a+
c
a+ b+
4abc
(a+ b)(b+ c)(c+ a)≥ 2.
Solution: Put x =a
b+ c, y =
b
c+ a, z =
c
a+ b, then we have
1
x+ 1+
1
y + 1+
1
z + 1= 2, or xy+yz+zx+2xyz = 1. And we need to prove x+y+z+4xyz ≥ 2,
or x+y+z ≥ 2(xy+yz+zx). Put p = x+y+z, q = xy+yz+zx, r = xyz then wehave q+2r = 1 and we need to prove p ≥ 2q. If p ≥ 2, then it is trivial since p ≥2 ≥ 2q. If p ≤ 2, then it is easy to verify that 2 ≥ p ≥ 3
2. By Schur’s inequality
for third degree, we have x(x−y)(x−z)+y(y−z)(y−x)+z(z−x)(z−y) ≥ 0,
or r ≥ p(4q − p2)9
. Hence q +2p(4q − p2)
9≤ 1, which implies us q ≤ 2p3 + 9
8p+ 9.
Therefore p− 2q ≥ p− 2(2p3 + 9)
8p+ 9=
(4p2 − 9)(2− p)8p+ 9
≥ 0.
Remark. Since (a+b)(b+c)(c+a) ≥ 8abc, this inequality implies us a stronger
version of the famous Nesbitt inequality
a
b+ c+
b
c+ a+
c
a+ b≥ 3
2.
Exercises
1. Let a, b, c be positive real numbers such that abc = 1. Prove that
1
(1 + a)2+
1
(1 + b)2+
1
(1 + c)2+
2
(1 + a)(1 + b)(1 + c)≥ 1.
Solution: Using the substitution as Example 3, we need to prove
(1 + x)2 + (1 + y)2 + (1 + z)2 + (1 + x)(1 + y)(1 + z) ≥ 4,
or
x2 + y2 + z2 + x2y2z2 ≥ 4xyz.
By AM-GM inequality, we have
x2 + y2 + z2 + x2y2z2 ≥ 4 4√x4y4z4 = 4 |xyz| ≥ 4xyz.
2. If a, b, c are distinct real numbers, then
a2 + ab+ b2
(a− b)2+b2 + bc+ c2
(b− c)2+c2 + ca+ a2
(c− a)2≥ 9
4.
180
Solution: We have a2 + ab+ b2 =3
4(a+ b)2 +
1
4(a− b)2, hence
a2 + ab+ b2
(a− b)2+b2 + bc+ c2
(b− c)2+c2 + ca+ a2
(c− a)2=
3
4
[∑cyc
(a+ b
a− b
)2
+ 1
].
Moreover, it is easy to verify that(a+ b
a− b+ 1
)(b+ c
b− c+ 1
)(c+ a
c− a+ 1
)=
=
(a+ b
a− b− 1
)(b+ c
b− c− 1
)(c+ a
c− a− 1
),
Hencea+ b
a− b· b+ c
b− c+b+ c
b− c· c+ a
c− a+c+ a
c− a· a+ b
a− b= −1
Therefore(a+ b
a− b
)2
+
(b+ c
b− c
)2
+
(c+ a
c− a
)2
≥ −2∑cyc
a+ b
a− b· b+ c
b− c= 2.
This inequality implies
a2 + ab+ b2
(a− b)2+b2 + bc+ c2
(b− c)2+c2 + ca+ a2
(c− a)2≥ 3
4(2 + 1) =
9
4.
3. If a, b, c are distinct real numbers, then(a2 + b2 + c2
)( 1
(a− b)2+
1
(b− c)2+
1
(c− a)2
)≥ 9
2.
Solution: According to the proof above, we have(a+ b
a− b
)2
+
(b+ c
b− c
)2
+
(c+ a
c− a
)2
≥ 2.
Hence
2(a2 + b2)
(a− b)2+
2(b2 + c2)
(b− c)2+
2(c2 + a2)
(c− a)2=
∑cyc
(a+ b)2 + (a− b)2
(a− b)2
=∑cyc
(a+ b
a− b
)2
+ 3 ≥ 5.
Also, we have (a
b− c+ 1
)(b
c− a+ 1
)(c
a− b+ 1
)=
=
(a
b− c− 1
)(b
c− a− 1
)(c
a− b− 1
),
Hencea
b− c· b
c− a+
b
c− a· c
a− b+
c
a− b· a
b− c= −1.
181
Thus(a
b− c
)2
+
(b
c− a
)2
+
(c
a− b
)2
≥ −2∑cyc
a
b− c· b
c− a= 2.
Hence (a2 + b2 + c2
)( 1
(a− b)2+
1
(b− c)2+
1
(c− a)2
)=
=∑cyc
a2 + b2
(a− b)2+∑cyc
(a
b− c
)2
≥ 5
2+ 2 =
9
2.
4. Let a, b, c be positive real numbers, show that(a
b+ c
)2
+
(b
c+ a
)2
+
(c
a+ b
)2
+10abc
(a+ b)(b+ c)(c+ a)≥ 2.
Solution: By the same substitution as Example 4, we need to prove
x2 + y2 + z2 + 10xyz ≥ 2.
Put p = x + y + z, q = xy + yz + zx, r = xyz then we have q + 2r = 1and we need to prove p2 − 2q + 10r ≥ 2, or p2 − 7q + 3 ≥ 0. If p ≥ 2,then it is trivial since p2 + 3 ≥ 7 ≥ 7q. If 2 ≥ p, then we can easily check
that 2 ≥ p ≥ 3
2. And from the proof of Example 4, we get q ≤ 2p3 + 9
8p+ 9.
Hence
p2 − 7q + 3 ≥ p2 + 3− 7(2p3 + 9)
8p+ 9=
3(2p− 3)(4− p2)8p+ 9
≥ 0.
5. Let a, b, c be distinct real numbers. Prove that
(a− 2b)2 + (a− 2c)2
(b− c)2+
(b− 2c)2 + (b− 2a)2
(c− a)2+
(c− 2a)2 + (c− 2b)2
(a− b)2≥ 22.
Solution: We have
(a− 2b)2 + (a− 2c)2
(b− c)2− 2 = 2
(b+ c− ab− c
)2
,
Hence, our inequality is equivalent to(b+ c− ab− c
)2
+
(c+ a− bc− a
)2
+
(a+ b− ca− b
)2
≥ 8.
We have (b+ c− ab− c
− 2
)(c+ a− bc− a
− 2
)(a+ b− ca− b
− 2
)=
=(3c− a− b)(3a− b− c)(3b− c− a)
(a− b)(b− c)(c− a)
=3b− c− ab− c
· 3c− a− bc− a
· 3a− b− ca− b
=
(b+ c− ab− c
+ 2
)(c+ a− bc− a
+ 2
)(a+ b− ca− b
+ 2
).
182
Hence
b+ c− ab− c
· c+ a− bc− a
+c+ a− bc− a
· a+ b− ca− b
+a+ b− ca− b
· b+ c− ab− c
= −4.
Thus, our inequality is equivalent to x2 + y2 + z2 ≥ 8, or x2 + y2 + z2 ≥−2(xy + yz + zx), which is just simply to (x+ y + z)2 ≥ 0.
6. Let a, b, c be distinct real numbers. Prove that
(1− a2)(1− b2)(a− b)2
+(1− b2)(1− c2)
(b− c)2+
(1− c2)(1− a2)(c− a)2
≥ −1.
Solution: Since
(1− a2)(1− b2)(a− b)2
+ 1 =
(1− aba− b
)2
,
Our inequality is equivalent to(1− aba− b
)2
+
(1− bcb− c
)2
+
(1− cac− a
)2
≥ 2.
This inequality has been already proved in Example 2.
7. Let a, b, c be positive real numbers such that abc = 1. Prove that
a+ 3
(a+ 1)2+
b+ 3
(b+ 1)2+
c+ 3
(c+ 1)2≥ 3.
Solution: Using the substitution as Example 3, we have
a+ 3
(a+ 1)2=
1− x1 + x
+ 3(1− x1 + x
+ 1
)2 =1
2(x2 + 3x+ 2).
Hence, our inequality is equivalent to x2 + y2 + z2 + 3(x+ y+ z) ≥ 0, orx2 + y2 + z2 ≥ 3xyz. By AM-GM inequality, we have
x2 + y2 + z2 ≥ 3 3√x2y2z2 ≥ 3 |xyz| (since |xyz| ≤ 1)
≥ 3xyz.
8. Let a, b, c be real numbers such that a, b, c 6= 1 and abc = 1. Prove that
a2
(a− 1)2+
b2
(b− 1)2+
c2
(c− 1)2≥ 1.
Solution: By the same subsitution as Example 3, put
a
a− 1=
1
1− 1
a
=1
1− 1 + x
1− x
=1
2
(1− 1
x
),
183
Hence, our inequality is equivalent to(1− 1
x
)2
+
(1− 1
y
)2
+
(1− 1
z
)2
≥ 4,
or1
x2+
1
y2+
1
z2− 2
(1
x+
1
y+
1
z
)− 1 ≥ 0.
Since x + y + z + xyz = 0 and x, y, z 6= 0 (since a, b, c 6= 1), we have1
xy+
1
yz+
1
zx= −1. Hence
1
x2+
1
y2+
1
z2=
(1
x+
1
y+
1
z
)2
− 2
(1
xy+
1
yz+
1
zx
)=
(1
x+
1
y+
1
z
)2
+ 2.
It suffices to show that(1
x+
1
y+
1
z
)2
− 2
(1
x+
1
y+
1
z
)+ 1 ≥ 0.
which is just simply to (1
x+
1
y+
1
z− 1
)2
≥ 0.
184
Chapter 2
Problems
83.1. Let a, b, c be the sidelengths of a triangle. Prove that
a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.
(IMO 1983)83.2. Show that for any positive reals a, b, c, d, e, f , we have
ab
a+ b+
cd
c+ d+
ef
e+ f≤ (a+ c+ e)(b+ d+ f)
a+ b+ c+ d+ e+ f.
(United Kingdom 1983)84.1. Let a1, a2, . . . , an > 0, n ≥ 2. Prove that
a21a2
+a22a3
+ · · ·+a2n−1an
+a2na1≥ a1 + a2 + · · ·+ an.
(China 1984)84.2. Let x, y, z be nonnegative real numbers such that x+ y + z = 1. Provethat
0 ≤ xy + yz + zx− 2xyz ≤ 7
27.
(IMO 1984)84.3. Prove that for any a, b > 0, we have that
(a+ b)2
2+a+ b
4≥ a√b+ b
√a.
(Russia 1984)85.1. Let x1, x2, . . . , xn be real numbers from the interval [0, 2]. Prove that
n∑i,j=1
|xi − xj | ≤ n2.
When do we have equality?(United Kingdom 1985)
86.1. Find the maximum value of the constant c such that for any x1, x2, . . . , xn >0 for which xk+1 ≥ x1 + x2 + · · ·+ xk for any k, the inequality
√x1 +
√x2 + · · ·+
√xn ≤ c
√x1 + x2 + · · ·+ xn
185
also holds for any n.(IMO Shorlist 1986)
86.2. Let n be a positive integer. Prove that
|sin 1|+ |sin 2|+ · · ·+ |sin 3n| > 8
5n.
(Russia 1986)86.3. Show that for all positive real numbers a1, a2, . . . , an, we have
1
a1+
2
a1 + a2+ · · ·+ n
a1 + a2 + · · ·+ an≤ 4
(1
a1+
1
a2+ · · ·+ 1
an
).
(Russia 1986)86.4. Find the maximum value of
x2y + y2z + z2x
for reals x, y, z with x+ y + z = 0 and x2 + y2 + z2 = 6.(United Kingdom 1986)
87.1. Prove that if x, y, z are real numbers such that x2 + y2 + z2 = 2, then
x+ y + z ≤ xyz + 2.
(IMO Shorlist 1987)88.1. Show that
(1 + x)n ≥ (1− x)n + 2nx(1− x2)n−12
for all 0 ≤ x ≤ 1 and all positive integer n.(Ireland 1988)
88.2. Given a triangle ABC. Prove that
2
(sinA
A+
sinB
B+
sinC
C
)≤(
1
B+
1
C
)sinA+
(1
C+
1
A
)sinB+
(1
A+
1
B
)sinC.
(Russia 1988)89.1. Let x1, x2, . . . , xn be positive real numbers, and let S = x1+x2+· · ·+xn.Prove that
(1 + x1)(1 + x2) · · · (1 + xn) ≤ 1 + S +S2
2!+ · · ·+ Sn
n!.
(APMO 1989)
89.2. Let x, y, z be real numbers such that 0 < x < y < z <π
2. Prove that
π
2+ 2 sinx cos y + 2 sin y cos z > sin 2x+ sin 2y + sin 2z.
(Iberoamerica 1989)89.3. Let a, b, c be the sidelengths of a triangle. Prove that∣∣∣∣a− ba+ b
+b− cb+ c
+c− ac+ a
∣∣∣∣ < 1
16.
186
(Iberoamerica 1989)89.4. Find the least possible value of
(x+ y)(y + z)
for positive reals x, y, z satisfying xyz(x+ y + z) = 1.(Russia 1989)
90.1. Let a, b, c, d be positive real numbers such that ab + bc + cd + da = 1.Prove that
a3
b+ c+ d+
b3
c+ d+ a+
c3
d+ a+ b+
d3
a+ b+ c≥ 1
3.
(IMO Shortlist 1990)90.2. Show that
x4 > x− 1
2
for all real x.(Russia 1990)
90.3. Let x1, x2, . . . , xn be positive reals with sum 1. Show that
x21x1 + x2
+x22
x2 + x3+ · · ·+ x2n
xn + x1≥ 1
2.
(Russia 1990)90.4. Show that√
x2 − xy + y2 +√y2 − yz + z2 ≥
√z2 + zx+ x2
for any positive real numbers x, y, z.(United Kingdom 1990)
91.1. Let a1, a2, . . . , an and b1, b2, . . . , bn be positive real numbers such thata1 + a2 + · · ·+ an = b1 + b2 + · · ·+ bn. Prove that
a21a1 + b1
+a22
a2 + b2+ · · ·+ a2n
an + bn≥ a1 + a2 + · · ·+ an
2.
(APMO 1991)91.2. Given positive real numbers a, b, c satisfying a+ b+ c = 1, show that
(1 + a) (1 + b) (1 + c) ≥ 8 (1− a) (1− b) (1− c) .
(Russia 1991)91.3. Show that
(x+ y + z)2
3≥ x√yz + y
√zx+ z
√xy
for all nonnegative reals x, y, z.(Russia 1991)
91.4. The real numbers x1, x2, . . . , x1991 satisfy
|x1 − x2|+ |x2 − x3|+ · · ·+ |x1990 − x1991| = 1991.
187
Denote sn =x1 + x2 + . . .+ xn
n. What is the maximum possible value of
|s1 − s2|+ |s2 − s3|+ . . .+ |s1990 − s1991|?
(Russia 1991)91.5. A triangle has sides a, b, c with sum 2. Show that
a2 + b2 + c2 + 2abc < 2.
(United Kingdom 1991)91.6. Prove the inequality
x2y
z+y2z
x+z2x
y≥ x2 + y2 + z2
for any positive real numbers x, y, z with x ≥ y ≥ z.(Vietnam 1991)
92.1. For every integer n ≥ 2, find the smallest positive number λ = λ(n)
such that if 0 ≤ a1, a2, . . . , an ≤1
2, b1, b2, . . . , bn > 0, and a1 + a2 + · · ·+ an =
b1 + b2 + · · ·+ bn = 1, then
b1b2 · · · bn ≤ λ(a1b1 + a2b2 + · · ·+ anbn).
(China 1992)92.2. Show that
x4 + y4 + z2 ≥ 2√
2xyz
for all positive reals x, y, z.(Russia 1992)
92.3. Show that for any real numbers x, y > 1, we have
x2
y − 1+
y2
x− 1≥ 8.
(Russia 1992)92.4. Positive real numbers a, b, c satisfy a ≥ b ≥ c. Prove that
a2 − b2
c+c2 − b2
a+a2 − c2
b≥ 3a− 4b+ c.
(Ukraine 1992)92.5. Let x, y, z, w be positive real numbers. Prove that
12
x+ y + z + w≤∑sym
1
x+ y≤ 3
4
(1
x+
1
y+
1
z+
1
w
).
(United Kingdom 1992)93.1. Let a, b, c, d be positive real numbers. Prove that
a
b+ 2c+ 3d+
b
c+ 2d+ 3a+
c
d+ a+ 3b+
d
a+ 2b+ 3c≥ 2
3.
(IMO Shortlist 1993)
188
93.2. Let a, b, c ∈ [0, 1]. Prove that
a2 + b2 + c2 ≤ a2b+ b2c+ c2a+ 1.
(Italia 1993)93.3. Let x, y, u, v be positive real numbers. Prove that
xy + xv + yu+ uv
x+ y + u+ v≥ xy
x+ y+
uv
u+ v.
(Poland 1993)93.4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root,then a2 + b2 ≥ 8.
(Tournament of the Towns 1993)93.5. Let x1, x2, x3, x4 be real numbers such that
1
2≤ x21 + x22 + x23 + x24 ≤ 1.
Find the largest and the smallest values of the expression
A = (x1 − 2x2 + x3)2 + (x2 − 2x3 + x4)
2 + (x2 − 2x1)2 + (x3 − 2x4)
2.
(Vietnam 1993)94.1. Let a1, a2, . . . , an be a sequence of positive real numbers satisfyinga1 + · · ·+ ak ≥
√k for all k = 1, 2, . . . , n. Prove that
a21 + a22 + · · ·+ a2n >1
4
(1 +
1
2+ · · ·+ 1
n
).
(USA 1994)95.1. Prove that for any positive real numbers x, y and any positive integersm,n,
(m−1)(n−1)(xm+n + ym+n
)+(m+ n− 1) (xmyn + xnym) ≥ mn
(xm+n−1y + xym+n−1) .
(Austrian-Polish Competition 1995)95.2. Let a, b, c, d be positive real numbers. Prove that
a+ c
a+ b+b+ d
b+ c+c+ a
c+ d+d+ b
d+ a≥ 4.
(Baltic Way 1995)95.3. Let x, y, z be positive real numbers. Prove that
xxyyzz ≥ (xyz)x+y+z
3 .
(Canada 1995)95.4. If a, b, c are positive real numbers such that abc = 1, then
1
a3 (b+ c)+
1
b3 (c+ a)+
1
c3 (a+ b)≥ 3
2.
(IMO 1995)
189
95.5. Suppose that x1, x2, . . . , xn are real numbers satisfying |xi − xi+1| < 1and xi ≥ 1 for i = 1, 2, . . . , n (xn+1 = x1). Prove the following inequality
x1x2
+x2x3
+ · · ·+ xnx1
< 2n− 1.
(India 1995)95.6. (a) Find the maximum value of the expression x2y−y2x when 0 ≤ x ≤ 1and 0 ≤ y ≤ 1.(b) Find the maximum value of the expression x2y+y2z+z2x−y2x−z2y−x2zwhen 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1.
(United Kingdom 1995)95.7. Let a, b, c be real numbers satisfying a < b < c, a + b + c = 6 andab+ bc+ ca = 9. Prove that
0 < a < 1 < b < 3 < c < 4.
(United Kingdom 1995)95.8. Find the greatest constant k such that for any positive integer n whichis not a square, ∣∣(1 +
√n)
sin(π√n)∣∣ > k.
(Vietnam (IMO training camp) 1995)96.1. Let m and n be positive integers such that n ≤ m. Prove that
2n · n! ≤ (m+ n)!
(m− n)!≤ (m2 +m)n.
(APMO 1996)96.2. Let a, b, c be the lengths of the sides of a triangle. Prove that
√a+ b− c+
√b+ c− a+
√c+ a− b ≤
√a+√b+√c,
and determine when equality occurs.(APMO 1996)
96.3. The real numbers x, y, z, t satisfy the equalities x + y + z + t = 0 andx2 + y2 + z2 + t2 = 1. Prove that
−1 ≤ xy + yz + zt+ tx ≤ 0.
(Austrian-Polish Competition 1996)96.4. Let a, b, c be positive real numbers, such that a+ b+ c =
√abc. Prove
thatab+ bc+ ca ≥ 9(a+ b+ c).
(Belarus 1996)96.5. Suppose n ∈ N, x0 = 0, xi > 0 for i = 1, 2, . . . , n, and x1+x2+· · ·+xn =1. Prove that
1 ≤n∑i=1
xi√1 + x0 + · · ·+ xi−1 ·
√xi + · · ·+ xn
<π
2.
(China 1996)
190
96.6. (a) Find the minimum value of xx for x a positive real number.(b) If x and y are positive real numbers, show that xy + yx > 1.
(France 1996)96.7. Let a, b, c be positive real numbers such that abc = 1. Prove that
ab
a5 + b5 + ab+
bc
b5 + c5 + bc+
ca
c5 + a5 + ca≤ 1.
(IMO Shortlist 1996)96.8. Let a and b be positive real numbers with a+ b = 1. Prove that
a2
a+ 1+
b2
b+ 1≥ 1
3.
(Hungary 1996)96.9. Prove the following inequality for positive real numbers x, y, z,
(xy + yz + zx)
[1
(y + z)2+
1
(z + x)2+
1
(x+ y)2
]≥ 9
4.
(Iran 1996)96.10. Let n ≥ 2 be a fixed natural number and let a1, a2, . . . , an be positivenumbers whose sum is 1. Prove that for any positive numbers x1, x2, . . . , xnwhose sum is 1,
2∑
1≤i<j≤nxixj ≤
n− 2
n− 1+
n∑i=1
aix2i
1− ai,
and determine when equality holds.(Poland 1996)
96.11. Let a, b, c be real numbers such that a+ b+ c = 1. Prove that
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 9
10.
(Poland 1996)96.12. Let x1, x2, . . . , xn, xn+1 be positive reals such that x1 +x2 + · · ·+xn =xn+1. Prove that
n∑i=1
√xi(xn+1 − xi) ≤
√√√√ n∑i=1
xn+1(xn+1 − xi).
(Romania 1996)96.13. Let x, y, z be real numbers. Prove that the following conditions areequivalent
(i) x > 0, y > 0, z > 0 and1
x+
1
y+
1
z≤ 1;
(ii) for every quadrilateral with sides a, b, c, d, a2x+ b2y + c2z > d2.(Romania 1996)
96.14. Let a, b, c be positive real numbers.(a) Prove that 4(a3 + b3) ≥ (a+ b)3.(b) Prove that 9(a3 + b3 + c3) ≥ (a+ b+ c)3.
191
(United Kingdom 1996)96.15. Let a, b, c, d be positive real numbers such that
2(ab+ ac+ ad+ bc+ bd+ cd) + abc+ bcd+ cda+ dab = 16.
Prove that
a+ b+ c+ d ≥ 2
3(ab+ ac+ ad+ bc+ bd+ cd).
(Vietnam 1996)96.16. Prove that for any real numbers a, b, c,
(a+ b)4 + (b+ c)4 + (c+ a)4 ≥ 4
7(a4 + b4 + c4).
(Vietnam 1996)97.1. a, b, c be positive numbers such that abc = 1. Prove that
1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a≤ 1
a+ 2+
1
b+ 2+
1
c+ 2.
(Bulgaria 1997)97.2. Prove that
1
1999<
1
2· 3
4· · · 1997
1998<
1
44.
(Canada 1997)97.3. Let x1, x2, . . . , x1997 be real numbers satisfying the following conditions
(a) − 1√3≤ xi ≤
√3 for i = 1, 2, . . . , 1997;
(b) x1 + x2 + · · ·+ x1997 = −318√
3.Determine the maximum value of x121 + x122 + · · ·+ x121997.
(China 1997)97.4. For each natural number n ≥ 2, determine the largest possible value ofthe expression
Vn = sinx1 cosx2 + sinx2 cosx3 + · · ·+ sinxn cosx1,
where x1, x2, . . . , xn are arbitrary real numbers.(Czech-Slovak 1997)
97.5. Let x, y, z be positive real numbers. Prove that
xyz(x+ y + z +
√x2 + y2 + z2
)(x2 + y2 + z2) (xy + yz + zx)
≤ 3 +√
3
9.
(Hong Kong 1997)97.6. Given x1, x2, x3, x4 are positive real numbers such that x1x2x3x4 = 1.Prove that
x31 + x32 + x33 + x34 ≥ max
{x1 + x2 + x3 + x4,
1
x1+
1
x2+
1
x3+
1
x4
}.
(Iran 1997)
192
97.7. Let a, b, c be nonnegative real numbers such that a+ b+ c ≥ abc. Provethat
a2 + b2 + c2 ≥ abc.
(Ireland 1997)97.8. Let a, b, c be positive real numbers. Prove that
(b+ c− a)2
(b+ c)2 + a2+
(c+ a− b)2
(c+ a)2 + b2+
(a+ b− c)2
(a+ b)2 + c2≥ 3
5.
(Japan 1997)97.9. Let a1, . . . , an be positive numbers, and define
A =a1 + · · ·+ an
n, G = n
√a1 · · · an, H =
n
a−11 + · · ·+ a−1n.
(a) If n is even, show thatA
H≤ −1 + 2
(A
G
)n.
(b) If n is odd, show thatA
H≤ −n− 2
n+
2(n− 1)
n
(A
G
)n.
(Korea 1997)97.10. For any positive real numbers x, y, z such that xyz = 1, prove theinequality
x9 + y9
x6 + x3y3 + y6+
y9 + z9
y6 + y3z3 + z6+
z9 + x9
z6 + z3z3 + x6≥ 2.
(Romania 1997)97.11. Let a, b, c be positive real numbers. Prove that
a2
a2 + 2bc+
b2
b2 + 2ca+
c2
c2 + 2ab≥ 1 ≥ bc
a2 + 2bc+
ca
b2 + 2ca+
ab
c2 + 2ab.
(Romania 1997)97.12. Show that if 1 < a < b < c, then
loga(loga b) + logb(logb c) + logc(logc a) > 0.
(Russia 1997)97.13. Prove that for x, y, z ≥ 2,
(y3 + x)(z3 + y)(x3 + z) ≥ 125xyz.
(Saint Petersburg 1997)97.14. Given an integer n ≥ 2, find the minimal value of
x51x2 + x3 + · · ·+ xn
+x52
x3 + · · ·+ xn + x1+ · · ·+ x5n
x1 + x2 + · · ·+ xn−1,
for positive real numbers x1, . . . , xn subject to the condition x21 + · · ·+x2n = 1.(Turkey 1997)
97.15. Let x, y and z be positive real numbers.
193
(a) If x+ y + z ≥ 3, is it necessarily true that1
x+
1
y+
1
z≤ 3?
(b) If x+ y + z ≤ 3, is it necessarily true that1
x+
1
y+
1
z≥ 3?
(United Kingdom 1997)
97.16. Prove that, for all positive real numbers a, b, c,
1
a3 + b3 + abc+
1
b3 + c3 + abc+
1
c3 + a3 + abc≤ 1
abc.
(USA 1997)
98.1. Let a, b, c be positive real numbers. Prove that(1 +
a
b
)(1 +
b
c
)(1 +
c
a
)≥ 2
(1 +
a+ b+ c3√abc
).
(APMO 1998)
98.2. Let x1, x2, y1, y2 be real numbers such that x21 + x22 ≤ 1. Prove theinequality
(x1y1 + x2y2 − 1)2 ≥ (x21 + x22 − 1)(y21 + y22 − 1).
(Austrian-Polish Competition 1998)
98.3. If n ≥ 2 is an integer and 0 < a1 < a2 < . . . < a2n+1 are real numbers,prove the inequality
n√a1− n√a2+ n√a3−· · ·− n
√a2n+ n
√a2n+1 <
n√a1 − a2 + a3 − . . .− a2n + a2n+1.
(Balkan 1998)
98.4. Let a, b, c be positive real numbers. Prove that
a
b+b
c+c
a≥ a+ b
b+ c+b+ c
a+ b+ 1.
(Belarus 1998)
98.5. Let n be a natural number such that n ≥ 2. Prove that
1
n+ 1
(1 +
1
3+ . . .+
1
2n− 1
)>
1
n
(1
2+
1
4+ . . .+
1
2n
).
(Canada 1998)
98.6. Let n ≥ 2 be a positive integer and let x1, x2, . . . , xn be real numberssuch that
n∑i=1
x2i +n−1∑i=1
xixi+1 = 1.
For every positive integer k, 1 ≤ k ≤ n, determine the maximum value of |xk| .(China 1998)
98.7. Let a, b, c ≥ 1. Prove that
√a− 1 +
√b− 1 +
√c− 1 ≤
√c (ab+ 1).
(Hong Kong 1998)
194
98.8. Let a1, a2, . . . , an > 0 such that a1 + a2 + · · ·+ an < 1. Prove that
a1a2 · · · an (1− a1 − a2 − · · · − an)
(a1 + a2 + · · ·+ an) (1− a1) (1− a2) · · · (1− an)≤ 1
nn+1.
(IMO Shortlist 1998)98.9. Let a, b, c be positive real numbers such that abc = 1. Prove that
a3
(1 + b) (1 + c)+
b3
(1 + a) (1 + c)+
c3
(1 + a) (1 + b)≥ 3
4.
(IMO Shortlist 1998)
98.10. Let x, y, z > 1 such that1
x+
1
y+
1
z= 2. Prove that
√x+ y + z ≥
√x− 1 +
√y − 1 +
√z − 1.
(Iran 1998)98.11. Suppose that a1 < a2 < · · · < an are real numbers. Prove that
a1a42 + a2a
43 + · · ·+ an−1a
4n + ana
41 ≥ a2a41 + a3a
42 + · · ·+ ana
4n−1 + a1a
4n.
(Iran 1998)98.12. Show that if x is a nonzero real number, then
x8 − x5 − 1
x+
1
x4≥ 0.
(Ireland 1998)98.13. Prove that if a, b, c are positive real numbers, then
9
a+ b+ c≤ 2
(1
a+ b+
1
b+ c+
1
c+ a
),
and1
a+ b+
1
b+ c+
1
c+ a≤ 1
2
(1
a+
1
b+
1
c
).
(Ireland 1998)98.14. If x, y, z are positive real numbers such that x+ y + z = xyz, then
1√1 + x2
+1√
1 + y2+
1√1 + z2
≤ 3
2.
(Korea 1998)98.15. Let a, b, c, d, e, f be positive real numbers such that
a+ b+ c+ d+ e+ f = 1, ace+ bdf ≥ 1
108.
Prove that
abc+ bcd+ cde+ def + efa+ fab ≤ 1
36.
(Poland 1998)
195
98.16. Let a be a positive real numbers and let x1, x2, . . . , xn be positive realnumbers such that x1 + x2 + · · ·+ xn = 1. Prove that
ax1−x2
x1 + x2+
ax2−x3
x2 + x3+ · · ·+ axn−x1
xn + x1≥ n2
2.
(Serbia 1998)98.17. Find the minimum of the expression
F (x, y) =√
(x+ 1)2 + (y − 1)2+√
(x− 1)2 + (y + 1)2+√
(x+ 2)2 + (y + 2)2,
where x, y are real numbers.(Vietnam 1998)
98.18. Let n ≥ 2 and x1, x2, . . . , xn be positive real numbers satisfying
1
x1 + 1998+
1
x2 + 1998+ · · ·+ 1
xn + 1998=
1
1998.
Prove thatn√x1x2 · · ·xnn− 1
≥ 1998.
(Vietnam 1998)99.1. Let {an} be a sequence of real numbers such that ai+j ≤ ai + aj for alli, j. Prove that the following inequality holds
a1 +a22
+ · · ·+ ann≥ an.
(APMO 1999)99.2. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Provethat
1
1 + ab+
1
1 + bc+
1
1 + ca≥ 3
2.
(Belarus 1999)99.3. For any nonnegative real numbers x, y, z such that x+ y+ z = 1, provethe following inequality
x2y + y2z + z2x ≤ 4
27.
(Canada 1999)99.4. Let a, b, c be positive real numbers. Prove that
a
b+ 2c+
b
c+ 2a+
c
a+ 2b≥ 1.
(Czech-Slovak 1999)99.5. Let n ≥ 2 be a fixed integer. Find the least constant C such that theinequality ∑
1≤i<j≤nxixj(x
2i + x2j ) ≤ C(x1 + x2 + · · ·+ xn)4
holds for every x1, . . . , xn ≥ 0. For this constant C, characterize the instancesof equality.
196
(IMO 1999)99.6. For real numbers x1, x2, . . . , x6 ∈ [0, 1], prove the inequality
x31x52 + x53 + x54 + x55 + x56 + 5
+ · · ·+ x36x51 + x52 + x53 + x54 + x55 + 5
≤ 3
5.
(Ukraine 1999)99.7. Nonnegative real numbers p, q and r satisfy p+ q + r = 1. Prove that
7(pq + qr + rp) ≤ 2 + 9pqr.
(United Kingdom 1999)99.8. Let n > 3 and a1, a2, . . . , an be real numbers such that a1+a2+· · ·+an ≥n and a21 + a22 + · · ·+ a2n ≥ n2. Prove that max {a1, a2, . . . , an} ≥ 2.
(USA 1999)
99.9. Let a0, a1, . . . , an be numbers from the interval(
0,π
2
)such that
tan(a0 −
π
4
)+ tan
(a1 −
π
4
)+ · · ·+ tan
(an −
π
4
)≥ n− 1.
Prove thattan a0 tan a1 · · · tan an ≥ nn+1.
(USA 1999)99.10. Let x, y, z > 1. Prove that
xx2+2yzyy
2+2zxzz2+2xy ≥ (xyz)xy+yz+zx .
(USA (Shortlist) 1999)00.1. Let a, b be real numbers and a 6= 0. Prove that
a2 + b2 +1
a2+b
a≥√
3.
(Austria 2000)00.2. For all real numbers a, b, c ≥ 0 such that a+ b+ c = 1, prove that
2 ≤ (1− a2)2 + (1− b2)2 + (1− c2)2 ≤ (1 + a)(1 + b)(1 + c).
(Austrian-Polish Competition 2000)00.3. Let a, b, c, x, y, z be positive real numbers. Prove that
a3
x+b3
y+c3
z≥ (a+ b+ c)3
3 (x+ y + z).
(Belarus 2000)00.4. Suppose that the real numbers a1, a2, . . . , a100 satisfy(i) a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0;(ii) a1 + a2 ≤ 100;(iii) a3 + a4 + · · ·+ a100 ≤ 100.Determine the maximum possible value of a21 + a22 + · · · + a2100, and find allpossible sequences a1, a2, . . . , a100 which achieve this maximum.
(Canada 2000)
197
00.5. Show that
3
√2 (a+ b)
(1
a+
1
b
)≥ 3
√a
b+
3
√b
a
for all positive real numbers a and b, and determine when the equality occurs.
(Czech-Slovak 2000)
00.6. Let a, b, c be positive real numbers such that abc = 1. Prove that
1 + ab2
c3+
1 + bc2
a3+
1 + ca2
b3≥ 18
a3 + b3 + c3.
(Hong Kong 2000)
00.7. Let x, y, and z denote positive real numbers, each less than 4. Prove
that at least one of the numbers1
x+
1
4− y,
1
y+
1
4− z, and
1
z+
1
4− xis greater
than or equal to 1.
(Hungary 2000)
00.8. Let a, b, c be positive real numbers such that abc = 1. Prove the in-equality (
a+1
b− 1
)(b+
1
c− 1
)(c+
1
a− 1
)≤ 1.
(IMO 2000)
00.9. Let x, y ≥ 0 with x+ y = 2. Prove that
x2y2(x2 + y2) ≤ 2.
(Ireland 2000)
00.10. The real numbers a, b, c, x, y, z satisfy a ≥ b ≥ c > 0 and x ≥ y ≥ z >0. Prove that
a2x2
(by + cz)(bz + cy)+
b2y2
(cz + ax)(cx+ az)+
c2z2
(ax+ by)(ay + bx)≥ 3
4.
(Korea 2000)
00.11. Let x, y, z be real numbers. Prove that
x2 + y2 + z2 ≥√
2 (xy + yz) .
(Macedonia 2000)
00.12. Let a, b, x, y, z be positive real numbers. Prove that
x
ay + bz+
y
az + bx+
z
ax+ by≥ 3
a+ b.
(MOSP 2000)
00.13. Let ABC be an acute triangle. Prove that(cosA
cosB
)2
+
(cosB
cosC
)2
+
(cosC
cosA
)2
+ 8 cosA cosB cosC ≥ 4.
(MOSP 2000)
198
00.14. Let a, b, c be positive real numbers such that min {a, b} ≥ c. Provethat √
c(a− c) +√c(b− c) ≤
√ab.
(MOSP 2000)00.15. Let a1, a2, . . . , an be nonnegative real numbers. Prove that
a1 + a2 + · · ·+ ann
− n√a1a2 · · · an ≥
≥ 1
2min
{(√a1 −
√a2)
2 , (√a2 −
√a3)
2 , . . . , (√an −
√a1)
2}.
(MOSP 2000)00.16. Let (an) be an infinite sequence of positive numbers such that
a11
+a42
+ · · ·+ ak2
k≤ 1
for all k. Prove thata11
+a22
+ · · ·+ ann< 2,
for all n.(MOSP 2000)
00.17. Let a, b, c be nonnegative real numbers such that ab + bc + ca = 1.Prove that
1
b+ c+
1
c+ a+
1
a+ b≥ 5
2.
(MOSP 2000)00.18. Let a1, a2, . . . , an be positive real numbers such that
1
a1+
1
a2+ · · ·+ 1
an≤ 1.
Prove that for any positive integer k,
(ak1 − 1)(ak2 − 1) · · · (akn − 1) ≥ (nk − 1)n.
(MOSP 2000)00.19. Let a, b, c be positive real numbers. Prove that
1
a+ b+
1
b+ c+
1
c+ a+
1
2 3√abc≥
(a+ b+ c+ 3
√abc)2
(a+ b)(b+ c)(c+ a).
(MOSP 2000)00.20. For any integer n ≥ 2, consider n−1 positive real numbers a1, a2, . . . , an−1having sum 1, and n real numbers b1, b2, . . . , bn. Prove that
b21 +b22a1
+b23a2
+ . . .+b2nan−1
≥ 2b1(b2 + b3 + . . .+ bn).
(Romania 2000)00.21. Positive real numbers x, y, z satisfy xyz = 1. Prove that the followinginequality holds
x2 + y2 + z2 + x+ y + z ≥ 2 (xy + yz + zx) .
199
(Russia 2000)00.22. Let x1, x2, . . . , xn be real numbers (n ≥ 2), satisfying the conditions−1 < x1 < x2 < · · · < xn < 1 and
x131 + x132 + · · ·+ x13n = x1 + x2 + · · ·+ xn.
Prove that
x131 y1 + x132 y2 + · · ·+ x13n yn < x1y1 + x2y2 + · · ·+ xnyn
for any real numbers y1 < y2 < · · · < yn.(Russia 2000)
00.23. Show that for all n ∈ N and x ∈ R,
sinn 2x+ (sinn x− cosn x)2 ≤ 1.
(Russia 2000)00.24. Let n ≥ 3 be a positive integer. Prove that for all positive real numbersa1, a2, . . . , an, we have
a1 + a22
· a2 + a32
· · · an + a12
≤ a1 + a2 + a3
2√
2· · · an + a1 + a2
2√
2.
(Saint Petersburg 2000)00.25. Let n ≥ 3 be an integer. Prove that for positive numbers x1 ≤ x2 ≤· · · ≤ xn,
xnx1x2
+x1x2x3
+ · · ·+ xn−1xnx1
≥ x1 + x2 + · · ·+ xn.
(Saint Petersburg 2000)
00.26. Let a, b, c, d be positive real numbers such that c2 + d2 =(a2 + b2
)3.
Prove thata3
c+b3
d≥ 1.
(Singapore 2000)00.27. Given that x, y, z are positive real numbers satisfying xyz = 32, findthe minimum value of
x2 + 4xy + 4y2 + 2z2.
(United Kingdom 2000)00.28. Prove that for any nonnegative real numbers a, b, c, the followinginequality holds
a+ b+ c
3− 3√abc ≤ max
{(√a−√b)2,(√
b−√c)2,(√c−√a)2}
.
(USA 2000)01.1. Let a, b, c ≥ 0 such that a + b + c ≥ abc. Prove that the followinginequality holds
a2 + b2 + c2 ≥√
3abc.
(Balkan 2001)
200
01.2. Let x1, x2, x3 be real numbers in [−1, 1], and let y1, y2, y3 be real num-bers in [0, 1). Find the maximum possible value of the expression
1− x11− x2y3
· 1− x21− x3y1
· 1− x31− x1y2
.
(Belarus 2001)01.3. Let x and y be any two real numbers. Prove that
3(x+ y + 1)2 + 1 ≥ 3xy.
Under what conditions does equality hold?(Colombia 2001)
01.4. Let n (n ≥ 2) be an integer and let a1, a2, . . . , an be positive realnumbers. Prove the inequality
(a31 + 1)(a32 + 1) · · · (a3n + 1) ≥ (a21a2 + 1) · · · (a2na1 + 1).
(Czech-Slovak-Polish 2001)01.5. Prove that for any positive real numbers a, b, c, the following inequalityholds
a√a2 + 8bc
+b√
b2 + 8ca+
c√c2 + 8ab
≥ 1.
(IMO 2001)01.6. Let a, b, c be positive real numbers such that abc = 1. Prove that
ab+cbc+aca+b ≤ 1.
(India 2001)01.7. Let x, y, z be positive real numbers such that xyz ≥ xy+yz+zx. Provethat
xyz ≥ 3 (x+ y + z) .
(India 2001)01.8. Prove that for any real numbers a, b, c the following inequality holds
(b+ c− a)2(c+ a− b)2(a+ b− c)2 ≥ (b2 + c2 − a2)(c2 + a2 − b2)(a2 + b2 − c2).
(Japan 2001)01.9. Let a, b, c be the sidelengths of an acute-angled triangle. Prove that
(a+ b+ c)(a2 + b2 + c2)(a3 + b3 + c3) ≥ 4(a6 + b6 + c6).
(Japan 2001)01.10. Prove that for any real numbers x1, x2, . . . , xn, y1, y2, . . . , yn such thatx21 + x22 + · · ·+ x2n = y21 + y22 + · · ·+ y2n = 1,
(x1y2 − x2y1)2 ≤ 2
(1−
n∑k=1
xkyk
).
(Korea 2001)
201
01.11. Prove that if a, b, c are positive real numbers, then√a4 + b4 + c4 +
√a2b2 + b2c2 + c2a2 ≥
√a3b+ b3c+ c3a+
√ab3 + bc3 + ca3.
(Korea 2001)01.12. Prove that for all a, b, c > 0,√
(a2b+ b2c+ c2a) (ab2 + bc2 + ca2) ≥ abc+ 3√
(a3 + abc) (b3 + abc) (c3 + abc).
(Korea 2001)01.13. Prove that if a, b, c > 0 have product 1, then
(a+ b)(b+ c)(c+ a) ≥ 4(a+ b+ c− 1).
(MOSP 2001)01.14. Show that the inequality
n∑i=1
ixi ≤(n
2
)+
n∑i=1
xii
holds for every integer n ≥ 2 and all real numbers x1, x2, . . . , xN ≥ 0.(Poland 2001)
01.15. Let a and b be positive real numbers in the interval (0, 1]. Prove that
1√1 + a2
+1√
1 + b2≤ 2√
1 + ab.
(Russia 2001)01.16. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1.Prove that
ax+ by + cz + 2√
(xy + yz + zx)(ab+ bc+ ca) ≤ a+ b+ c.
(Ukraine 2001)01.17. Let a, b, c be positive real numbers such that a + b + c ≥ abc. Provethat at least two of the inequalities
2
a+
3
b+
6
c≥ 6,
2
b+
3
c+
6
a≥ 6,
2
c+
3
a+
6
b≥ 6
are true.(USA 2001)
01.18. Prove that for any nonnegative real numbers a, b, c such that a2 + b2 +c2 + abc = 4, we have
0 ≤ ab+ bc+ ca− abc ≤ 2.
(USA 2001)01.19. Let x, y, z be positive real numbers satisfying
(i)1√2≤ z ≤ 1
2min
{x√
2, y√
3}
;
(ii) x+ z√
3 ≥√
6;(iii) y
√3 + z
√10 ≥ 2
√5.
202
Find the maximum of P (x, y, z) =1
x2+
2
y2+
3
z2.
(Vietnam 2001)
01.20. Let x, y, z be positive real numbers such that
(i)2
5≤ z ≤ min {x, y} ;
(ii) xz ≥ 4
15;
(iii) yz ≥ 1
5.
Determine the maximum possible value of
P (x, y, z) =1
x+
2
y+
3
z.
(Vietnam 2001)
01.21. Find the minimum value of the expression1
a+
2
b+
3
cwhere a, b, c are
positive real numbers such that 21ab+ 2bc+ 8ca ≤ 12.
(Vietnam 2001)
02.1. Let x, y, z be positive real numbers such that
1
x+
1
y+
1
z= 1.
Prove that
√x+ yz +
√y + zx+
√z + xy ≥ √xyz +
√x+√y +√z.
(APMO 2002)
02.2. Let a, b, c be positive real numbers. Prove that
a3
b2+b3
c2+c3
a2≥ a2
b+b2
c+c2
a.
(Balkan (Shortslit) 2002)
02.3. If a, b, c are positive real numbers such that abc = 2, then
a3 + b3 + c3 ≥ a√b+ c+ b
√c+ a+ c
√a+ b.
(Balkan (Shortlist) 2002)
02.4. Let a, b, c be real numbers such that a2+b2+c2 = 1. Prove the inequality
a2
1 + 2bc+
b2
1 + 2ca+
c2
1 + 2ab≥ 3
5.
(Bosnia and Herzegovina 2002)
02.5. Show that for any positive real numbers a, b, c, we have
a3
bc+b3
ca+c3
ab≥ a+ b+ c.
(Canada 2002)
203
02.6. Assume (P1, P2, . . . , Pn) (n ≥ 2) is an arbitrary permutation of (1, 2, . . . , n).Prove that
1
P1 + P2+
1
P2 + P3+ . . .+
1
Pn−1 + Pn>n− 1
n+ 2.
(China 2002)02.7. Let x, y be positive real numbers such that x+ y = 2. Prove that
x3y3(x3 + y3) ≤ 2.
(India 2002)02.8. For any positive real numbers a, b, c, show that the following inequalityholds
a
b+b
c+c
a≥ c+ a
c+ b+a+ b
a+ c+b+ c
b+ a.
(India 2002)02.9. Let x1, x2, . . . , xn be positive real numbers. Prove that
x11 + x21
+x2
1 + x21 + x22+ · · ·+ xn
1 + x21 + x22 + · · ·+ x2n<√n.
(India 2002)02.10. Let a, b, c, d be the positive real numbers such that
1
1 + a4+
1
1 + b4+
1
1 + c4+
1
1 + d4= 1.
Prove that abcd ≥ 3.(Latvia 2002)
02.11. Prove that for any positive real numbers a, b, c, we have
a
2a+ b+
b
2b+ c+
c
2c+ a≤ 1.
(Moldova 2002)02.12. Positive numbers α, β, x1, x2, . . . , xn (n ≥ 1) satisfy the conditionx1 + x2 + · · ·+ xn = 1. Prove that
x31αx1 + βx2
+x32
αx2 + βx3+ · · ·+ x3n
αxn + βx1≥ 1
n(α+ β).
(Moldova 2002)02.13. Let a, b, c be positive real numbers. Prove that(
2a
b+ c
) 23
+
(2b
c+ a
) 23
+
(2c
a+ b
) 23
≥ 3.
(MOSP 2002)02.14. If a, b, c ∈ (0, 1), prove that
√abc+
√(1− a)(1− b)(1− c) < 1.
(Romania 2002)
204
02.15. Given positive real numbers a, b, c and x, y, z, for which a+x = b+y =c+ z = 1. Prove that
(abc+ xyz)
(1
ay+
1
bz+
1
cx
)≥ 3.
(Russia 2002)02.16. Let x, y, z be positive real numbers with sum 3. Prove that
√x+√y +√z ≥ xy + yz + zx.
(Russia 2002)02.17. Let a1, a2, . . . , an and b1, b2, . . . , bn be real numbers between 1001 and2002 inclusive. Suppose a21 + · · ·+ a2n = b21 + · · ·+ b2n. Prove that
n∑i=1
a3ibi≤ 17
10
n∑i=1
a2i .
Determine when equality holds.(Singapore 2002)
02.18. Let a, b, c, d be real numbers contained in the interval
(0,
1
2
). Prove
that
a4 + b4 + c4 + d4
abcd≥ (1− a)4 + (1− b)4 + (1− c)4 + (1− d)4
(1− a)(1− b)(1− c)(1− d).
(Taiwan 2002)02.19. Let x, y, z be positive real numbers such that x2 + y2 + z2 = 1. Provethat
x2yz + y2zx+ z2xy ≤ 1
3.
(United Kingdom 2002)02.20. Let a, b, c be real numbers satisfying a2 + b2 + c2 = 9. Prove theinequality
2(a+ b+ c)− abc ≤ 10.
(Vietnam 2002)03.1. If a, b, c > −1, then
1 + a2
1 + b+ c2+
1 + b2
1 + c+ a2+
1 + c2
1 + a+ b2≥ 2.
(Laurentiu Panaitopol, Balkan 2003)03.2. Prove that if a, b and c are positive real numbers with sum 3, then
a
b2 + 1+
b
c2 + 1+
c
a2 + 1≥ 3
2.
(Bulgaria 2003)03.3. Let a, b, c, d be positive real numbers such that ab + cd = 1 and letx1, x2, x3, x4, y1, y2, y3, y4 be real numbers such that x21 + y21 = x22 + y22 =x23 + y23 = x24 + y24 = 1. Prove that the following inequality holds
(ay1 + by2 + cy3 + dy4)2 + (ax4 + bx3 + cx2 + dx1)
2 ≤ 2
(a2 + b2
ab+c2 + d2
cd
).
205
(China 2003)03.4. Let a1, a2, . . . , a2n be real numbers such that
2n−1∑i=1
(ai − ai+1)2 = 1.
Determine the maximum value of
(an+1 + an+2 + · · ·+ a2n)− (a1 + a2 + . . .+ an).
(China 2003)
03.5. Suppose x be a real number in the interval
[3
2, 5
]. Prove that
2√x+ 1 +
√2x− 3 +
√15− 3x < 2
√19.
(China 2003)03.6. Let x1, x2, . . . , x5 be nonnegative real numbers such that
5∑i=1
1
1 + xi= 1.
Prove that5∑i=1
xix2i + 4
≤ 1.
(China 2003)03.7. Find the greatest real number k such that, for any positive a, b, c witha2 > bc,
(a2 − bc)2 > k(b2 − ca)(c2 − ab).
(Japan 2003)03.8. Prove that in any acute triangle ABC,
cot3A+ cot3B + cot3C + 6 cotA cotB cotC ≥ cotA+ cotB + cotC.
(MOSP 2003)03.9. Let ai be positive real numbers, for all i = 1, 2, . . . , n, satisfying
a1 + a2 + · · ·+ an =1
a1+
1
a2+ · · ·+ 1
an.
Prove that
1
n− 1 + a1+
1
n− 1 + a2+ · · ·+ 1
n− 1 + an≤ 1.
(Vasile Cirtoaje, MOSP 2003)03.10. Let a, b, c be nonnegative real numbers satisfying a2 + b2 + c2 = 1.Prove that
1 ≤ a
1 + bc+
b
1 + ca+
c
1 + ab≤√
2.
(Faruk Zejnulahi, MOSP 2003)
206
03.11. Let a, b, c be positive real numbers so that abc = 1. Prove that
1 +3
a+ b+ c≥ 6
ab+ bc+ ca.
(Romania 2003)03.12. Let a, b, c be positive real numbers. Prove that
(2a+ b+ c)2
2a2 + (b+ c)2+
(2b+ c+ a)2
2b2 + (c+ a)2+
(2c+ a+ b)2
2c2 + (a+ b)2≤ 8.
(USA 2003)
03.13. Prove that for any a, b, c ∈(
0,π
2
), the following inequality holds
∑ sin a · sin(a− b) · sin(a− c)sin(b+ c)
≥ 0.
(USA 2003)04.1. Prove that
(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab+ bc+ ca)
for any positive real numbers a, b, c.(APMO 2004)
04.2. Let x, y, z, t be positive real numbers such that xyzt = 1. Prove that
1
(1 + x)2+
1
(1 + y)2+
1
(1 + z)2+
1
(1 + t)2≥ 1.
(China 2004)04.3. If a, b, c are positive real numbers, then
1 <a√
a2 + b2+
b√b2 + c2
+c√
c2 + a2≤ 3√
2
2.
(China 2004)04.4. Let n be a positive integer with n greater than one, and let a1, a2, . . . , anbe positive integers such that a1 < a2 < · · · < an and
1
a1+
1
a2+ · · ·+ 1
an≤ 1.
Prove that, for any real number x, the following inequality holds(1
a21 + x2+
1
a22 + x2+ · · ·+ 1
a2n + x2
)2
≤ 1
2· 1
a1(a1 − 1) + x2.
(China 2004)04.5. Find all real numbers k such that the following inequality
a3 + b3 + c3 + d3 + 1 ≥ k(a+ b+ c+ d)
holds for all real numbers a, b, c, d ≥ −1.
207
(China 2004)04.6. Determine the maximum constant λ such that
x+ y + z ≥ λ,
where x, y, z are positive real numbers with x√yz + y
√zx+ z
√xy ≥ 1.(China 2004)
04.7. Let a, b, c be positive real numbers. Determine the minimal value of thefollowing expression
a+ 3c
a+ 2b+ c+
4b
a+ b+ 2c− 8c
a+ b+ 3c.
(China 2004)04.8. Determine the largest constant M such that the following inequalityholds for all real numbers x, y, z,
x4 + y4 + z4 + xyz (x+ y + z) ≥M (xy + yz + zx)2 .
(Hellenic 2004)04.9. Let n ≥ 3 be an integer and let t1, t2, . . . , tn be positive real numberssuch that
n2 + 1 > (t1 + t2 + · · ·+ tn)
(1
t1+
1
t2+ · · ·+ 1
tn
).
Show that ti, tj , tk are the side lengths of a triangle for all i, j, k with 1 ≤ i <j < k ≤ n.
(IMO 2004)04.10. For a, b, c positive reals, find the minimum value of
b2 + c2
a2 + bc+c2 + a2
b2 + ca+a2 + b2
c2 + ab.
(India 2004)
04.11. Let x1, x2, . . . , xn be real numbers in the interval
(0,
1
2
). Prove that
n∏i=1
xi(n∑i=1
xi
)n ≤n∏i=1
(1− xi)[n∑i=1
(1− xi)
]n .
(India 2004)04.12. Prove that for all positive real numbers a, b, the following inequalityholds √
2a(a+ b)3 + b√
2(a2 + b2) ≤ 3(a2 + b2).
(Ireland 2004)04.13. If a, b, c are positive reals such that a+ b+ c = 1, prove that
1 + a
1− a+
1 + b
1− b+
1 + c
1− c≤ 2
(b
a+c
b+a
c
).
208
(Japan 2004)04.14. Let a, b be real numbers in the interval [0, 1]. Prove that
a√2b2 + 5
+b√
2a2 + 5≤ 2√
7.
(Lithuania 2004)04.15. Prove that for any positive real numbers a, b, c,∣∣∣∣a3 − b3a+ b
+b3 − c3
b+ c+c3 − a3
c+ a
∣∣∣∣ ≤ (a− b)2 + (b− c)2 + (c− a)2
4.
(Moldova 2004)04.16. Prove that if n > 3 and x1, x2, . . . , xn > 0 have product 1, then
1
1 + x1 + x1x2+
1
1 + x2 + x2x3+ · · ·+ 1
1 + xn + xnx1> 1.
(Russia 2004) 04.17. (a) Given real numbers a, b, c with a+ b+ c = 0, provethat
a3 + b3 + c3 > 0 if and only if a5 + b5 + c5 > 0.
(b) Given real numbers a, b, c, d with a+ b+ c+ d = 0, prove that
a3 + b3 + c3 + d3 > 0 if and only if a5 + b5 + c5 + d5 > 0.
(United Kingdom 2004)04.18. Let a, b, c be positive real numbers. Prove that
(a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a+ b+ c)3.
(USA 2004)04.19. Let x, y, z > 0 such that (x+ y+ z)3 = 32xyz. Find the minimum and
maximum ofx4 + y4 + z4
(x+ y + z)4.
(Vietnam 2004)05.1. For any positive real numbers a, b, c such that abc = 8, then
a2√(a3 + 1) (b3 + 1)
+b2√
(b3 + 1) (c3 + 1)+
c2√(c3 + 1) (a3 + 1)
≥ 4
3.
(APMO 2005)05.2. Let a, b, c, d be positive real numbers. Show that
1
a3+
1
b3+
1
c3+
1
d3≥ a+ b+ c+ d
abcd.
(Austria 2005)05.3. Let a, b, c be positive real numbers. Prove that
a2
b+b2
c+c2
a≥ a+ b+ c+
4(a− b)2
a+ b+ c.
(Balkan 2005)
209
05.4. If a, b, c are positive real numbers such that abc = 1, then the inequalityholds
a
a2 + 2+
b
b2 + 2+
c
c2 + 2≤ 1.
(Baltic Way 2005)05.5. Let a, b, c be positive real numbers. Prove that(
a2 + b+3
4
)(b2 + a+
3
4
)≥(
2a+1
2
)(2b+
1
2
).
(Belarus 2005)05.6. Given positive real numbers a, b, c such that a+ b+ c = 1. Prove that
a√b+ b
√c+ c
√a ≤ 1√
3.
(Bosnia and Hercegovina 2005)05.7. Let x, y be positive real numbers such that x3 + y3 = x− y. Prove that
x2 + 4y2 < 1.
(China 2005)
05.8. Let a, b, c be positive real numbers such that ab + bc + ca =1
3. Prove
that1
a2 − bc+ 1+
1
b2 − ca+ 1+
1
c2 − ab+ 1≤ 3.
(China 2005)05.9. Given positive real numbers a, b, c such that a + b + c = 1. Prove thatthe following inequality holds
10(a3 + b3 + c3)− 9(a5 + b5 + c5) ≥ 1.
(China 2005)05.10. Let ABC be an acute triangle. Determine the least value of thefollowing expression
P =cos2A
cosA+ 1+
cos2B
cosB + 1+
cos2C
cosC + 1.
(China 2005)
05.11. Let a, b, c be positive real numbers such that1
a+
1
b+
1
c= 1. Prove
that
(a− 1) (b− 1) (c− 1) ≥ 8.
(Croatia 2005)05.12. Let a, b, c > 0 such that abc = 1. Prove that
a
(a+ 1) (b+ 1)+
a
(b+ 1) (c+ 1)+
a
(c+ 1) (a+ 1)≥ 3
4.
(Czech-Slovak 2005)
210
05.13. If a, b, c are three positive real numbers such that ab + bc + ca = 1,prove that
3
√1
a+ 6b+
3
√1
b+ 6c+
3
√1
c+ 6a ≤ 1
abc.
(Germany 2005)05.14. Given positive real numbers x, y, z such that xyz ≥ 1. Prove that
x5 − x2
x5 + y2 + z2+
y5 − y2
y5 + z2 + x2+
z5 − z2
z5 + x2 + y2≥ 0.
(IMO 2005)05.15. Let a1 ≤ a2 ≤ · · · ≤ an be positive real numbers such that
a21 + a22 + · · ·+ a2nn
= 1,a1 + a2 + · · ·+ an
n= m,
where 1 ≥ m > 0. Prove that for all i satisfying ai ≤ m, we have
n− i ≥ n(m− ai)2.
(Iran 2005)05.16. If three nonnegative real numbers a, b, c satisfy the condition
1
a2 + 1+
1
b2 + 1+
1
c2 + 1= 2,
prove that
ab+ bc+ ca ≤ 3
2.
(Iran 2005)05.17. If x, y, z are real numbers satisfying xyz = −1, prove that
x4 + y4 + z4 + 3(x+ y + z) ≥ y2 + z2
x+z2 + x2
y+x2 + y2
z.
(Iran 2005)05.18. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
a3√
1 + b− c+ b 3√
1 + c− a+ c3√
1 + a− b ≤ 1.
(Japan 2005)05.19. For any real numbers x1, x2, . . . , xn with x21 + x22 + · · ·+ x2n = 1, provethat
x11 + x21
+x2
1 + x21 + x22+ · · ·+ xn
1 + x21 + x22 + · · ·+ x2n<
√n
2.
(Korea 2005)05.20. Given positive real numbers a, b, c such that a4 + b4 + c4 = 3. Provethat
1
4− ab+
1
4− bc+
1
4− ca≤ 1.
(Moldova 2005)
211
05.21. Let a, b, c ∈ [0, 1]. Prove that
a
bc+ 1+
b
ca+ 1+
c
ab+ 1≤ 2.
(Poland 2005)05.22. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
√ab (1− c) +
√bc (1− a) +
√ca (1− b) ≤
√2
3.
(Republic of Srpska 2005)05.23. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that
1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a≤ 1.
(Romania 2005)05.24. Let n is a positive integer. Prove that if x is a positive real numbers,then
1 + xn+1 ≥ (2x)n
(1 + x)n−1.
(Russia 2005)05.25. Given positive real numbers x, y, z such that x2 + y2 + z2 = 1. Provethe following inequality
x
x3 + yz+
y
y3 + zx+
z
z3 + xy> 3.
(Russia 2005)05.26. Let a, b, c be positive real numbers. Prove that
a√b+ c
+b√c+ a
+c√a+ b
≥√
3
2(a+ b+ c).
(Serbia and Montenegro 2005)05.27. If a, b, c are positive real numbers such that ab + bc + ca = 1. Provethat
33
√1
abc+ 6 (a+ b+ c) ≤
3√
3
abc.
(Slovenia 2005)05.28. Let a1, a2, . . . , a95 be positive real numbers. Prove that
95∑k=1
ak ≤ 94 +
95∏k=1
max {1, ak} .
(Taiwan 2005)05.29. Let a, b, c be positive real numbers. Prove that(
a
b+b
c+c
a
)2
≥ (a+ b+ c)
(1
a+
1
b+
1
c
).
212
(United Kingdom 2005)05.30. Let a, b, c be positive real numbers. Prove that(
a
a+ b
)3
+
(b
b+ c
)3
+
(c
c+ a
)3
≥ 3
8.
(Vietnam 2005)06.1. Let x1, x2, . . . , xn be positive real numbers such that x1+x2+ · · ·+xn =1. Prove that (
n∑i=1
√xi
)(n∑i=1
1√1 + xi
)≤ n2√
n+ 1.
(China 2006)06.2. Let a, b, c be positive real numbers satisfying a+ b+ c = 1. Prove that
ab√ab+ bc
+bc√
bc+ ca+
ca√ca+ ab
≤√
2
2.
(China 2006)06.3. Suppose that a1, a2, . . . , an are real numbers with sum 0. Prove that thefollowing inequality holds
max1≤i≤n
a2i ≤n
3
n−1∑i=1
(ai − ai+1)2.
(China 2006)06.4. Let a, b, c be the sidelengths of a triangle. Prove that
√b+ c− a√
b+√c−√a
+
√c+ a− b
√c+√a−√b
+
√a+ b− c
√a+√b−√c≤ 3.
(IMO Shortlist 2006)06.5. Determine the least real number M such that the inequality∣∣ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)
∣∣ ≤M(a2 + b2 + c2)2
holds for all real numbers a, b, c.(IMO 2006)
06.6. Let x1, x2, x3, y1, y2, y3, z1, z2, z3 be positive real numbers. Find themaximum value of real number A if
M = (x31 + x32 + x33 + 1)(y31 + y32 + y33 + 1)(z31 + z32 + z33 + 1),
and
N = A(x1 + y1 + z1)(x2 + y2 + z2)(x3 + y3 + z3),
then M ≥ N always holds, and find the condition that the equality holds.(Japan 2006)
06.7. Let a, b, c, d be real numbers with sum 0. Prove the inequality
(ab+ ac+ ad+ bc+ bd+ cd)2 + 12 ≥ 6(abc+ abd+ acd+ bcd).
213
(Kazakhstan 2006)06.8. Let a, b, c be sides of the triangle. Prove that
a2(b
c− 1
)+ b2
( ca− 1)
+ c2(ab− 1)≥ 0.
(Moldova 2006)06.9. Let a, b, c be positive real numbers with ab+ bc+ ca = abc. Prove that
a4 + b4
ab(a3 + b3)+
b4 + c4
bc(b3 + c3)+
c4 + a4
ca(c3 + a3)≥ 1.
(Poland 2006)06.10. Find the maximum value of
(x3 + 1)(y3 + 1),
for all real numbers x, y, satisfying the condition that x+ y = 1.(Romania 2006)
06.11. Let a, b, c be three positive real numbers with sum 3. Prove that
1
a2+
1
b2+
1
c2≥ a2 + b2 + c2.
(Romania 2006)
06.12. Consider real numbers a, b, c contained in the interval
[1
2, 1
]. Prove
that
2 ≤ a+ b
1 + c+b+ c
1 + a+c+ a
1 + b≤ 3.
(Romania 2006)06.13. Let a, b be positive real numbers. Determine the largest constant Msuch that for all k ∈ [0, π] , we have
1
ka+ b+
1
kb+ a≥ M
a+ b.
(Thailand 2006)06.14. If x, y, z are positive numbers satisfying the condition xy+yz+zx = 1,show that
27
4(x+ y)(y + z)(z + x) ≥
(√x+ y +
√y + z +
√z + x
)2 ≥ 6√
3.
(Turkey 2006)06.15. Let a, b, c be real numbers. Prove that the following inequality holds∑√
(a2 − ab+ b2)(b2 − bc+ c2) ≥ a2 + b2 + c2.
(VMEO 2006)07.1. Let x, y, z be positive real numbers such that
√x+√y+√z = 1. Prove
that the following inequality holds
x2 + yz√2x2(y + z)
+y2 + zx√2y2(z + x)
+z2 + xy√2z2(x+ y)
≥ 1.
214
(APMO 2007)
07.2. If a, b, c ∈ R such that abc = 1, then
a2+b2+c2+1
a2+
1
b2+
1
c2+2
(a+ b+ c+
1
a+
1
b+
1
c
)≥ 6+2
(b+ c
a+c+ a
b+a+ b
c
).
(Brazil 2007) 07.3. Given an integer n ≥ 2, find the largest constant C(n)for which the inequality
n∑i=1
xi ≥ C(n)∑
1≤j<i≤n
(2xixj +
√xixj
)
holds for all real numbers xi ∈ (0, 1) satisfying (1 − xi)(1 − xj) ≥1
4for
1 ≤ j < i ≤ n.
(Bulgaria 2007)
07.4. Let a, b, c be the sidelengths of a triangle such that a + b + c = 3.Determine the minimum value of
a2 + b2 + c2 +4abc
3.
(China 2007)
07.5. Let α, β be acute angles. Determine the maximal value of(1−√
tanα tanβ)2
cotα+ cotβ.
(China 2007)
07.6. Let a, b, c be positive real numbers such that abc = 1. Prove that for allk ≥ 2, we have
ak
a+ b+
bk
b+ c+
ck
c+ a≥ 3
2.
(China 2007)
07.7. Let a, b, c > 0 such that a+ b+ c = 1. Prove that
a2
b+b2
c+c2
a≥ 3(a2 + b2 + c2).
(Croatia 2007)
07.8. Let a, b, c, d be positive real numbers such that a+ b+ c+ d = 1. Provethat
6(a3 + b3 + c3 + d3) ≥ a2 + b2 + c2 + d2 +1
8.
(France 2007)
07.9. Let a, b, c be sides of a triangle, show that
(c+ a− b)4
a(a+ b− c)+
(a+ b− c)4
b(b+ c− a)+b(b+ c− a)
c(c+ a− b)≥ ab+ bc+ ca.
(Greece 2007)
215
07.10. Let a, b, c, d be real numbers such that
a2 ≤ 1, a2 + b2 ≤ 5, a2 + b2 + c2 ≤ 14, a2 + b2 + c2 + d2 ≤ 30.
Prove thata+ b+ c+ d ≤ 10.
(Hungary-Isarel 2007)07.11. Let a1, a2, . . . , a100 be nonnegative eral numbers such that a21 + a22 +· · ·+ a2100 = 1. Prove that
a21a2 + a22a3 + · · ·+ a2100a1 <12
25.
(IMO Shortlist 2007)07.12. Let n be a positive integer, and let x and y be positive real numberssuch that xn + yn = 1. Prove that(
n∑k=1
1 + x2k
1 + x4k
)(n∑k=1
1 + y2k
1 + y4k
)<
1
(1− x)(1− y).
(IMO Shortlist 2007)07.13. Real numbers a1, a2, . . . , an are given. For each i (1 ≤ i ≤ n) define
di = max {aj : 1 ≤ j ≤ i} −min {aj : i ≤ j ≤ n},
and letd = max {di : 1 ≤ i ≤ n}.
(a) Prove that for any real numbers x1 ≤ x2 ≤ · · · ≤ xn, we have
max {|xi − ai| : 1 ≤ i ≤ n} ≥ d
2.
(b) Show that there are real numbers x1 ≤ x2 ≤ · · · ≤ xn such that we haveequality in (a).
(IMO 2007)07.14. If x, y, z are positive real numbers, prove that the following inequalityholds
(x+ y + z)2(yz + zx+ xy)2 ≤ 3(y2 + yz + z2)(z2 + zx+ x2)(x2 + xy + y2).
(India 2007)07.15. Let a, b, c be distinct positive real numbers. Show that∣∣∣∣a+ b
a− b+b+ c
b− c+c+ a
c− a
∣∣∣∣ > 1.
(Iran 2007)07.16. Find the largest constant T such that for all nonnegative real numbersa, b, c, d, e satisfying a+ b = c+ d+ e, we have√
a2 + b2 + c2 + d2 + e2 ≥ T(√
a+√b+√c+√d+√e)2.
216
(Iran 2007)
07.17. Prove that for any positive real numbers a, b, c, we have
a+ b+ c
3≤√a2 + b2 + c2
3≤ 1
3
(bc
a+ca
b+ab
c
).
(Ireland 2007)
07.18. Let n ≥ 2 be a given integer. Determine
(a) the largest real cn such that
1
1 + a1+
1
1 + a2+ · · ·+ 1
1 + an≥ cn
holds for any positive numbers a1, a2, . . . , an with a1a2 · · · an = 1.
(b) the largest real dn such that
1
1 + 2a1+
1
1 + 2a2+ · · ·+ 1
1 + 2an≥ dn
holds for any positive numbers a1, a2, . . . , an with a1a2 · · · an = 1.
(Italy 2007)
07.19. If a, b are positive real numbers such that ab ≥ 1, then
1
(2a+ 3)2+
1
(2b+ 3)2≥ 2
5(2ab+ 3).
(Kiev 2007)
07.20. For all positive real numbers a, b, c, find all values of positive numberk such that the following inequality holds
a
c+ kb+
b
a+ kc+
c
b+ ka≥ 1
2007.
(Korea 2007)
07.21. Let a, b, c be positive real numbers. Prove that
1 +3
ab+ bc+ ca≥ 6
a+ b+ c.
(Macedonia 2007)
07.22. Let a, b, c, d be nonnegative real numbers such that a+ b+ c+ d = 4.Prove that
a2bc+ b2cd+ c2da+ d2ab ≤ 4.
(Middle Europe 2007)
07.23. Let a, b, c, d be positive real numbers in the interval
[1
2, 2
]and abcd =
1. Find the maximum value of(a+
1
b
)(b+
1
c
)(c+
1
d
)(d+
1
a
).
(Middle Europe 2007)
217
07.24. Let a1, a2, . . . , an be positive real numbers such that ai ≥1
ifor all
i = 1, 2, . . . , n. Prove the inequality
(a1 + 1)
(a2 +
1
2
)· · ·(an +
1
n
)≥ 2n
(n+ 1)!(1 + a1 + 2a2 + · · ·+ nan).
(Moldova 2007)07.25. Let a1, a2, . . . , an ∈ [0, 1]. If S = a31 + a32 + · · ·+ a3n, then prove that
a12n+ 1 + S − a31
+a2
2n+ 1 + S − a32+ · · ·+ an
2n+ 1 + S − a3n≤ 1
3.
(Moldova 2007)07.26. Let a, b, c be positive real numbers such that
a+ b+ c ≥ 1
a+
1
b+
1
c.
Prove that
a+ b+ c ≥ 3
a+ b+ c+
2
abc.
(Peru 2007)07.27. a, b, c, d are positive real numbers satisfying the following condition
1
a+
1
b+
1
c+
1
d= 4.
Prove that
3
√a3 + b3
2+
3
√b3 + c3
2+
3
√c3 + d3
2+
3
√d3 + a3
2≤ 2(a+ b+ c+ d)− 4.
(Poland 2007)07.28. Let x, y, z be nonnegative real numbers. Prove that the followinginequality holds
x3 + y3 + z3
3≥ xyz +
3
4|(x− y)(y − z)(z − x)| .
(Romania 2007)07.29. For n ∈ N, n ≥ 2, determine
maxn∏i=1
(1− xi), for xi ∈ R+, 1 ≤ i ≤ n,n∑i=1
x2i = 1.
(Romania 2007)07.30. Let a, b, c be positive real numbers such that
1
a+ b+ 1+
1
b+ c+ 1+
1
c+ a+ 1≥ 1.
Show that
a+ b+ c ≥ ab+ bc+ ca.
218
(Romania 2007)07.31. For n ∈ N, n ≥ 2, ai, bi ∈ R, 1 ≤ i ≤ n, such that
n∑i=1
a2i = 1,n∑i=1
b2i = 1, andn∑i=1
aibi = 0,
prove that (n∑i=1
ai
)2
+
(n∑i=1
bi
)2
≤ n.
(C. Lupu and T. Lupu, Romania 2007)07.32. Positive real numbers a, b, c satisfy a+ b+ c = 1. Show that
1
ab+ 2c2 + 2c+
1
bc+ 2a2 + 2a+
1
ca+ 2b2 + 2b≥ 1
ab+ bc+ ca.
(Turkey 2007)07.33. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that(
a+1
a+ 1
)(b+
1
b+ 1
)(c+
1
c+ 1
)≥ 27
8,
and
27(a3+a2+a+1)(b3+b2+b+1)(c3+c2+c+1) ≥ 64(a2+a+1)(b2+b+1)(c2+c+1).
(Ukraine 2007)07.34. Show that for any real numbers a, b, c, then
(a2 + b2)2 ≥ (a+ b+ c)(b+ c− a)(c+ a− b)(a+ b− c).
(United Kingdom 2007)07.35. Given a triangle ABC. Determine the minimum value of
cos2A
2cos2
B
2
cos2C
2
+cos2
B
2cos2
C
2
cos2A
2
+cos2
C
2cos2
A
2
cos2B
2
.
(Vietnam 2007)08.1. Let a, b, c be real numbers such that a2 + b2 + c2 = 3. Prove that
a2
2 + b+ c2+
b2
2 + c+ a2+
c2
2 + a+ b2≥ (a+ b+ c)2
12.
(Baltic Way 2008)08.2. Suppose that a, b, c are positive real numbers with a2 + b2 + c2 = 1.Prove that
a5 + b5
ab(a+ b)+
b5 + c5
bc(b+ c)+
c5 + a5
ca(a+ b)≥ 3(ab+ bc+ ca)− 2.
(Bosnia 2008)
219
08.3. Let x, y, z be real numbers. Show that the following inequality holds
x2 + y2 + z2 − xy − yz − zx ≥ max
{3(x− y)2
4,3(y − z)2
4,3(z − x)2
4
}.
(Bosnia 2008)
08.4. Let a, b, c be positive real numbers. Prove that(1 +
4a
b+ c
)(1 +
4b
a+ c
)(1 +
4c
a+ b
)> 25.
(Bosnia 2008)
08.5. Let x, y, z be real numbers such that x+y+z = xy+yz+zx. Determinethe least value of the following expression
P =x
x2 + 1+
y
y2 + 1+
z
z2 + 1.
(Brazil 2008)
08.6. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
a− bca+ bc
+b− cab+ ca
+c− abc+ ab
≤ 3
2.
(Canada 2008)
08.7. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Provethat √
a+(b− c)2
4+√b+√c ≤√
3.
(China 2008)
08.8. Let x, y, z be positive numbers. Find the minimal value of
(a)x2 + y2 + z2
xy + yz;
(b)x2 + y2 + 2z2
xy + yz.
(Croatia 2008)
08.9. Determine the smallest constant C such that the following inequality
1 + (x+ y)2 ≤ C(1 + x2)(1 + y2)
holds for any real numbers x, y.
(Germany 2008)
08.10. Prove that if a1, a2, . . . , an are positive integers, then the followinginequality (
a21 + a22 + · · ·+ a2na1 + a2 + · · ·+ an
) knt
≥ a1a2 · · · an
holds for k = max {a1, a2, . . . , an} and t = min {a1, a2, . . . , an} . When do wehave the equality?
(Greece 2008)
220
08.11. Let x, y, z be positive real numbers such that x2 + y2 + z2 = 3. Provethe inequality
3
2<
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z< 3.
(Greece 2008)08.12. Prove that for any positive real numbers a, b, c, d, we have the followinginequality
(a− b)(a− c)a+ b+ c
+(b− c)(b− d)
b+ c+ d+
(c− d)(c− a)
c+ d+ a+
(d− a)(d− b)d+ a+ b
≥ 0.
(Darij Grinberg, IMO Shortlist 2008)08.13. (i) If x, y, z are three real numbers, all different from 1, such thatxyz = 1, then prove that
x2
(x− 1)2+
y2
(y − 1)2+
z2
(z − 1)2≥ 1.
(ii) Prove that equality is achieved for infinitely many triples of rational num-bers x, y, z.
(IMO 2008)08.14. Let n ≥ 3 is an integer and let x1, x2, . . . , xn be real numbers suchthat xi > 1 for all i. Prove the following inequality
x1x2x3 − 1
+ · · ·+ xn−1xnx1 − 1
+xnx1x2 − 1
≥ 4n.
(Indonesia 2008)08.15. Let a, b, c be nonnegative real numbers, from which at least two arenonzero and satisfying the condition ab+ bc+ ca = 1. Prove that√
a3 + a+√b3 + b+
√c3 + c ≥ 2
√a+ b+ c.
(Iran 2008)08.16. Let x, y, z be positive real numbers such that x+y+z = 3. Prove that
x3
y3 + 8+
y3
z3 + 8+
z3
x3 + 8≥ 1
9+
2
27(xy + xz + yz).
(Iran 2008)08.17. Find the smallest real k such that for each x, y, z > 0, we have theinequality
x√y + y
√z + z
√x ≤ k
√(x+ y)(y + z)(z + x).
(Iran 2008)08.18. For any positive real numbers a, b, c, d such that a2 + b2 + c2 + d2 = 1,prove the inequality
a2b2cd+ b2c2da+ c2d2ab+ d2a2bc+ c2a2db+ d2b2ac ≤ 3
32.
(Ireland 2008)
221
08.19. Let a, b, c be positive real numbers satisfying abc = 1. Prove that
1
b(a+ b)+
1
c(b+ c)+
1
a(c+ a)≥ 3
2.
(Kazakhstan 2008)08.20. Let a, b, c be positive real numbers such that (a+ b) (b+ c) (c+ a) = 8.Prove the inequality
a+ b+ c
3≥ 27
√a3 + b3 + c3
3.
(Macedonia 2008)08.21. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d andabcd = 1. Prove that
1
a3 + 1+
1
b3 + 1+
1
c3 + 1≥ 3
abc+ 1.
(MathLinks Contest 2008)08.22. Let a, b, c be nonnegative real numbers satisfying ab + bc + ca = 3.Prove that
1
1 + a2(b+ c)+
1
1 + b2(c+ a)+
1
1 + c2(a+ b)≤ 3
1 + 2abc.
(MathLinks Contest 2008)08.23. Determine the least value of the expression
P = abc+1
abc
where a, b, c are positive real numbers satisfying a+ b+ c ≤ 3
2.
(Moldova 2008)08.24. Let a1, a2, . . . , an be positive real numbers such that a1+a2+· · ·+an ≤n
2. Determine the smallest value of the following expression
A =
√a21 +
1
a22+
√a22 +
1
a23+ · · ·+
√a2n +
1
a21.
(Moldova 2008)08.25. Find the maximum value of constant C such that the inequality
x3 + y3 + z3 + C(xy2 + yz2 + zx2) ≥ (C + 1)(x2y + y2z + z2x)
holds for any nonnegative real numbers x, y, z.(Mongolia 2008)
08.26. If a, b, c are nonnegative real numbers, then the inequality holds
4(√
a3b3 +√b3c3 +
√c3a3
)≤ 4c3 + (a+ b)3.
(Poland 2008)
222
08.27. For real numbers xi > 1, 1 ≤ i ≤ n, n ≥ 2, such that
x2ixi − 1
≥ S =
n∑j=1
xj , for all i = 1, 2, . . . , n,
find, with proof, supS.
(Romania 2008)
08.28. Show that for all integers n ≥ 1, we have
n
(1 +
1
2+ · · ·+ 1
n
)≥ (n+ 1)
(1
2+
1
3+ · · ·+ 1
n+ 1
).
(Romania 2008)
08.29. Let a, b, c be positive real numbers with ab+ bc+ ca = 3. Prove that
1
1 + a2(b+ c)+
1
1 + b2(c+ a)+
1
1 + c2(a+ b)≤ 1
abc.
(Romania 2008)
08.30. Determine the maximum value of real number k such that
(a+ b+ c)
(1
a+ b+
1
b+ c+
1
c+ a− k)≥ k
for all real numbers a, b, c ≥ 0 with a+ b+ c = ab+ bc+ ca.
(Romania 2008)
08.31. Let a, b ∈ [0, 1]. Prove the inequality
1
1 + a+ b≤ 1− a+ b
2+ab
3.
(Romania 2008)
08.32. Let n ≥ 3 is an odd integer. Determine the maximum value of thecyclic sum, for 0 ≤ xi ≤ 1, i = 1, 2, . . . , n,
E =√|x1 − x2|+
√|x2 − x3|+ · · ·+
√|xn − x1|.
(Romania 2008)
08.33. Let a, b, c be three positive real numbers satisfying abc = 8. Provethat
a− 2
a+ 1+b− 2
b+ 1+c− 2
c+ 1≤ 0.
(Romania 2008)
08.34. Let a1, a2, . . . , an be positive real numbers satisfying the conditionthat a1 + a2 + · · ·+ an = 1. Prove that
n∑j=1
aj1 + a1 + · · ·+ aj
<1√2.
(Romania 2008)
223
08.35. Let x, y, z be positive real numbers such that x+ y+ z = 1. Prove theinequality
1
yz + x+1
x
+1
xz + y +1
y
+1
xy + z +1
z
≤ 27
31.
(Serbia 2008)08.36. Determine the least value of the expression x2 + y2 + z2, where x, y, zare real numbers such that x3 + y3 + z3 − 3xyz = 1.
(United Kingdom 2008)08.37. If a, b, c, d are positive real numbers, then
(a+ b)(b+ c)(c+ d)(d+ a)(
1 +4√abcd
)4≥ 16abcd(1 + a)(1 + b)(1 + c)(1 + d).
(Ukraine 2008)08.38. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Provethe inequality√
a2
a2 + b+ c+
√b2
b2 + c+ a+
√c2
c2 + a+ b≤√
3.
(Ukraine 2008)08.39. Let x, y, z be distinct nonnegative real numbers. Prove that
1
(x− y)2+
1
(y − z)2+
1
(z − x)2≥ 4
xy + yz + zx.
(Vietnam 2008)
224
Chapter 3
Solutions
83.1. Let a, b, c be the sidelengths of a triangle. Prove that
a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0.
(IMO 1983)
First solution: After setting a = y+ z, b = z + x, c = x+ y for x, y, z > 0, itbecomes
x3z + y3x+ z3y ≥ x2yz + xy2z + xyz2 orx2
y+y2
z+z2
x≥ x+ y + z.
However, an application of the Cauchy Schwarz Inequality gives
(y + z + x)
(x2
y+y2
z+z2
x
)≥ (x+ y + z)2.
Note that the equality holds if and only if a = b = c.
Second solution: Without loss of generality, we may assume that b is thesmallest number between a and c. Then we have two cases to consider: a ≥c ≥ b and c ≥ a ≥ b.For the former case, we have
a2b(a− b) = b(a− b)(a− c)2 + bc(2a− c)(a− b) ≥ bc(2a− c)(a− b).
Therefore, it suffices to prove that
b(2a− c)(a− b) + b2(b− c) + ca(c− a) ≥ 0,
or(2b− c)a2 − (2b− c)(b+ c)a+ b3 ≥ 0.
If c ≥ 2b then we are done because
(2b− c)a2 − (2b− c)(b+ c)a+ b3 = a(c− 2b)(b+ c− a) + b3 ≥ 0.
Conversely, if 2b ≥ c, then we have
(2b− c)a2 = (2b− c)(a− c)2 + c(2b− c)(2a− c) ≥ c(2b− c)(2a− c).
225
It now remains to check that
c(2b− c)(2a− c)− (2b− c)(b+ c)a+ b3 ≥ 0.
However, this is true because
c(2b− c)(2a− c)− (2b− c)(b+ c)a+ b3 = (2b− c)(c− b)a+ c3 − 2bc2 + b3
≥ (2b− c)(c− b)c+ c3 − 2bc2 + b3
= b(b− c)2 ≥ 0.
For the latter case, we have
a3b+ b3c+ c3a− (ab3 + bc3 + ca3) = (a− b)(c− b)(c− a)(a+ b+ c) ≥ 0,
and hence
2(a3b+b3c+c3a) ≥ (a3b+ab3)+(b3c+bc3)+(c3a+ca3) ≥ 2(a2b2+b2c2+c2a2),
from which we deduce that
a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0,
as desired.?F?
83.2. Show that for any positive reals a, b, c, d, e, f , we have
ab
a+ b+
cd
c+ d+
ef
e+ f≤ (a+ c+ e)(b+ d+ f)
a+ b+ c+ d+ e+ f.
(United Kingdom 1983)
First solution: The inequality is equivalent to(a+ b
4− ab
a+ b
)+
(c+ d
4− cd
c+ d
)+
(e+ f
4− ef
e+ f
)≥
≥ a+ c+ e+ b+ d+ f
4− (a+ c+ e)(b+ d+ f)
a+ b+ c+ d+ e+ f,
or(a− b)2
a+ b+
(c− d)2
c+ d+
(e− f)2
e+ f≥ (a+ c+ e− b− d− f)2
a+ b+ c+ d+ e+ f.
Now, using the Cauchy Schwarz Inequality, we get
(a− b)2
a+ b+
(c− d)2
c+ d+
(e− f)2
e+ f≥ [(a− b) + (c− d) + (e− f)]2
(a+ b) + (c+ d) + (e+ f)
=(a+ c+ e− b− d− f)2
a+ b+ c+ d+ e+ f.
Therefore, the last inequality is valid and so, the problem is solved.
226
Second solution: According to the Cauchy Schwarz Inequality, we have[(b+ d+ f)2a+ (a+ c+ e)2b
](b+ a) ≥
[(b+ d+ f)
√ab+ (a+ c+ e)
√ba]2
= ab(a+ b+ c+ d+ e+ f)2.
This implies that
ab
a+ b≤ (b+ d+ f)2a+ (a+ c+ e)2b
(a+ b+ c+ d+ e+ f)2.
Similarly, we have
cd
c+ d≤ (b+ d+ f)2c+ (a+ c+ e)2d
(a+ b+ c+ d+ e+ f)2,
andef
e+ f≤ (b+ d+ f)2e+ (a+ c+ e)2f
(a+ b+ c+ d+ e+ f)2.
Adding up these three inequalities, we get
ab
a+ b+
cd
c+ d+
ef
e+ f≤ (b+ d+ f)2(a+ c+ e) + (a+ c+ e)2(b+ d+ e)
(a+ b+ c+ d+ e+ f)2
=(a+ c+ e)(b+ d+ f)
a+ b+ c+ d+ e+ f,
as desired.?F?
84.1. Let a1, a2, . . . , an > 0, n ≥ 2. Prove that
a21a2
+a22a3
+ · · ·+a2n−1an
+a2na1≥ a1 + a2 + · · ·+ an.
(China 1984)
Solution: From the AM-GM Inequality, for each i (an+1 = a1), we get
a2iai+1
+ ai+1 ≥ 2
√a2iai+1
· ai+1 = 2ai.
Setting i = 1, 2, . . . , n, we get the n similar inequalities and summing themup, we can get the desired result. Note that the equality holds if and only ifa1 = a2 = · · · = an.
?F?
84.2. Let x, y, z be nonnegative real numbers such that x+ y + z = 1. Provethat
0 ≤ xy + yz + zx− 2xyz ≤ 7
27.
(IMO 1984)
227
Solution: Let f(x, y, z) = xy + yz + zx − 2xyz. Without loss of generality,we may assume that 0 ≤ x ≤ y ≤ z ≤ 1. Since x + y + z = 1, this implies
that x ≤ 1
3. It follows that f(x, y, z) = (1− 3x)yz+ xyz+ zx+ xy ≥ 0, which
proves the left inequality. For the right part, applying the AM-GM Inequality,
we obtain yz ≤(y + z
2
)2
=
(1− x
2
)2
. Since 1− 2x ≥ 0, this implies that
f(x, y, z) = x(y+z)+yz(1−2x) ≤ x(1−x)+
(1− x
2
)2
(1−2x) =−2x3 + x2 + 1
4.
Our job is now to maximize a one-variable function F (x) =1
4(−2x3 +x2 + 1),
where x ∈[0,
1
3
]. Since F ′(x) =
3
2x
(1
3− x)≥ 0 on
[0,
1
3
], we conclude that
F (x) ≤ F(
1
3
)=
7
27for all x ∈
[0,
1
3
].
?F?
84.3. Prove that for any a, b > 0, we have that
(a+ b)2
2+a+ b
4≥ a√b+ b
√a.
(Russia 1984)
Solution: The AM-GM Inequality gives us(2a2 +
b
2
)+(
2b2 +a
2
)≥ 2a
√b+ 2b
√a,
and (2ab+
b
2
)+(
2ab+a
2
)≥ 2b√a+ 2a
√b.
By summing up these two inequalities, we get
2(a+ b)2 + (a+ b) ≥ 4a√b+ 4b
√a,
and then, we deduce that
(a+ b)2
2+a+ b
4≥ a√b+ b
√a,
as desired. Note that the equality holds if and only if a = b =1
4.
?F?
85.1. Let x1, x2, . . . , xn be real numbers from the interval [0, 2]. Prove that
n∑i,j=1
|xi − xj | ≤ n2.
When do we have equality?
228
(United Kingdom 1985)
Solution: Since the inequality is symmetric, we can assume that 0 ≤ x1 ≤x2 ≤ · · · ≤ xn ≤ 2. Accordingly, we find that
n∑i,j=1
|xi − xj | = 2
n∑1≤i<j≤n
|xi − xj | = 2
n∑1≤i<j≤n
(xj − xi) = 2
n∑i=1
(2i− 1− n)xi.
And so, we are left to prove
n∑i=1
(2i− 1− n)xi ≤n2
2.
If n = 2k (k ∈ N, k ≥ 1) , then we have
n∑i=1
(2i− 1− n)xi =
2k∑i=1
(i− 1− 2k)xi
=2k∑
i=k+1
(2i− 1− 2k)xi +k∑i=1
(2i− 1− 2k)xi
≤2k∑
i=k+1
(2i− 1− 2k)xi ≤ 22k∑
i=k+1
(2i− 1− 2k) =n2
2,
If n = 2k + 1 (k ∈ N) , we proceed the same way as the preceding case andobtain
n∑i=1
(2i− 1− n)xi = 2
2k+1∑i=1
(i− 1− k)xi
= 22k+1∑i=k+2
(i− 1− k)xi + 2k+1∑i=1
(i− 1− k)xi
≤ 22k+1∑i=k+2
(i− 1− k)xi ≤ 42k+1∑i=k+2
(i− 1− k)
= 2k(k + 1) <(2k + 1)2
2=n2
2.
Since the inequality holds for both cases n even and n odd, we conclude thatit holds for any value of n. In addition, from the two cases above, one can findthat the inequality can become equality only for n = 2k, and in this case, theequality is attained for x1 = x2 = · · · = xk = 0 and xk+1 = xk+2 = · · · =x2k = 2.
?F?
86.1. Find the maximum value of the constant c such that for any x1, x2, . . . , xn >0 for which xk+1 ≥ x1 + x2 + · · ·+ xk for any k, the inequality
√x1 +
√x2 + · · ·+
√xn ≤ c
√x1 + x2 + · · ·+ xn
229
also holds for any n.
(IMO Shorlist 1986)
Solution: First, let us see what happens if xk+1 and x1 + x2 + · · · + xk areequal for any k. For example, we can let n ≥ 2 and take x1 = 1, xk = 2k−2 forany k ≥ 2. In this case, we have xk+1 = x1 + x2 + · · · + xk for all k ≥ 1, andthus, we find that
c ≥
1 +
n∑k=2
(√2)k−2
√√√√1 +n∑k=2
2k−2
for any n ≥ 2. Taking the limit, we find that c ≥ 1 +√
2. Now, let us provethat 1 +
√2 works. We will prove the inequality
√x1 +
√x2 + · · ·+
√xn ≤
(1 +√
2)√
x1 + x2 + · · ·+ xn
by induction. For n = 1 and n = 2, it is clear. Suppose that it is true for nand we will prove that
√x1 +
√x2 + · · ·+
√xn +
√xn+1 ≤
(1 +√
2)√
x1 + x2 + · · ·+ xn + xn+1.
Of course, it is enough to prove that
√xn+1 ≤
(1 +√
2) (√
x1 + x2 + · · ·+ xn + xn+1 −√x1 + x2 + · · ·+ xn
),
which is equivalent to
√x1 + x2 + · · ·+ xn + xn+1 +
√x1 + x2 + · · ·+ xn ≤
(1 +√
2)√
xn+1.
But this one is obvious because x1 + x2 + · · ·+ xn ≤ xn+1.?F?
86.2. Let n be a positive integer. Prove that
|sin 1|+ |sin 2|+ · · ·+ |sin 3n| > 8
5n.
(Russia 1986)
Solution: We will prove first that the following inequality holds for any x ≥ 0
f(x) = |sinx|+ |sin(x+ 1)|+ |sin(x+ 2)| − 8
5> 0.
Since f(x) = f(x + π), it suffices to prove it for x ∈ [0, π], then for x > πwe can put x = kπ + r where k ≥ 1, k ∈ N and 0 < r < π, and thusf(x) = f(kπ + r) = f(r) > 0. Now, from the assumption that x ∈ [0, π], wewill consider two cases
230
The first case is when 0 ≤ x ≤ π−2. In this case, we have sinx ≥ 0, sin(x+2) ≥0 and sin(x+ 1) ≥ sin 1 > 0. Therefore
|sinx|+ |sin(x+ 1)|+ |sin(x+ 2)| = sinx+ sin(x+ 1) + sin(x+ 2)
= (1 + 2 cos 1) sin(x+ 1) ≥ (1 + 2 cos 1) sin 1 >8
5.
The second one is when π−2 ≤ x ≤ π. In this case, sinx ≥ 0 and sin(x+2) ≤ 0,our inequality is equivalent to
sinx+ |sin(x+ 1)| − sin(x+ 2) >8
5.
Denote t = cos(x+ 1) then −1 ≤ t ≤ − cos 1. We have
sinx+ |sin(x+ 1)| − sin(x+ 2) = |sin(x+ 1)| − 2 sin 1 cos(x+ 1)
=√
1− t2 − 2t sin 1 = g(t)
Since g(t) is concave, we have
g(t) ≥ min {g(−1), g (− cos 1)} = min {2 sin 1, sin 1 + 2 sin 1 cos 1} > 8
5.
Now, we will prove the original inequality by induction on n. For n = 1, it isclear. Suppose that the inequality holds for n and let us prove it for n + 1,that is to prove
|sin 1|+|sin 2|+· · ·+|sin 3n|+| sin(3n+1)|+| sin(3n+2)|+| sin(3n+3)| > 8
5(n+1).
However, this is trivial since from the inductive hypothesis, we have
|sin 1|+ |sin 2|+ · · ·+ |sin 3n| > 8
5n,
and from what we have shown above,
| sin(3n+ 1)|+ | sin[(3n+ 1) + 1]|+ | sin[(3n+ 1) + 2]| ≥ 8
5.
This completes our proof.?F?
86.3. Show that for all positive real numbers a1, a2, . . . , an, we have
1
a1+
2
a1 + a2+ · · ·+ n
a1 + a2 + · · ·+ an≤ 4
(1
a1+
1
a2+ · · ·+ 1
an
).
(Russia 1986)
Solution: We will show that the stronger inequality holds
n∑k=1
kk∑j=1
aj
≤ 2n∑k=1
1
ak.
231
Indeed, by the Cauchy Schwarz Inequality, we have
(a1 + · · ·+ ak)
(b21a1
+ · · ·+b2kak
)≥ (b1 + · · ·+ bk)
2,
where bi are arbitrary positive real numbers for all i = 1, 2, . . . , n. Therefore
kk∑j=1
aj
≤ k
(b1 + · · ·+ bk)2
(b21a1
+ · · ·+b2kak
),
from which we deduce that
n∑k=1
kk∑j=1
aj
≤n∑i=1
ciai,
where
ck =kb2k
(b1 + · · ·+ bk)2+
(k + 1)b2k(b1 + · · ·+ bk+1)2
+ · · ·+nb2k
(b1 + · · ·+ bn)2,
for all k = 1, 2, . . . , n. Choosing now bk = k, we get
ck =k3(k∑i=1
i
)2 +k2(k + 1)(k+1∑i=1
i
)2 + · · ·+ k2n(n∑i=1
i
)2
= 4k2n∑j=k
1
j(j + 1)2= 4k2
n∑j=k
[1
j(j + 1)− 1
(j + 1)2
]
≤ 4k2n∑j=k
[1
2j2+
1
2(j + 1)2− 1
(j + 1)2
]
= 2k2n∑j=k
[1
j2− 1
(j + 1)2
]= 2k2
[1
k2− 1
(n+ 1)2
]< 2,
for all k = 1, 2, . . . , n. In this case, it follows that
n∑k=1
kk∑j=1
aj
≤ 2
n∑i=1
1
ai,
and so our proof is completed.?F?
86.4. Find the maximum value of
x2y + y2z + z2x
232
for reals x, y, z with x+ y + z = 0 and x2 + y2 + z2 = 6.(United Kingdom 1986)
First solution: From the given condition, we have xy + yz + zx = −3, andthen, it follows that
x2y2 + y2z2 + z2x2 = (xy + yz + zx)2 − 2xyz(x+ y + z) = 9.
Now, let us expand the following nonnegative expression (x− xy − 1)2 + (y −yz − 1)2 + (z − zx− 1)2. The work is very simple, we find that it is equal to
3 +∑
x2 +∑
x2y2 + 2∑
xy − 2∑
x− 2∑
x2y.
Replacing∑
x2 by 6,∑
x2y2 by 9,∑
xy by −3 and∑
x by 0, respectively,
we conclude that
(x− xy − 1)2 + (y − yz − 1)2 + (z − zx− 1)2 = 12− 2(x2y + y2z + z2x).
Accordingly, we infer that
x2y + y2z + z2x ≤ 6.
Note that the equality can occur, for example when x = 2 cos2π
9,y = 2 cos
4π
9,z =
2 cos8π
9. This allows us to conclude that the maximum of x2y + y2z + z2x is
6.
Second solution: We will give another way to prove that x2y+y2z+z2x ≤ 6.Denote with p, q, r the values of x+ y+ z, xy+ yz + zx and xyz, respectively.By this substitution, one can easily check that
(x− y)2(y − z)2(z − x)2 = p2q2 − 4q3 + 2(9q − 2p2)pr − 27r2,
and(x2y + y2z + z2x) + (xy2 + yz2 + zx2) = pq − 3r.
On the other hand, from the given hypothesis, we have p = 0 and q = −3.Therefore, the above identities imply
(x− y)2(y − z)2(z − x)2 = −4 · (−3)3 − 27r2 = 27(4− r2),
and(x2y + y2z + z2x) + (xy2 + yz2 + zx2) = −3r.
Accordingly, we have
2∑
x2y =∑
x2y +∑
xy2 +(∑
x2y −∑
xy2)
= −3r + (x− y)(x− z)(y − z) ≤ −3r +√
(x− y)2(y − z)2(z − x)2
= −3r + 3√
3(4− r2) ≤ 3
√[12 +
(√3)2] [
(−r)2 +(√
4− r2)2]
= 12,
233
from which it deduces that x2y + y2z + z2x ≤ 6, as desired.
Remark: In addition, we can prove that the following inequality holds (inthe same manner with the two solutions above)
−2
9
(3n2 −m2
2
)3/2
+m3
9≤ x2y + y2z + z2x ≤ 2
9
(3n2 −m2
2
)3/2
+m3
9
for all real numbers x, y, z satisfying x + y + z = m and x2 + y2 + z2 = n2
(3n2 > m2).?F?
87.1. Prove that if x, y, z are real numbers such that x2 + y2 + z2 = 2, then
x+ y + z ≤ xyz + 2.
(IMO Shorlist 1987)
First solution: If one of x, y, z is negative, for example z < 0. Then, werewrite the inequality in the form
−z(1− xy) + 2− x− y ≥ 0,
which is true because
2− x− y ≥ 2−√
2(x2 + y2) ≥ 2−√
2(x2 + y2 + z2) = 0,
and
−z(1− xy) ≥ −z(
1− x2 + y2
2
)= −z
3
2≥ 0.
Let us consider the now the case x, y, z ≥ 0. Without loss of generality, we
may assume that z = max {x, y, z} . By this assumption, we have z >1√3.
Denote t = x+ y, we find that
xy =t2 − x2 − y2
2=t2 + z2 − 2
2.
Therefore, our inequality is equivalent to
2 + z · t2 + z2 − 2
2≥ t+ z,
orf(t) = zt2 − 2t+ z3 − 4z + 4 ≥ 0.
We see that f(t) is a quadratic function of t with the largest coefficient isz > 0. In addtion, its discrimimant is
∆′f = 1−z(z3−4z+4) = −(z2+2z−1)(1−z)2 < −(
1
3+
2√3− 1
)(1−z)2 ≤ 0.
Therefore f(t) ≥ 0, and this ends our proof. Note that the equality holds ifand only if x = y = 1, z = 0 or y = z = 1, x = 0 or z = x = 1, y = 0.
234
Second solution: Using the Cauchy Schwarz Inequality, we find that
x+ y + z − xyz = x(1− yz) + (y + z) · 1 ≤√
[x2 + (y + z)2][(1− yz)2 + 1].
So, it is enough to prove that this last quantity is at most 2, which is equivalentto the inequality (2 + 2yz)(2 − 2yz + y2z2) ≤ 4, or 2y3z3 ≤ 2y2z2, which isclearly true, because 2 ≥ y2 + z2 ≥ 2yz.
Third solution: By the same arguments with the first solution, we see thatit suffices to prove the inequality for x, y, z ≥ 0. Because of symmetry, we mayassume that 0 ≤ x ≤ y ≤ z. If z ≤ 1, then
2 + xyz − x− y − z = (1− z)(1− xy) + (1− x)(1− y) ≥ 0.
Now, if z > 1, we have
z + (x+ y) ≤√
2[z2 + (x+ y)2] = 2√
1 + xy ≤ 1 + (1 + xy) ≤ 2 + xyz.
This ends the proof.?F?
88.1. Show that
(1 + x)n ≥ (1− x)n + 2nx(1− x2)n−12
for all 0 ≤ x ≤ 1 and all positive integer n.(Ireland 1988)
Solution: For n = 1, the inequality becomes equality. Suppose that n ≥ 2,
then since (1− x2)n−12 ≤ 1, it suffices to prove that
f(x) = (1 + x)n − (1− x)n − 2nx ≥ 0.
We have f ′(x) = n[(1 + x)n−1 + (1− x)n−1 − 2
]. Since n−1 ≥ 1, the Bernoulli’s
Inequality implies that
(1 + x)n−1 + (1− x)n−1 ≥ 1 + (n− 1)x+ 1 + (n− 1)(−x) = 2.
This means that f ′(x) ≥ 0. Therefore f(x) is increasing on [0, 1], and wededuce that f(x) ≥ f(0) = 0, as claimed. Note that the equality holds if andonly if n = 1 or x = 0.
?F?
88.2. Given a triangle ABC. Prove that
2
(sinA
A+
sinB
B+
sinC
C
)≤(
1
B+
1
C
)sinA+
(1
C+
1
A
)sinB+
(1
A+
1
B
)sinC.
(Russia 1988)
Solution: Rewrite the inequality as(1
B+
1
C− 2
A
)sinA+
(1
C+
1
A− 2
B
)sinB +
(1
A+
1
B− 2
C
)sinC ≥ 0.
235
Now, since this inequality is symmetric, we can assume that A ≥ B ≥ C. Bythis assumption, we have a ≥ b ≥ c, and since a = 2R sinA, we deduce thatsinA ≥ sinB ≥ sinC. Also, it is clear from the assumption that
1
B+
1
C− 2
A≥ 1
C+
1
A− 2
B≥ 1
A+
1
B− 2
C.
Thefore, by applying the Chebyshev’s Inequality, we deduce that(1
B+
1
C− 2
A
)sinA+
(1
C+
1
A− 2
B
)sinB +
(1
A+
1
B− 2
C
)sinC ≥
≥ 1
3
(1
B+
1
C− 2
A+
1
C+
1
A− 2
B+
1
A+
1
B− 2
C
)(sinA+sinB+sinC) = 0.
Note that the equality holds if and only if ABC is a equilateral triangle.?F?
89.1. Let x1, x2, . . . , xn be positive real numbers, and let S = x1+x2+· · ·+xn.Prove that
(1 + x1)(1 + x2) · · · (1 + xn) ≤ 1 + S +S2
2!+ · · ·+ Sn
n!.
(APMO 1989)
Solution: By the AM-GM Inequality, we have
(1 + x1)(1 + x2) . . . (1 + xn) ≤[
(1 + x1) + (1 + x2) + · · ·+ (1 + xn)
n
]n=
(1 +
S
n
)n.
Therefore, it suffices to prove that(1 +
S
n
)n≤ 1 + S +
S2
2!+ . . .+
Sn
n!,
and since
(1 +
S
n
)n= 1 +
n∑k=1
(n
k
)nk
Sk, we can rewrite this inequality as
n∑k=1
1
k!−
(n
k
)nk
Sk ≥ 0.
This is true since for any k = 1, . . . , n, we have
nk − k!
(n
k
)= nk − k! · n!
(n− k)!k!= nk − n!
(n− k)!
= nk − (n− k + 1) · · ·n ≥ nk − n · · ·n︸ ︷︷ ︸k numbers
= 0.
236
?F?
89.2. Let x, y, z be real numbers such that 0 < x < y < z <π
2. Prove that
π
2+ 2 sinx cos y + 2 sin y cos z > sin 2x+ sin 2y + sin 2z.
(Iberoamerica 1989)
Solution: Put a = sinx, b = sin y, c = sin z, then 1 > c > b > a > 0. Ourinequality becomes
π
4+ a√
1− b2 + b√
1− c2 > a√
1− a2 + b√
1− b2 + c√
1− c2,
orπ
4> a
√1− a2 + (b− a)
√1− b2 + (c− b)
√1− c2.
By the AM-GM Inequality, we get
a√
1− a2 + (b− a)√
1− b2 + (c− b)√
1− c2 ≤
≤ a(
1− a2 +1
4
)+ (b− a)
(1− b2 +
1
4
)+ (c− b)
(1− c2 +
1
4
)= −1
4(4a3 + 4b3 + 4c3 − 4ab2 − 4bc2 − 5c).
Therefore, it suffices to prove that
4a3 + 4b3 + 4c3 − 4ab2 − 4bc2 − 5c+ π > 0.
Using the AM-GM Inequality again, we haveπ
2+π
2+
500
27π2c3 ≥ 5c, and hence
π − 5c ≥ − 500
27π2c3 > −2c3. Consequently, it is enough to check that
2a3 + 2b3 + c3 − 2ab2 − 2bc2 ≥ 0,
or (c3 − 2bc2 +
32
27b3)
+
(2a3 − 2ab2 +
22
27b3)≥ 0.
This is true since by the AM-GM Inequality, we have
c3 − 2bc2 +32
27b3 =
c3
2+c3
2+
32
27b3 − 2bc2
≥ 33
√c3
2· c
3
2· 32
27b3 − 2bc2 = 0,
and
2a3 − 2ab2 +22
27b3 = 2a3 +
11
27b3 +
11
27b3 − 2ab2
≥ 33
√2a3 · 11
27b3 · 11
27b3 − 2ab2
=
(3√
242
3− 2
)ab2 ≥ 0.
237
?F?
89.3. Let a, b, c be the sidelengths of a triangle. Prove that∣∣∣∣a− ba+ b+b− cb+ c
+c− ac+ a
∣∣∣∣ < 1
16.
(Iberoamerica 1989)
Solution: Using the identity
a− ba+ b
+b− cb+ c
+c− ac+ a
=(a− b)(b− c)(a− c)(a+ b)(b+ c)(c+ a)
,
we can rewrite the original inequality as
(a+ b)(b+ c)(c+ a) > 16 |(a− b)(b− c)(a− c)| .
Now, since a, b, c are the sidelengths of a triangle, we may put a = y + z, b =z + x, c = x+ y where x, y, z > 0, and hence, our inequality becomes
(2x+ y + z)(2y + z + x)(2z + x+ y) > 16 |(x− y)(y − z)(z − x)| .
Without loss of generality, we may assume that x ≥ y ≥ z, then we have
(2x+ y + z)(2y + z + x)(2z + x+ y) > (2x+ y)(x+ 2y)(x+ y),
and|(x− y)(y − z)(z − x)| = (x− y)(x− z)(y − z) < xy(x− y).
Therefore, it suffices to prove that
(2x+ y)(x+ 2y)(x+ y) > 16xy(x− y).
Now, we apply the AM-GM Inequality to get
(2x+ y)(x+ 2y) = 2(x− y)2 + 9xy ≥ 2√
2(x− y)2 · 9xy = 6√
2xy(x− y),
andx+ y ≥ 2
√xy.
Multiplying these two inequalities, we deduce that
(2x+ y)(x+ 2y)(x+ y) ≥ 12√
2xy(x− y) ≥ 16xy(x− y).
However, it is clear that the equality cannot occur, so we must have
(2x+ y)(x+ 2y)(x+ y) > 16xy(x− y).
The proof is completed.?F?
89.4. Find the least possible value of
(x+ y)(y + z)
238
for positive reals x, y, z satisfying xyz(x+ y + z) = 1.(Russia 1989)
Solution: From the AM-GM Inequality, we have
(x+ y)(y + z) = y(x+ y + z) + xz ≥ 2√xyz(x+ y + z) = 2.
The equality holds iff y(x + y + z) = xz and xyz(x + y + z) = 1, which canbe attained for x = z = 1, y =
√2 − 1. This allows us to conclude that the
minimum value of (x+ y)(y + z) is 2.?F?
90.1. Let a, b, c, d be positive real numbers such that ab + bc + cd + da = 1.Prove that
a3
b+ c+ d+
b3
c+ d+ a+
c3
d+ a+ b+
d3
a+ b+ c≥ 1
3.
(IMO Shortlist 1990)
Solution: From the AM-GM Inequality, we get
2 (ab+ ac+ ad+ bc+ bd+ cd) ≤ 3(a2 + b2 + c2 + d2
).
Furthermore, the Cauchy Schwarz Inequality gives us
a2 + b2 + c2 + d2 =√a2 + b2 + c2 + d2
√b2 + c2 + d2 + a2
≥ ab+ bc+ cd+ da = 1.
Therefore, from these two inequalities, we deduce that
(a2 + b2 + c2 + d2)2 ≥ 2
3(ab+ ac+ ad+ bc+ bd+ cd).
Now, using the Cauchy Schwarz Inequality and this inequality, we obtain
a3
b+ c+ d+
b3
c+ d+ a+
c3
d+ a+ b+
d3
a+ b+ c≥
≥(a2 + b2 + c2 + d2
)22 (ab+ ac+ ad+ bc+ bc+ cd)
≥ 1
3,
as desired. Note that the equality holds if and only if a = b = c = d =1
2.
?F?
90.2. Show that
x4 > x− 1
2
for all real x.(Russia 1990)
Solution: According to the AM-GM Inequality, we have
x4 +1
2= x4 +
1
6+
1
6+
1
6≥ 4
4
√x4
63=
44√
63|x| ≥ |x| ≥ x.
239
It is easy to see that the equality cannot be attained. Therefore
x4 > x− 1
2,
and our proof is completed.?F?
90.3. Let x1, x2, . . . , xn be positive reals with sum 1. Show that
x21x1 + x2
+x22
x2 + x3+ · · ·+ x2n
xn + x1≥ 1
2.
(Russia 1990)
Solution: According to the Cauchy Schwarz Inequality, we have
x21x1 + x2
+x22
x2 + x3+ · · ·+ x2n
xn + x1≥ (x1 + x2 + · · ·+ xn)2
(x1 + x2) + (x2 + x3) + · · ·+ (xn + x1)
=x1 + x2 + · · ·+ xn
2=
1
2,
as desired. Note that the equality holds if and only if x1 = x2 = · = xn =1
n.
?F?
90.4. Show that√x2 − xy + y2 +
√y2 − yz + z2 ≥
√z2 + zx+ x2
for any positive real numbers x, y, z.(United Kingdom 1990)
Solution: By applying the Minkowsky’s Inequality, we have√x2 − xy + y2 +
√y2 − yz + z2 ≥
=
√√√√(√3
2x
)2
+(x
2− y)2
+
√√√√(√3
2z
)2
+(y − z
2
)2
≥
√√√√(√3
2x+
√3
2z
)2
+(x
2− y + y − z
2
)2=√x2 + xz + z2,
as desired.?F?
91.1. Let a1, a2, . . . , an and b1, b2, . . . , bn be positive real numbers such thata1 + a2 + · · ·+ an = b1 + b2 + · · ·+ bn. Prove that
a21a1 + b1
+a22
a2 + b2+ · · ·+ a2n
an + bn≥ a1 + a2 + · · ·+ an
2.
(APMO 1991)
240
First solution: By the Cauchy Schwarz Inequality, we have
a21a1 + b1
+a22
a2 + b2+ · · ·+ a2n
an + bn≥ (a1 + a2 + · · ·+ an)2
(a1 + b1) + (a2 + b2) + · · ·+ (an + bn)
=(a1 + a2 + · · ·+ an)2
(a1 + a2 + · · ·+ an) + (b1 + b2 + · · ·+ bn)
=(a1 + a2 + · · ·+ an)2
(a1 + a2 + · · ·+ an) + (a1 + a2 + · · ·+ an)
=a1 + a2 + · · ·+ an
2,
as desired. Note that the equality holds if and only if ai = bi for all i =1, 2, . . . , n.
Second solution: From the AM-GM Inequality, we see that for all i =1, 2, . . . , n
a2iai + bi
= ai −aibiai + bi
≥ ai −(ai + bi)
2
4(ai + bi)=
3ai − bi4
.
Therefore
a21a1 + b1
+a22
a2 + b2+ · · ·+ a2n
an + bn≥ 3a1 − b1
4+
3a2 − b24
+ · · ·+ 3an − bn4
=3(a1 + a2 + · · ·+ an)− (b1 + b2 + · · ·+ bn)
4
=a1 + a2 + · · ·+ an
2,
as desired.?F?
91.2. Given positive real numbers a, b, c satisfying a+ b+ c = 1, show that
(1 + a) (1 + b) (1 + c) ≥ 8 (1− a) (1− b) (1− c) .
(Russia 1991)
Solution: By applying the AM-GM Inequality, we have that
1 + a = (1− b) + (1− c) ≥ 2√
(1− b) (1− c).
Similarly, we also have
1 + b ≥ 2√
(1− c) (1− a), 1 + c ≥ 2√
(1− a) (1− b).
Multiplying these three inequalities, we get the desired result. Note that the
equality holds if and only if a = b = c =1
3.
?F?
91.3. Show that
(x+ y + z)2
3≥ x√yz + y
√zx+ z
√xy
241
for all nonnegative reals x, y, z.(Russia 1991)
Solution: By the AM-GM Inequality, we get
x√yz + y
√zx+ z
√xy ≤ x · y + z
2+ y · z + x
2+ z · x+ y
2
= xy + yz + zx ≤ (x+ y + z)2
3,
as desired. The equality occurs iff x = y = z.?F?
91.4. The real numbers x1, x2, . . . , x1991 satisfy
|x1 − x2|+ |x2 − x3|+ · · ·+ |x1990 − x1991| = 1991.
Denote sn =x1 + x2 + . . .+ xn
n. What is the maximum possible value of
|s1 − s2|+ |s2 − s3|+ . . .+ |s1990 − s1991|?
(Russia 1991)
Solution: We denote with E the desired expression. For all 1 ≤ i ≤ 1990, wehave
si+1 − si =
i+1∑k=1
xk
i+ 1−
i∑k=1
xk
i=
ixi+1 −i∑
k=1
xk
i(i+ 1)=
i∑k=1
k(xk+1 − xk)
i(i+ 1).
This implies that
|si+1 − si| ≤
i∑k=1
k |xk+1 − xk|
i(i+ 1).
According to this inequality, we deduce that
E =1990∑i=1
|si+1 − si| ≤1990∑i=1
i∑k=1
k |xk+1 − xk|
i(i+ 1)=
1990∑i=1
i
(1990∑k=i
1
k(k + 1)
)|xi+1 − xi|
=
1990∑i=1
(1− i
1991
)|xi+1 − xi| ≤
1990∑i=1
(1− 1
1991
)|xi+1 − xi|
=1990
1991
1990∑i=1
|xi+1 − xi| = 1990.
On ther other hand, let x1 = 1991, x2 = x3 = · · · = x1991 = 0, we have|x1 − x2| + |x2 − x3| + · · · + |x1990 − x1991| = 1991 and E = 1990. Therefore,the maximum value of E is 1990.
242
?F?
91.5. A triangle has sides a, b, c with sum 2. Show that
a2 + b2 + c2 + 2abc < 2.
(United Kingdom 1991)
Solution: Since a, b, c are the sidelengths of a triangle and a+ b+ c = 2, wesee that 1 > a, b, c > 0. Hence (1 − a)(1 − b)(1 − c) > 0, which implies thatabc < 1− (a+ b+ c) + ab+ bc+ ca = ab+ bc+ ca− 1. And thus,
a2 + b2 + c2 + 2abc < a2 + b2 + c2 + 2(ab+ bc+ ca− 1)
= (a+ b+ c)2 − 2 = 2,
as desired.?F?
91.6. Prove the inequality
x2y
z+y2z
x+z2x
y≥ x2 + y2 + z2
for any positive real numbers x, y, z with x ≥ y ≥ z.(Vietnam 1991)
First solution: By expanding, we see that the original inequality is equivalentto
x3y2 + y3z2 + z3x2 ≥ x3yz + y3zx+ z3xy,
or
x3y(y − z) + y2z2(y − z) + z3(y2 − 2yx+ x2)− xyz(y2 − z2) ≥ 0,
that is(y − z)(x− z)[x2y + yz(x− y)] + z3(x− y)2 ≥ 0.
The last inequality is obviously true since x ≥ y ≥ z. So, our problem is solved.Note that the equality holds if and only if x = y = z.
Second solution: From the Cauchy Schwarz Inequality, we have√x2y
z+y2z
x+z2x
y·
√x2z
y+y2x
z+z2y
x≥ x2 + y2 + z2.
Hence, it suffices to prove that
x2y
z+y2z
x+z2x
y≥ x2z
y+y2x
z+z2y
x.
However, we find that(x2y
z+y2z
x+z2x
y
)−(x2z
y+y2x
z+z2y
x
)=
(x− y)(y − z)(x− z)(xy + yz + zx)
xyz,
243
which is obviously nonnegative because x ≥ y ≥ z > 0. Thus, the lastinequality is true and our proof is completed.
?F?
92.1. For every integer n ≥ 2, find the smallest positive number λ = λ(n)
such that if 0 ≤ a1, a2, . . . , an ≤1
2, b1, b2, . . . , bn > 0, and a1 + a2 + · · ·+ an =
b1 + b2 + · · ·+ bn = 1, then
b1b2 · · · bn ≤ λ(a1b1 + a2b2 + · · ·+ anbn).
(China 1992)
Solution: If n = 2, then from 0 ≤ a1, a2 ≤ 12 and a1 +a2 = 1, we deduce that
a1 = a2 =1
2. Therefore
a1b1 + a2b2b1b2
=b1 + b22b1b2
=1
2
(1
b1+
1
b2
)≥ 2
b1 + b2= 2,
and the equality holds for b1 = b2 =1
2. This allows us to conclude that
λ(2) =1
2=
1
2(2− 1)2−1. Now, suppose that n ≥ 3, and let b1 = b2 =
1
2(n− 1),
b3 = · · · = bn =1
n− 1, a1 = a2 =
1
2and a3 = · · · = an = 0, we get
λ(n) ≥ 1
2(n− 1)n−1.
We will show that it is the value we need to find, i.e.
a1b1 + a2b2 + · · ·+ anbn ≥ 2(n− 1)n−1b1b2 · · · bn,
ora1b1 + a2b2 + · · ·+ anbn
a1 + a2 + · · ·+ an≥ 2(n− 1)n−1b1b2 · · · bn.
Without loss of generality, we may assume that b1 ≤ b2 ≤ · · · ≤ bn. By thisassumption, we have
a1b1 + a2b2 + · · ·+ anbna1 + a2 + · · ·+ an
− (a2 + · · ·+ an)b1 + a2b2 + · · ·+ anbn2(a2 + · · ·+ an)
=
=(a2 + · · ·+ an − a1) [a2b2 + · · ·+ anbn − (a2 + · · ·+ an)b1]
2(a2 + · · ·+ an)(a1 + a2 + · · ·+ an)
=(a2 + · · ·+ an − a1) [a2(b2 − b1) + · · ·+ an(bn − b1)]
2(a2 + · · ·+ an)(a1 + a2 + · · ·+ an).
The last quantity is nonnegative since a2(b2− b1) + · · ·+ an(bn− b1) ≥ 0 fromthe assumption and a2 + · · ·+ an− a1 = 1− 2a1 ≥ 0. Therefore, from this, we
244
deduce that
a1b1 + a2b2 + · · ·+ anbna1 + a2 + · · ·+ an
≥ (a2 + · · ·+ an)b1 + a2b2 + · · ·+ anbn2(a2 + · · ·+ an)
=(b2 + b1)a2 + · · ·+ (bn + b1)an
2(a2 + · · ·+ an)
≥ (b2 + b1)a2 + · · ·+ (b2 + b1)an2(a2 + · · ·+ an)
=(b2 + b1)(a2 + · · ·+ an)
2(a2 + · · ·+ an)
=b1 + b2
2≥√b1b2.
The inequality is reduced to proving that√b1b2 ≥ 2(n− 1)n−1b1b2 · · · bn,
or equivalently,
b1b2b23 · · · b2n ≤
1
4(n− 1)2(n−1).
This is true since by the AM-GM Inequality, we have
b1b2b23 · · · b2n =
1
4· (2b1) · (2b2) · b3 · b3 · · · bn · bn
≤ 1
4
(2b1 + 2b2 + b3 + b3 + · · ·+ bn + bn
2(n− 1)
)2(n−1)
=1
4
(b1 + b2 + · · ·+ bn
n− 1
)2(n−1)=
1
4(n− 1)2(n−1).
Our statement is proved. And thus, we conclude that λ(n) =1
2(n− 1)n−1for
every integer n.?F?
92.2. Show thatx4 + y4 + z2 ≥ 2
√2xyz
for all positive reals x, y, z.(Russia 1992)
Solution: Using the AM-GM Inequality, we get
x4 + y4 + z2 ≥ 2x2y2 + z2 ≥ 2√
2x2y2 · z2 = 2√
2xyz,
as desired. Note that the equality holds if and only if x = y and z =√
2x2.?F?
92.3. Show that for any real numbers x, y > 1, we have
x2
y − 1+
y2
x− 1≥ 8.
245
(Russia 1992)
Solution: From the AM-GM Inequality, we have
x2
y − 1+
y2
x− 1≥ 4x2
[1 + (y − 1)]2+
4y2
[1 + (x− 1)]2=
4x2
y2+
4y2
x2≥ 8,
as desired. Note that the equality holds if and only if x = y = 2.?F?
92.4. Positive real numbers a, b, c satisfy a ≥ b ≥ c. Prove that
a2 − b2
c+c2 − b2
a+a2 − c2
b≥ 3a− 4b+ c.
(Ukraine 1992)
Solution: From a ≥ b ≥ c > 0, we havea+ b
c≥ 2,
c+ b
a≤ 2, and
a+ c
b≥ 1.
Now, we get (with noting that a− b ≥ 0, c− b ≤ 0 and a− c ≥ 0)
a2 − b2
c≥ 2(a− b), c2 − b2
a≥ 2(c− b), a2 − c2
b≥ a− c.
After addition of these inequalities, we have
a2 − b2
c+c2 − b2
a+a2 − c2
b≥ 2(a− b) + 2(c− b) + (a− c) = 3a− 4b+ c,
as claimed. Note that the equality holds if and only if a = b = c.?F?
92.5. Let x, y, z, w be positive real numbers. Prove that
12
x+ y + z + w≤∑sym
1
x+ y≤ 3
4
(1
x+
1
y+
1
z+
1
w
).
(United Kingdom 1992)
Solution: By the Cauchy Schwarz Inequality, we have∑sym
1
x+ y≥ 62∑
sym
(x+ y)=
12
x+ y + z + w,
and ∑sym
1
x+ y≤∑sym
(1
4x+
1
4y
)=
3
4
(1
x+
1
y+
1
z+
1
w
).
From these two inequalities, we infer that
12
x+ y + z + w≤∑sym
1
x+ y≤ 3
4
(1
x+
1
y+
1
z+
1
w
),
246
as desired. Note that the equality holds (in both parts) if and only if x = y =z = w.
?F?
93.1. Let a, b, c, d be positive real numbers. Prove that
a
b+ 2c+ 3d+
b
c+ 2d+ 3a+
c
d+ a+ 3b+
d
a+ 2b+ 3c≥ 2
3.
(IMO Shortlist 1993)
Solution: From the AM-GM Inequality, we have
2 (ab+ ac+ ad+ bc+ bd+ cd) ≤ 3(a2 + b2 + c2 + d2
),
and hence, it follows that
(a+ b+ c+ d)2 = a2 + b2 + c2 + d2 + 2 (ab+ ac+ ad+ bc+ bd+ cd)
≥ 8
3(ab+ ac+ ad+ bc+ bd+ cd) .
Now, using the Cauchy Schwarz Inequality in combination with this inequality,we get
a
b+ 2c+ 3d+
b
c+ 2d+ 3a+
c
d+ a+ 3b+
d
a+ 2b+ 3c≥
≥ (a+ b+ c+ d)2
4 (ab+ ac+ ad+ bc+ bd+ cd)≥ 2
3,
as desired. The equality holds if and only if a = b = c = d.?F?
93.2. Let a, b, c ∈ [0, 1]. Prove that
a2 + b2 + c2 ≤ a2b+ b2c+ c2a+ 1.
(Italia 1993)
Solution: The desired inequality is equivalent to
a2 (1− b) + b2 (1− c) + c2 (1− a) ≤ 1.
Since a, b, c ∈ [0, 1], we have
a2 (1− b) + b2 (1− c) + c2 (1− a) ≤ a (1− b) + b (1− c) + c (1− a)
= (a− 1)(b− 1)(c− 1) + 1− abc ≤ 1.
?F?
93.3. Let x, y, u, v be positive real numbers. Prove that
xy + xv + yu+ uv
x+ y + u+ v≥ xy
x+ y+
uv
u+ v.
247
(Poland 1993)
Solution: Put a = x+ y, b = u+ v. Then we have to prove that
xy + xv + uy + uv
a+ b≥ xy
a+uv
b,
or equivalently,
ab (xy + xv + uy + uv)− (a+ b) (bxy + auv) ≥ 0.
The left hand side of the last inequality is equal to (au− bx)(by − av) whichis clearly nonnegative because
(au− bx) (by − av) = [(x+ y)u− (u+ v)x] [(u+ v) y − (x+ y) v]
= (yu− xv)2 ≥ 0.
The proof is completed.?F?
93.4. If the equation x4 + ax3 + 2x2 + bx + 1 = 0 has at least one real root,then a2 + b2 ≥ 8.
(Tournament of the Towns 1993)
First solution: Let x be the real root of the equation. Using the CauchySchwarz Inequality, we infer that
a2 + b2 ≥ (x4 + 2x2 + 1)2
x6 + x2≥ 8,
because the last inequality is equivalent to (x2 − 1)2 ≥ 0.
Second solution: Assume that x0 is the root of the desired equation, then
x40 + ax30 + 2x20 + bx0 + 1 = 0.
According to the two trivial inequalities x20(a+ 2x0)2 ≥ 0 and (bx0 + 2)2 ≥ 0,
we find that
x40 + ax30 ≥ −1
4a20x
20, and b0x0 + 1 ≥ −1
4b20x
20.
Therefore
0 = x40 + ax30 + 2x20 + bx0 + 1 ≥ −1
4a20x
20 + 2x20 −
1
4b20x
20 =
1
4x20(8− a2 − b2).
On the other hand, it is easy to see that x0 6= 0. Thus, from the aboveinequality, we can deduce that
a2 + b2 ≥ 8,
as desired.?F?
248
93.5. Let x1, x2, x3, x4 be real numbers such that
1
2≤ x21 + x22 + x23 + x24 ≤ 1.
Find the largest and the smallest values of the expression
A = (x1 − 2x2 + x3)2 + (x2 − 2x3 + x4)
2 + (x2 − 2x1)2 + (x3 − 2x4)
2.
(Vietnam 1993)
Solution: (1) To find the maxA, we rewrite
A = 5(x21 + x24) + 6(x22 + x2 − 3)− 8(x1x2 + x2x3 + x3x4) + 2x1x3 + 2x2x4.
Now, we try to replace xixj by a sum of squares, so that at the end all squaresx2i appear with the same coefficient. This can be done using the followingremark: For any positive number r, the AM-GM Inequality implies that
• −8x1x2 ≤ rx21 +16
rx22 with equality iff rx1 = −4x2;
• −8x2x3 ≤ 4(x22 + x23) with equality iff x2 = −x3;• −8x3x4 ≤ rx24 +
16
rx23 with equality iff rx4 = −4x3;
• 2x1x3 ≤r
4x21 +
4
rx23 with equality iff rx1 = 4x3;
• 2x2x4 ≤4
rx22 +
r
4x24 with equality iff rx4 = 4x2.
Therefore
A ≤(
5 +5r
4
)(x21 + x24) +
(10 +
20
r
)(x22 + x23).
Now, by choosing r > 0 such that 5 +5r
4= 10 +
20
r, or equivalently r =
2(1 +√
5), we get
A ≤5(3 +√
5)
2.
The equality holds if and only if the above conditions on xi can be simulta-neously satisfied and x21 + x22 + x23 + x24 = 1. One can verify that A reaches itsmaximum at two points
x1 = ±1
2
√5−√
5
5, x2 = −1 +
√5
2x1, x3 =
1 +√
5
2x1, x4 = −x1.
(2) For finding the minimum, we follow the same idea (but more complicated).Fix a number r and rewrites
A = (x1 − rx2 + x3)2 + (x2 − rx3 + x4)
2 − 2(4− r)x1x2 − 4(2− r)x2x3− 2(4− r)x3x4 + 4(x21 + x24) + (5− r2)(x22 + x23).
By the freedom, we may assume r < 2. Then for any s > 0, we have
• −2(4− r)x1x2 ≥ −(4− r)(sx21 +
x22s
)with equality iff sx1 = x2;
249
• −2(4− r)x3x4 ≥ −(4− r)(sx24 +
x23s
)with equality iff sx4 = x3;
• −4(2− r)x2x3 ≥ −2(2− r)(x22 + x23) with equality iff x2 = x3.Hence
A ≥ −2(4− r)x1x2 − 4(2− r)x2x3 − 2(4− r)x3x4 + 4(x21 + x24) + (5− r2)(x22 + x23)
≥ [4− (4− r)s](x21 + x24) +
(1 + 2r − 4− r
s− r2
)(x22 + x23).
Choosing s such that
4− (4− r)s = 1 + 2r − 4− rs− r2, (∗)
we get
A ≥ 4− (4− r)s2
.
The equality holds if and only if
sx1 = x2
sx4 = x3x2 = x3x1 − rx2 + x3 = 0x2 − rx3 + x4 = 0
x21 + x22 + x23 + x24 =1
2
⇔
x2 = x3x1 = x4x1 = (r − 1)x2
x1 =e
2√
1 + s2(e = ±1)
r − 1 =1
s
.
From (∗) and r − 1 =1
s, we get
4 +
(1
s− 3
)s = 2− 1
s2+
1
s
(1
s− 3
),
which is equivalent to s2 − s− 1 = 0, i.e. s =1 +√
5
2. Then
x1 =e√
10 + 2√
5, x2 =
1 +√
5
2x1.
Thus minA =7− 3
√5
4and it is attained at
x1 = x4 = ± 1√10 + 2
√5, x2 = x3 = ± 1 +
√5
2√
10 + 2√
5.
?F?
94.1. Let a1, a2, . . . , an be a sequence of positive real numbers satisfyinga1 + · · ·+ ak ≥
√k for all k = 1, 2, . . . , n. Prove that
a21 + a22 + · · ·+ a2n >1
4
(1 +
1
2+ · · ·+ 1
n
).
250
(USA 1994)
Solution: Denote bi =√i−√i− 1 for i = 1, 2, . . . , n, then we have a1 + · · ·+
ak ≥ b1 + · · ·+ bk for every k = 1, 2, . . . , n. We claim that
a21 + a22 + · · ·+ a2n ≥ b21 + b22 + · · ·+ b2n.
Indeed, from the Cauchy Schwarz Inequality, we have(n∑i=1
a2i
)(n∑i=1
b2i
)≥
(n∑i=1
aibi
)2
.
Therefore, in order to prove our claim, it suffices to prove that
n∑i=1
aibi ≥n∑i=1
b2i .
Now, using the Abel’s Summation formula, we have
n∑i=1
aibi−n∑i=1
b2i =n∑i=1
bi(ai− bi) =n−1∑i=1
(bi− bi+1)i∑
k=1
(ak− bk) + bn
n∑i=1
(ai− bi).
Since
i∑k=1
(ak − bk) ≥ 0 for all i = 1, 2, . . . , n, and
bi − bi+1 = 2√i−(√
i+ 1 +√i− 1
)≥ 2√i−√
(1 + 1)(i+ 1 + i− 1) = 0,
we see that each element of the above sum is nonnegative and hence, it is clearthat
n∑i=1
aibi ≥n∑i=1
b2i .
This proves the claim that stated
n∑i=1
a2i ≥n∑i=1
b2i .
On the other hand, we have
n∑i=1
b2i =n∑i=1
(√i−√i− 1
)2=
n∑i=1
1(√i+√i− 1
)2>
n∑i=1
1(√i+√i)2 =
1
4
(1 +
1
2+ · · ·+ 1
n
).
Therefore, from this and the above inequality, we deduce that
a21 + a22 + · · ·+ a2n >1
4
(1 +
1
2+ · · ·+ 1
n
),
251
as desired.?F?
95.1. Prove that for any positive real numbers x, y and any positive integersm,n,
(m−1)(n−1)(xm+n + ym+n
)+(m+ n− 1) (xmyn + xnym) ≥ mn
(xm+n−1y + xym+n−1) .
(Austrian-Polish Competition 1995)
First solution: The inequality is equivalent to
(m− 1)(n− 1)(xm+n + ym+n − xmyn − xnym
)≥
≥ mn(xm+n−1y + xym+n−1 − xmyn − xnym
),
or
(m− 1)(n− 1)(xm − ym)(xn − yn) ≥ mnxy(xm−1 − ym−1)(xn−1 − yn−1).
Due to symmetry, we may assume that x ≥ y > 0. Then, we find that theabove inequality follows from multiplying the following inequalities
(m− 1)(xm − ym) ≥ m√xy(xm−1 − ym−1),
and(n− 1)(xn − yn) ≥ n√xy(xn−1 − yn−1).
So, it suffices to prove that these two inequalities holds. Let us first provethe former inequality. Denote x = t2y where t ≥ 1, then we can rewrite thisinequality as
f(t) = (m− 1)(t2m − 1)−mt(t2m−2 − 1) ≥ 0.
We have
f ′(t) = m[(2m− 2)t2m−1 − (2m− 1)t2m−2 + 1
]≥ 0,
since by the AM-GM Inequality, it follows that
(2m− 2)t2m−1 + 1 = (2m− 2) · t2m−1 + 1 · 1 ≥ (2m− 1)(t2m−1
) 2m−22m−1 · 1
12m−1
= (2m− 1)t2m−2.
This means that f(t) is increasing for all t ≥ 1 and thus, it is clear thatf(t) ≥ f(1) = 0 for any t ≥ 1. This proves the first inequality. Similarly, wecan prove that the second inequality holds, and it ends our proof. Note thatthe equality holds if and only if x = y or m = 1 or n = 1.
Second solution: If x = y, then it is trivial. Conversely, in case x 6= y, wetransform the inequality as follows
mn(x− y)(xm+n−1 − ym+n−1) ≥ (m+ n− 1)(xm − ym)(xn − yn),
252
or equivalently,
xm+n−1 − ym+n−1
(m+ n− 1)(x− y)≥ xm − ym
m(x− y)· x
n − yn
n(x− y).
(we have assumed that x > y). The last relation can also be written
(x− y)
x∫y
tm+n−2dt ≥x∫y
tm−1dt ·x∫y
tn−1dt,
and this follows from the Chebyshev’s Inequality for integrals.?F?
95.2. Let a, b, c, d be positive real numbers. Prove that
a+ c
a+ b+b+ d
b+ c+c+ a
c+ d+d+ b
d+ a≥ 4.
(Baltic Way 1995)
Solution: According to the AM-GM Inequality, we have
a+ c
a+ b+b+ d
b+ c+c+ a
c+ d+d+ b
d+ a=
=(a+ c) (a+ b+ c+ d)
(a+ b) (c+ d)+
(b+ d) (a+ b+ c+ d)
(b+ c) (d+ a)
≥ (a+ b+ c+ d)
[4 (a+ c)
[(a+ b) + (c+ d)]2+
4 (b+ d)
[(b+ c) + (d+ a)]2
]= 4 · a+ b+ c+ d
a+ b+ c+ d= 4,
as desired. Note that the equality holds if and only if a = c and b = d.?F?
95.3. Let x, y, z be positive real numbers. Prove that
xxyyzz ≥ (xyz)x+y+z
3 .
(Canada 1995)
Solution: The original inequality can be written as
x lnx+ y ln y + z ln z ≥ 1
3(x+ y + z)(lnx+ ln y + ln z).
Due to symmetry, we may assume that x ≥ y ≥ z. Since the function f(x) =lnx is increasing for any positive reals x, this assumption also implies thatlnx ≥ ln y ≥ ln z. Therefore, we can see that the above inequality is a directconsequence of the Chebyshev’s Inequality. Note that the equality holds ifand ony if x = y = z.
?F?
253
95.4. If a, b, c are positive real numbers such that abc = 1, then
1
a3 (b+ c)+
1
b3 (c+ a)+
1
c3 (a+ b)≥ 3
2.
(IMO 1995)
Solution: Applying the AM-GM Inequality, we have
1
a3 (b+ c)+b+ c
4bc≥ 2
√b+ c
4a3bc (b+ c)=
1
a.
Adding this to the two analogous inequalities, we get
1
a3 (b+ c)+
1
b3 (c+ a)+
1
c3 (a+ b)+
1
4
(b+ c
bc+c+ a
ca+a+ b
ab
)≥ 1
a+
1
b+
1
c.
Therefore
1
a3 (b+ c)+
1
b3 (c+ a)+
1
c3 (a+ b)≥ 1
2
(1
a+
1
b+
1
c
)≥ 3
23
√1
abc=
3
2.
The proof is completed. Note that the equality holds if and only if a = b =c = 1.
?F?
95.5. Suppose that x1, x2, . . . , xn are real numbers satisfying |xi − xi+1| < 1and xi ≥ 1 for i = 1, 2, . . . , n (xn+1 = x1). Prove the following inequality
x1x2
+x2x3
+ · · ·+ xnx1
< 2n− 1.
(India 1995)
First solution: Sincex1x2· x2x3· · · xn
x1= 1, there exists an index k such that
xkxk+1
< 1. On the other hand, since |xi−xi+1| < 1 and xi ≥ 1 for all i, we find
that xi < xi+1+1 ≤ 2xi+1, and thusxixi+1
< 2 for all i. From these arguments,
we get
x1x2
+x2x3
+ · · ·+ xnx1
=∑i 6=k
xixi+1
+xkxk+1
< 2(n− 1) + 1 = 2n− 1,
as desired.
Second solution: Similar to the preceding solution, one may prove thatxi < 2xi+1 for all i. Now, we rewrite the inequality as(
2− x1x2
)+
(2− x2
x3
)+ · · ·+
(2− xn
x1
)> 1,
orn∑i=1
2xi+1 − xixi+1
> 1.
254
Since 2xi+1 − xi > 0, the Cauchy Schwarz Inequality implies
n∑i=1
2xi+1 − xixi+1
≥
(n∑i=1
xi
)2
n∑i=1
xi+1(2xi+1 − xi)=
(n∑i=1
xi
)2
2
n∑i=1
x2i −n∑i=1
xixi+1
>
n∑i=1
x2i +
n∑i=1
xixi+1
2n∑i=1
x2i −n∑i=1
xixi+1
>
n∑i=1
x2i +
n∑i=1
xi ·xi2
2n∑i=1
x2i −n∑i=1
xi ·xi2
= 1.
This proves our inequality.?F?
95.6. (a) Find the maximum value of the expression x2y−y2x when 0 ≤ x ≤ 1and 0 ≤ y ≤ 1.
(b) Find the maximum value of the expression x2y+y2z+z2x−y2x−z2y−x2zwhen 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1.
(United Kingdom 1995)
Solution: (a) By the AM-GM Inequality, we have
x2y − y2x = xy(x− y) ≤ xy(1− y) ≤ y(1− y) ≤(y + 1− y
2
)2
=1
4,
and the equality holds when x = 1 and y =1
2. Therefore, the maximum value
of the desired expression is1
4.
(b) Due to the cyclicity, we may assume that x = max {x, y, z} . Then, applyingthe AM-GM Inequality, we find that
x2y + y2z + z2x− y2x− z2y − x2z = (x− y)(x− z)(y − z)≤ (x− y)(x− z)y ≤ xy(x− y)
≤ x ·(y + x− y
2
)2
=x3
4≤ 1
4.
The equality holds for example when x = 1, y =1
2and z = 0. Therefore, the
maximum value of the desired expression is1
4.
?F?
95.7. Let a, b, c be real numbers satisfying a < b < c, a + b + c = 6 andab+ bc+ ca = 9. Prove that
0 < a < 1 < b < 3 < c < 4.
255
(United Kingdom 1995)
Solution: From the given condition, we have a < 2 < c. Now, let us provethe following chain of identities
3(a− 1)(a− 3)− (a− b)(a− c) = 3(b− 1)(b− 3)− (b− c)(b− a)
= 3(c− 1)(c− 3)− (c− a)(c− b) = 0.
Indeed, we have
0 = 9− ab− bc− ca = 9 + a2 − 2a(b+ c)− (a− b)(a− c)= 9 + a2 − 2a(6− a)− (a− b)(a− c)= 3(a− 1)(a− 3)− (a− b)(a− c).
Since a < b < c, it is clear that
(a− b)(a− c) > 0, (b− c)(b− a) < 0, (c− a)(c− b) > 0.
And hence, the above identities imply that (a−1)(a−3) > 0, (b−1)(b−3) < 0and (c − 1)(c − 3) > 0, from which we deduce that (with noting a < 2 < c)a < 1 < b < 3 < c. This yields that (1− a)(1− b) < 0 and (3− b)(3− c) < 0.Therefore, according to the identities
0 = 9− ab− bc− ca = 10− (c+ 1)(a+ b)− (1− a)(1− b)= 10− (c+ 1)(6− c)− (1− a)(1− b)= (c− 1)(c− 4)− (1− a)(1− b),
and
0 = 9− ab− bc− ca = 18− (a+ 3)(b+ c)− (3− b)(3− c)= 18− (a+ 3)(6− a)− (3− b)(3− c)= a(a− 3)− (3− b)(3− c),
we infer that (c− 1)(c− 4) < 0 and a(a− 3) < 0, from which we obtain c < 4and a > 0. Thus, from these arguments, we conclude that
0 < a < 1 < b < 3 < c < 4,
as desired.?F?
95.8. Find the greatest constant k such that for any positive integer n whichis not a square, ∣∣(1 +
√n)
sin(π√n)∣∣ > k.
(Vietnam (IMO training camp) 1995)
Solution: We will prove thatπ
2is the best constant. For any positive integer
i, we must have
k <(
1 +√i2 + 1
) ∣∣∣sin(π√i2 + 1)∣∣∣ .
256
Because∣∣∣sin(π√i2 + 1
)∣∣∣ = sinπ√
i2 + 1 + i, we deduce that
π√i2 + 1 + i
≥∣∣∣sin(π√i2 + 1
)∣∣∣ > k√i2 + 1 + 1
,
from which it follows that k ≤ π
2. Now, let us prove that the constant
π
2is
good. Clearly, the inequality can be rewritten as
sin(π{√
n})
>π
2 (√n+ 1)
.
We have two cases
Case 1. {n} ≤ 1
2. Of course,
{√n}≥√n−√n− 1 =
1√n+√n− 1
,
and because sinx ≥ x− x3
6, we find that
sin(π{√
n})≥ sin
π√n+√n− 1
≥ π√n− 1 +
√n− 1
6
(π√
n− 1 +√n
)3
.
Let us prove that the last quantity is at leastπ
2 (1 +√n). This comes down to
2 +√n−√n− 1
1 +√n
>π2
3(√n+√n− 1
)2 ,or
6(√n+√n− 1
)2+ 3
(√n+√n− 1
)> π2
(1 +√n),
and it is clear.
Case 2. {n} > 1
2. Let x = 1 − {
√n} < 1
2and let n = k2 + p, 1 ≤ p ≤ 2k.
Because {n} > 1
2, we have p ≥ k + 1. Then it is easy to see that x ≥
1
k + 1 +√k2 + 2k
and so, it suffices to prove that
sinx
k + 1 +√k2 + 2k
≥ π
2(
1 +√k2 + k
) .Using again the inequality sinx ≥ x− x3
6, we infer that
2√k2 + k −
√k2 + 2k − k + 1
1 +√k2 + k
>π2
3(
1 + k +√k2 + 2k
)2 .But from the Cauchy Schwarz Inequality, we have 2
√k2 + k−
√k2 + 2k−k ≥ 0.
And because the inequality(
1 + k +√k2 + 2k
)2>π2
3
(1 +√k2 + k
)holds,
this case is also solved.
257
?F?
96.1. Let m and n be positive integers such that n ≤ m. Prove that
2n · n! ≤ (m+ n)!
(m− n)!≤ (m2 +m)n.
(APMO 1996)
Solution: The quantity in the middle is (m+n)(m+n−1) · · · (m−n+ 1). Ifwe pair of terms of the form (m+ x) and (m+ 1− x), we get products whichdo not exceed m(m + 1), since the function f(x) = (m + x)(m + 1 − x) is a
concave parabola with maximum at x =1
2. From this, the right inequality
follows. For the left, we need only show (m + x)(m + 1 − x) ≥ 2x for x ≤ n,this rearranges to (m− x)(m+ 1 + x) ≥ 0, which holds because m ≥ n ≥ x.
?F?
96.2. Let a, b, c be the lengths of the sides of a triangle. Prove that
√a+ b− c+
√b+ c− a+
√c+ a− b ≤
√a+√b+√c,
and determine when equality occurs.(APMO 1996)
Solution: By the triangle inequality, b+ c− a and c+ a− b are positive. Forany positive x, y, we have
2(x+ y) ≥ x+ y + 2√xy =
(√x+√y)2
by the AM-GM Inequality, with equality for x = y. Substituting x = a+ b−c, y = b+ c− a, we get
√a+ b− c+
√b+ c− a ≤ 2
√a,
which added to the two analogous inequalities yields the desired result. Theequality holds for a+ b− c = b+ c− a = c+ a− b, i.e. a = b = c.
?F?
96.3. The real numbers x, y, z, t satisfy the equalities x + y + z + t = 0 andx2 + y2 + z2 + t2 = 1. Prove that
−1 ≤ xy + yz + zt+ tx ≤ 0.
(Austrian-Polish Competition 1996)
Solution: The inner expression is (x + z)(y + t) = −(x + z)2, so the secondinequality is obvious. As for the first, note that
1 = (x2 + z2) + (y2 + t2) ≥ 1
2[(x+ z)2 + (y + t)2] ≥ |(x+ z)(y + t)|
by two applications of the power mean inequality.
258
?F?
96.4. Let a, b, c be positive real numbers, such that a+ b+ c =√abc. Prove
thatab+ bc+ ca ≥ 9(a+ b+ c).
(Belarus 1996)
Solution: Applying the well-known inequality (x+ y+ z)2 ≥ 3(xy+ yz+ zx)with x = ab, y = bc, z = ca, we get
(ab+ bc+ ca)2 ≥ 3abc(a+ b+ c) = 3(a+ b+ c)3,
and thusab+ bc+ ca ≥ (a+ b+ c)
√3(a+ b+ c).
On the other hand, the AM-GM Inequality implies
a+ b+ c =√abc ≤
√(a+ b+ c
3
)3
,
from which we deduce that√
3(a+ b+ c) ≥ 9. Therefore
ab+ bc+ ca ≥ (a+ b+ c)√
3(a+ b+ c) ≥ 9(a+ b+ c),
as claimed. Note that the equality holds if and only if a = b = c =√
3.?F?
96.5. Suppose n ∈ N, x0 = 0, xi > 0 for i = 1, 2, . . . , n, and x1+x2+· · ·+xn =1. Prove that
1 ≤n∑i=1
xi√1 + x0 + · · ·+ xi−1 ·
√xi + · · ·+ xn
<π
2.
(China 1996)
Solution: The left inequality follows from the fact that√1 + x0 + · · ·+ xi−1 ·
√xi + · · ·+ xn ≤
1
2(1 + x0 + · · ·+ xn) = 1,
so that the middle quantity is at least x1+· · ·+xn = 1. For the right inequality,let θi = arcsin(x0 + · · ·+ xi) for i = 0, . . . , n, so that√
1 + x0 + · · ·+ xi−1 ·√xi + · · ·+ xn = cos θi−1,
and the desired inequality is
n∑i=1
sin θi − sin θi−1cos θi−1
<π
2.
Now note that
sin θi − sin θi−1 = 2 cosθi + θi−1
2sin
θi − θi−12
< (θi − θi−1) cos θi−1,
259
using the facts that θi−1 < θi and that sinx < x for x > 0, so that
n∑i=1
sin θi − sin θi−1cos θi−1
<
n∑i=1
(θi − θi−1) = θn − θ0 <π
2,
as claimed.?F?
96.6. (a) Find the minimum value of xx for x a positive real number.
(b) If x and y are positive real numbers, show that xy + yx > 1.
(France 1996)
First solution: (a) Since xx = ex log x and ex is an increasing function of x, itsuffices to determine the minimum of x log x. This is easily done by setting its
derivative 1+log x to zero, yielding x =1
e. The second derivative
1
xis positive
for x > 0, so the function is everywhere convex, and the unique extremum isindeed a global minimum. Hence xx has minimum value e−
1e .
(b) If x ≥ 1, then xy ≥ 1 for y > 0, so we may assume 0 < x, y < 1. Withoutloss of generality, assume x ≤ y, now note that the function f(x) = xy + yx
has derivative f ′(x) = xy log x + yx−1. Since yx ≥ xx ≥ xy for x ≤ y and1
x≥ − log x, we see that f ′(x) > 0 for 0 ≤ x ≤ y and so the minimum of
f occurs with x = 0, in which case f(x) = 1, since x > 0, we have strictinequality.
Second solution: We present another way to prove (b). Firstly, we will
prove that ab ≥ a
a+ b− abfor any a, b ∈ (0, 1). Indeed, from the Bernoull’s
Inequality, it follows that a1−b = [1+(a−1)]1−b ≤ 1+(a−1)(1−b) = a+b−ab,and thus the conclusion follows. Now, if x or y is at least 1, we are done.Otherwise, let 0 < x, y < 1. In this case, we apply the above observation andfind that
xy + yx ≥ x
x+ y − xy+
y
x+ y − xy>
x
x+ y+
y
x+ y= 1.
?F?
96.7. Let a, b, c be positive real numbers such that abc = 1. Prove that
ab
a5 + b5 + ab+
bc
b5 + c5 + bc+
ca
c5 + a5 + ca≤ 1.
(IMO Shortlist 1996)
Solution: For any positive numbers x, y, we have (x3− y2)(x2− y2) ≥ 0, andhence x5 + y5 ≥ x2y2(x+ y). According to this inequality, we have
ab
a5 + b5 + ab=
a2b2c
a5 + b5 + a2b2c≤ a2b2c
a2b2 (a+ b) + a2b2c=
c
a+ b+ c.
260
Adding this to the two analogous inequalities, we get the desired result. Notethat the equality holds if and ony if a = b = c = 1.
?F?
96.8. Let a and b be positive real numbers with a+ b = 1. Prove that
a2
a+ 1+
b2
b+ 1≥ 1
3.
(Hungary 1996)
First solution: Using the condition a + b = 1, we can reduce the giveninequality to homogeneous one
1
3≤ a2
(a+ b)[a+ (a+ b)]+
b2
(a+ b)[b+ (a+ b)],
ora2b+ ab2 ≤ a3 + b3,
which follows from
(a3 + b3)− (a2b+ ab2) = (a− b)2(a+ b) ≥ 0.
The equality holds if and only if a = b =1
2.
Second solution: By the Cauchy Schwarz Inequality, we have that
a2
a+ 1+
b2
b+ 1≥ (a+ b)2
a+ 1 + b+ 1=
1
3,
as desired.?F?
96.9. Prove the following inequality for positive real numbers x, y, z,
(xy + yz + zx)
[1
(y + z)2+
1
(z + x)2+
1
(x+ y)2
]≥ 9
4.
(Iran 1996)
First solution: Without loss of generality, we may assume that x ≥ y ≥ z.Denote
P (x, y, z) =1
(y + z)2+
1
(z + x)2+
1
(x+ y)2− 9
4(xy + yz + zx).
We have to show that P (x, y, z) ≥ 0. Note that
P (t, t, z) =z(z − t)2
2t2(2z + t)(z + t)2≥ 0
for any positive reals t ≥ z > 0. Therefore, a key to solve this problem is tofind a suitable number t ≥ z such that
P (x, y, z) ≥ P (t, t, z).
261
We will prove that the number t =x+ y
2satisfies this property, i.e.
1
(x+ z)2+
1
(x+ z)2− 2
(t+ z)2≥ 9
4(xy + yz + zx)− 9
4(t2 + 2zt).
With noting that
1
(x+ z)2+
1
(x+ z)2− 2
(t+ z)2=
(1
x+ z− 1
y + z
)2
+2
(x+ z)(y + z)− 2
(t+ z)2
=(x− y)2
(x+ z)2(y + z)2+
(x− y)2
2(x+ z)(y + z)(t+ z)2,
and
9
4(xy + yz + zx)− 9
4(t2 + 2zt)=
9(x− y)2
16(t2 + 2tz)(xy + yz + zx),
we can rewrite the above inequality as
(x− y)2
(x+ z)2(y + z)2+
(x− y)2
2(x+ z)(y + z)(t+ z)2≥ 9(x− y)2
16(t2 + 2tz)(xy + yz + zx),
or equivalently,
1
(x+ z)2(y + z)2+
1
2(x+ z)(y + z)(t+ z)2≥ 9
16(xy + yz + zx)(t2 + 2tz).
We have 4(xy + yz + zx)− 3(x+ z)(y + z) = xy + yz + zx− 3z2 ≥ 0, then
(x+ z)(y + z) ≤ 4
3(xy + yz + zx),
and hence, it follows that
1
(x+ z)2(y + z)2≥ 9
16(xy + yz + zx)2≥ 9
16(xy + yz + zx)(t2 + 2zt).
Therefore, the last inequality is valid and our proof is completed. Note thatthe equality holds if and only if x = y = z.
Second solution: Due to symmetry, we can assume that x ≥ y ≥ z > 0.
Now, sincexy + yz + zx
(y + z)2=
x
y + z+
yz
(y + z)2, the original inequality can be
written as
x
y + z+
y
z + x+
z
x+ y+
xy
(x+ y)2+
yz
(y + z)2+
zx
(z + x)2≥ 9
4.
Sincex
y + z+
y
z + x= (x + y + z)
(1
x+ z+
1
y + z
)− 1, this inequality is
equivalent to
(x+ y+ z)
(1
x+ z+
1
y + z
)+
z
x+ y+
xy
(x+ y)2+
yz
(y + z)2+
zx
(z + x)2≥ 17
4,
262
or
(x+ y + z)
(1
x+ z+
1
y + z− 4
x+ y + 2z
)+
4(x+ y + z)
x+ y + 2z+
z
x+ y+
+xy
(x+ y)2+
yz
(y + z)2+
zx
(z + x)2≥ 17
4.
From this, it is not hard for us to write the inequality in the form
M(x− y)2 +Nz ≥ 0,
where
M =x+ y + z
(x+ z)(y + z)(x+ y + 2z)− 1
4(x+ y)2
≥ x+ y + z
(x+ y)(y + x)(x+ y + 2z)− 1
4(x+ y)2
=3x+ 3y + 2z
4(x+ y)2(x+ y + 2z)≥ 0,
and
N =1
x+ y+
x
(x+ z)2+
y
(y + z)2− 4
x+ y + 2z
=
(1
x+ z+
1
y + z− 4
x+ y + 2z
)+
1
x+ y− z
(x+ z)2− z
(y + z)2
=(x− y)2
(x+ z)(y + z)(x+ y + 2z)+
1
x+ y− z
(x+ z)2− z
(y + z)2.
Sincez
(x+ z)2− y
(x+ y)2= −(y − z)(x2 − yz)
(x+ y)2(x+ z)2≤ 0,
we deduce that
N ≥ (x− y)2
(x+ y)(y + y)(x+ y + 2y)+
1
x+ y− y
(x+ y)2− z
4yz
=(x− y)2
2y(x+ y)(x+ 3y)− (x− y)2
4y(x+ y)2=
(x− y)3
4y(x+ y)2(x+ 3y)≥ 0.
Therefore, in the last inequality, each element of it is nonnegative, and hence,the result follows.
?F?
96.10. Let n ≥ 2 be a fixed natural number and let a1, a2, . . . , an be positivenumbers whose sum is 1. Prove that for any positive numbers x1, x2, . . . , xnwhose sum is 1,
2∑
1≤i<j≤nxixj ≤
n− 2
n− 1+
n∑i=1
aix2i
1− ai,
and determine when equality holds.(Poland 1996)
263
Solution: The left side is 1−n∑i=1
x2i , so we can rewrite the desired result as
1
n− 1≤
n∑i=1
x2i1− ai
.
By the Cauchy Schwarz Inequality,(n∑i=1
x2i1− ai
)[n∑i=1
(1− ai)
]≥
(n∑i=1
xi
)2
= 1.
Sincen∑i=1
(1− ai) = n− 1, we have the desired result.
?F?
96.11. Let a, b, c be real numbers such that a+ b+ c = 1. Prove that
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 9
10.
(Poland 1996)
First solution: In order to prove this inequality, we shall consider two cases:
Case 1. If min {a, b, c} ≥ −3
4, then we have
36a+ 3
50− a
a2 + 1=
(4a+ 3)(3a− 1)2
50(a2 + 1)≥ 0,
from which it follows thata
a2 + 1≤ 36a+ 3
50, and so
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 36
50(a+ b+ c) +
9
50=
36
50+
9
50=
9
10.
Case 2. If min {a, b, c} < −3
4, without loss of generality, we can assume that
c < −3
4. From the AM-GM Inequality, we have
b
b2 + 1≤ 1
2, and thus, the
initial inequality is proved if we can show that
a
a2 + 1≤ 2
5.
If a ≤ 1
2or a ≥ 2, we are done. So, it suffices to consider the case 2 ≥ a ≥ 1
2.
Similarly, we can show separately that it is enough to show the case when
2 ≥ b ≥ 1
2. Therefore, we can combine these two remarks and can resume
to proving the problem under the condition 2 ≥ a, b ≥ 1
2. In this case,
−3
4> c = 1− a− b ≥ −3, which means that
c
c2 + 1≤ − 3
10. Hence
a
a2 + 1+
b
b2 + 1+
c
c2 + 1≤ 1
2+
1
2− 3
10=
7
10<
9
10.
264
This completes our proof. The equality holds if and only if a = b = c =1
3.
Second solution: Firstly, we note that at least two of the three numbers a, b,
and c are both greater than or equal to1
3or less than or equal to
1
3. Without
loss of generality, we assume that the numbers with this property are a and b.Then we have (3a− 1)(3b− 1) ≥ 0. According to this, we infer that
a2 + b2 = (a+ b)2 − 2ab = (a+ b)2 − 2
3
(a+ b− 1
3
)− 2
9(3a− 1) (3b− 1)
≤ (1− c)2 − 2
3
(1− c− 1
3
)=
9c2 − 12c+ 5
9.
Now, we rewrite our inequality as(1− 2a
a2 + 1
)+
(1− 2b
b2 + 1
)≥ 1
5+
2c
c2 + 1,
or(a− 1)2
a2 + 1+
(b− 1)2
b2 + 1≥ c2 + 10c+ 1
5(c2 + 1).
Applying the Cauchy Schwarz Inequality and the above note, we obtain
(a− 1)2
a2 + 1+
(b− 1)2
b2 + 1≥ (a+ b− 2)2
a2 + b2 + 2=
(c+ 1)2
9c2 − 12c+ 5
9+ 2
=9(c+ 1)2
9c2 − 12c+ 23.
Therefore, it suffices to prove that
9(c+ 1)2
9c2 − 12c+ 23≥ c2 + 10c+ 1
5(c2 + 1).
By some easy computations, we find that the last inequality can be simplifiedinto
2(2c2 + 2c+ 11)(3c− 1)2
5(9c2 − 12c+ 23)(c2 + 1)≥ 0,
which is true since 2c2+2c+11 > 0 and 9c2−12c+23 > 0 for any real numberc.
?F?
96.12. Let x1, x2, . . . , xn, xn+1 be positive reals such that x1 +x2 + · · ·+xn =xn+1. Prove that
n∑i=1
√xi(xn+1 − xi) ≤
√√√√ n∑i=1
xn+1(xn+1 − xi).
(Romania 1996)
First solution: First, note that
n∑i=1
xn+1(xn+1 − xi) = nx2n+1 − xn+1
n∑i=1
xi = (n− 1)x2n+1.
265
Hence, the given inequality may be rewritten as
n∑i=1
√√√√ xixn+1
·1− xi
xn+1
n− 1≤ 1.
On the other hand, by the AM-GM Inequality, the left side is at most
n∑i=1
xi2xn+1
+
1− xixn+1
2(n− 1)
=1
2+
n− 1
2(n− 1)= 1.
Second solution: From the Cauchy Schwarz Inequality, we see that
n∑i=1
√xi(xn+1 − xi) ≤
√√√√n
n∑i=1
xi(xn+1 − xi) =
√√√√nxn+1
n∑i=1
xi − nn∑i=1
x2i
≤
√√√√nxn+1
n∑i=1
xi −
(n∑i=1
xi
)2
=√
(n− 1)x2n+1.
On the other hand, we have√√√√ n∑i=1
xn+1(xn+1 − xi) =
√√√√nx2n+1 − xn+1
n∑i=1
xi =√
(n− 1)x2n+1.
Therefore, we conclude that
n∑i=1
√xi(xn+1 − xi) ≤
√√√√ n∑i=1
xn+1(xn+1 − xi),
as desired.?F?
96.13. Let x, y, z be real numbers. Prove that the following conditions areequivalent
(i) x > 0, y > 0, z > 0 and1
x+
1
y+
1
z≤ 1;
(ii) for every quadrilateral with sides a, b, c, d, a2x+ b2y + c2z > d2.(Romania 1996)
Solution: (i)⇒ (ii). Applying the Cauhy Schwarz Inequality, we have
a2x+ b2y + c2z ≥ (a2x+ b2y + c2z)
(1
x+
1
y+
1
z
)≥ (a+ b+ c)2 > d2.
(ii) ⇒ (i). If x ≤ 0, then by taking a quadrilateral with sides a = n, b =1, c = 1, d = n, we get y + z > n2(1 − x), which is impossible for large n.
266
Therefore x > 0, and in the same way y, z > 0. Using now a quadrilateral
with sides a =1
x, b =
1
y, c =
1
z, d =
1
x+
1
y+
1
z− 1
n(where n is sufficiently
large), one has1
x2· x+
1
y2· y+
1
z2· z >
(1
x+
1
y+
1
z− 1
n
)2
, that is1
x+
1
y+
1
z>
(1
x+
1
y+
1
z− 1
n
)2
for every sufficiently large n, whence1
x+
1
y+
1
z≥(
1
x+
1
y+
1
z
)2
, and therefore1
x+
1
y+
1
z≤ 1.
?F?
96.14. Let a, b, c be positive real numbers.(a) Prove that 4(a3 + b3) ≥ (a+ b)3.(b) Prove that 9(a3 + b3 + c3) ≥ (a+ b+ c)3.
(United Kingdom 1996)
Solution: Both parts follow from the Power Mean Inequality: for r > 1 andx1, . . . , xn positive, (
xr1 + · · ·+ xrnn
) 1r
≥ x1 + · · ·+ xnn
,
which in turn follows from the Jensen’s Inequality applied to the convex func-tion xr.
?F?
96.15. Let a, b, c, d be positive real numbers such that
2(ab+ ac+ ad+ bc+ bd+ cd) + abc+ bcd+ cda+ dab = 16.
Prove that
a+ b+ c+ d ≥ 2
3(ab+ ac+ ad+ bc+ bd+ cd).
(Vietnam 1996)
Solution: By applying the Rolle’s Theorem, we see that exist positive realnumbers x, y, z such that
a+ b+ c+ d =4
3(x+ y + z)
ab+ ac+ ad+ bc+ bd+ cd = 2(xy + yz + zx)abc+ bcd+ cda+ dab = 4xyz
.
Therefore, from the given hypothesis, we have xy + yz + zx + xyz = 4, andwe need to prove that
x+ y + z ≥ xy + yz + zx.
According to the Schur’s Inequality (applied for third degree), we have
(x+ y + z)3 + 9xyz ≥ 4(x+ y + z)(xy + yz + zx).
267
Multiplying both sides of this inequality by(xy + yz + zx)2
x+ y + z, we get
(x+ y + z)2(xy + yz + zx)2 +9xyz(xy + yz + zx)2
x+ y + z≥ 4(xy + yz + zx)3.
On the other hand, the AM-GM Inequality implies that
9xyz(xy + yz + zx)2
x+ y + z≤ xyz(x+ y + z)3.
Therefore, from the above inequality, we deduce that
(x+ y + z)2(xy + yz + zx)2 + xyz(x+ y + z)3 ≥ 4(xy + yz + zx)3.
Setting p = x+ y + z, q = xy + yz + zx, this inequality is equivalent to
p2q2 + (4− q)p3 ≥ 4q3,
or
(p− q)[(4− q)p2 + 4pq + 4q2
]≥ 0.
Sine 4 ≥ q, it is clear that (4−q)p2+4pq+4q2 > 0, and hence, we conclude thatp ≥ q, as desired. Note that the equality holds if and only if a = b = c = d = 1.
?F?
96.16. Prove that for any real numbers a, b, c,
(a+ b)4 + (b+ c)4 + (c+ a)4 ≥ 4
7(a4 + b4 + c4).
(Vietnam 1996)
First solution: Let us make the substitution a+b = 2z, c+a = 2y, b+c = 2x.
The inequality becomes P =∑
(y + z − x)4 ≤ 28∑
x4. Now, we have the
following chain of identities
P =∑(∑
x2 + 2yz − 2xy − 2xz)2
= 3(∑
x2)2
+ 4(∑
x2) [∑
(yz − xy − xz)]
+ 4∑
(xy + xz − yz)2
= 3(∑
x2)2− 4
(∑x2)(∑
xy)
+ 16∑
x2y2 − 4(∑
xy)2
= 4(∑
x2)2
+ 16∑
x2y2 −(∑
x)4≤ 28
∑x4,
because(∑
x2)2≤ 3
∑x4,∑
x2y2 ≤∑
x4.
?F?
97.1. a, b, c be positive numbers such that abc = 1. Prove that
1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a≤ 1
a+ 2+
1
b+ 2+
1
c+ 2.
268
(Bulgaria 1997)
Solution: Put x = a+ b+ c and y = ab+ bc+ ca. Then the given inequalitycan be rewritten
3 + 4x+ y + x2
2x+ y + x2 + xy≤ 12 + 4x+ y
9 + 4x+ 2y,
or3x2y + xy2 + 6xy − 5x2 − y2 − 24x− 3y − 27 ≥ 0.
This inequality is equivalent to
(3x2y − 5x2 − 12x) + (xy2 − y2 − 3x− 3y) + (6xy − 9x− 27) ≥ 0,
which is true because x, y ≥ 3. Note that the equality holds if and only ifa = b = c = 1.
?F?
97.2. Prove that1
1999<
1
2· 3
4· · · 1997
1998<
1
44.
(Canada 1997)
Solution: Let us first note that for every integer n ≥ 1,
n2
(n+ 1)2=
n2
n(n+ 2) + 1<
n2
n(n+ 2)=
n
n+ 2, (1)
andn2
(n+ 1)2>
n2 − 1
(n+ 1)2=n− 1
n+ 1. (2)
From (1), we deduce that(1
2· 3
4· · · 1997
1998
)2
<1
3· 3
5· · · 1997
1999=
1
1999,
and hence1
2· 3
4· · · 1997
1998<
1√1999
<1
44.
This proves the right inequality. On the hand, from (2), we infer that(1
2· 3
4· · · 1997
1998
)2
>1
4· 2
4· 4
6· · · 1996
1998=
1
4· 2
1998=
1
3996>
1
19992.
Therefore1
2· 3
4· · · 1997
1998>
1
1999.
The left inequality is also proved and so, our proof is completed.?F?
97.3. Let x1, x2, . . . , x1997 be real numbers satisfying the following conditions
(a) − 1√3≤ xi ≤
√3 for i = 1, 2, . . . , 1997;
269
(b) x1 + x2 + · · ·+ x1997 = −318√
3.Determine the maximum value of x121 + x122 + · · ·+ x121997.
(China 1997)
Solution: Since x12 is a convex function of x, the sum of the twelfth powersof the xi is maximized by having all but perhaps one of the xi at the endpoints
of the prescribed interval. Suppose n of the xi equal − 1√3, 1996−n equal
√3
and the last one equals
−318√
3 +n√3− (1996− n)
√3.
This number must be in the range as well, so
−1 ≤ −318 · 3 + n− 3(1996− n) ≤ 3.
Equivalently −1 ≤ 4n− 6942 ≤ 3. The only such integer is n = 1736, the last
value is2√3
, and the maximum is 1736 · 3−6 + 260 · 36 +
(4
3
)6
.
?F?
97.4. For each natural number n ≥ 2, determine the largest possible value ofthe expression
Vn = sinx1 cosx2 + sinx2 cosx3 + · · ·+ sinxn cosx1,
where x1, x2, . . . , xn are arbitrary real numbers.(Czech-Slovak 1997)
Solution: By the inequality 2ab ≤ a2 + b2, we get
Vn ≤sin2 x1 + cos2 x2
2+ · · ·+ sin2 xn + cos2 x1
2=n
2,
with equality for x1 = · · · = xn =π
4.
?F?
97.5. Let x, y, z be positive real numbers. Prove that
xyz(x+ y + z +
√x2 + y2 + z2
)(x2 + y2 + z2) (xy + yz + zx)
≤ 3 +√
3
9.
(Hong Kong 1997)
Solution: We apply first the Cauchy Schwarz Inequality and then the AM-GM Inequality, and get
xyz(x+ y + z +
√x2 + y2 + z2
)(x2 + y2 + z2) (xy + yz + zx)
≤xyz
(√3 (x2 + y2 + z2) +
√x2 + y2 + z2
)(x2 + y2 + z2) (xy + yz + zx)
=
(√3 + 1
)xyz√
x2 + y2 + z2 (xy + yz + zx)
≤(√
3 + 1)xyz√
3 3√x2y2z23 3
√x2y2z2
=3 +√
3
9,
270
as desired. Note that the equality holds if and only if x = y = z.?F?
97.6. Given x1, x2, x3, x4 are positive real numbers such that x1x2x3x4 = 1.Prove that
x31 + x32 + x33 + x34 ≥ max
{x1 + x2 + x3 + x4,
1
x1+
1
x2+
1
x3+
1
x4
}.
(Iran 1997)
Solution: The problem requires us to prove that
x31 + x32 + x33 + x34 ≥ x1 + x2 + x3 + x4,
and
x31 + x32 + x33 + x34 ≥1
x1+
1
x2+
1
x3+
1
x4.
The proof of these two inequalities are very easy. Indeed, for the first inequal-ity, applying the AM-GM Inequality, for each i, we have
x3i + 1 + 1 ≥ 3xi,
and we deduce that
x31 + x32 + x33 + x34 + 8 ≥ (x1 + x2 + x3 + x4) + 2 (x1 + x2 + x3 + x4)
≥ x1 + x2 + x3 + x4 + 2 · 4 4√x1x2x3x4
= x1 + x2 + x3 + x4 + 8.
This inequality implies that
x31 + x32 + x33 + x34 ≥ x1 + x2 + x3 + x4,
which is the first one. For the second one, we apply the AM-GM Inequalityagain and get
x31 + x32 + x33 + x34 =
=1
3
[(x31 + x32 + x33
)+(x32 + x33 + x34
)+(x33 + x34 + x31
)+(x34 + x31 + x32
)]≥ 1
3
(3 3
√x31x
32x
33 + 3 3
√x32x
33x
34 + 3 3
√x33x
34x
31 + 3 3
√x34x
31x
32
)=
1
x4+
1
x1+
1
x2+
1
x3.
The proof is completed. Note that the equality holds if and only if x1 = x2 =x3 = x4 = 1.
?F?
97.7. Let a, b, c be nonnegative real numbers such that a+ b+ c ≥ abc. Provethat
a2 + b2 + c2 ≥ abc.
271
(Ireland 1997)
Solution: We may assume a, b, c > 0. Suppose by way of contradiction thata2 + b2 + c2 < abc, then abc > a2 and so a < bc, and likewise b < ca, c < ab.Then
abc ≥ a2 + b2 + c2 ≥ ab+ bc+ ca,
by the AM-GM Inequality, and the right side exceeds a+ b+ c, contradiction.?F?
97.8. Let a, b, c be positive real numbers. Prove that
(b+ c− a)2
(b+ c)2 + a2+
(c+ a− b)2
(c+ a)2 + b2+
(a+ b− c)2
(a+ b)2 + c2≥ 3
5.
(Japan 1997)
Solution: Without loss of generality, we may assume that c ≥ b ≥ a. Then,by applying the Cauchy Schwarz Inequality, we deduce that[∑ (b+ c− a)2
(b+ c)2 + a2
] [∑b2[(b+ c)2 + a2
]]≥[∑
b(b+ c− a)]2
=(∑
a2)2.
The inequality is reduced to proving that
5(∑
a2)2≥ 3
∑b2[(b+ c)2 + a2
].
After some simple computations, we find that it is equivalent to
2(a4 + b4 + c4) + 4(a2b2 + b2c2 + c2a2) ≥ 6(a3b+ b3c+ c3a).
Now, since c ≥ b ≥ a > 0, we have
a3b+ b3c+ c3a− (ab3 + bc3 + ca3) = −(b− a)(c− b)(c− a)(a+ b+ c) ≤ 0,
and hence
6(a3b+ b3c+ c3a) ≤ 3(a3b+ b3c+ c3a+ ab3 + bc3 + ca3
).
From this, we can see that the above inequality is deduced from
2(a4 + b4 + c4) + 4(a2b2 + b2c2 + c2a2) ≥ 3(a3b+ b3c+ c3a+ ab3 + bc3 + ca3
).
Now, note that
a4 + b4 + 4a2b2 − 3a3b− 3ab3 = (a2 − ab+ b2)(a− b)2 ≥ 0.
Adding this to the analogous inequalities, the result follows immediately. Theequality holds if and only if a = b = c.
?F?
97.9. Let a1, . . . , an be positive numbers, and define
A =a1 + · · ·+ an
n, G = n
√a1 · · · an, H =
n
a−11 + · · ·+ a−1n.
272
(a) If n is even, show thatA
H≤ −1 + 2
(A
G
)n.
(b) If n is odd, show thatA
H≤ −n− 2
n+
2(n− 1)
n
(A
G
)n.
(Korea 1997)
Solution: Note that
Gn
H=a1 · · · an(a−11 + · · ·+ a−1n )
n
=1
n
n∑i=1
a1 · · · anai
≤
(1
n
n∑i=1
ai
)n−1= An−1.
by the Maclaurin’s Inequality, soA
H≤(A
G
)n. Since A ≥ G,
(A
G
)n≥ 1, so
A
H≤ −1 + 2
(A
G
)n, proving part (a). For part (b),
A
H≤(A
G
)n≤(A
G
)n+n− 2
n
[(A
G
)n− 1
]= −n− 2
n+
2(n− 1)
n
(A
G
)n.
?F?
97.10. For any positive real numbers x, y, z such that xyz = 1, prove theinequality
x9 + y9
x6 + x3y3 + y6+
y9 + z9
y6 + y3z3 + z6+
z9 + x9
z6 + z3z3 + x6≥ 2.
(Romania 1997)
Solution: Setting a = x3, b = y3, c = z3. Then our inequality becomes
a3 + b3
a2 + ab+ b2+
b3 + c3
b2 + bc+ c2+
c3 + a3
c2 + ca+ a2≥ 2.
Now, notice that for any u, v > 0, we have
u3 + v3
u2 + uv + v2− u+ v
3=
2(u+ v)(u− v)2
3(u2 + uv + v2)≥ 0.
Applying this into the left hand side of the above inequality, we deduce that
a3 + b3
a2 + ab+ b2+
b3 + c3
b2 + bc+ c2+
c3 + a3
c2 + ca+ a2≥ a+ b
3+b+ c
3+c+ a
3
=2
3· 3 3√abc = 2.
The equality holds if and only if x = y = z = 1.?F?
97.11. Let a, b, c be positive real numbers. Prove that
a2
a2 + 2bc+
b2
b2 + 2ca+
c2
c2 + 2ab≥ 1 ≥ bc
a2 + 2bc+
ca
b2 + 2ca+
ab
c2 + 2ab.
273
(Romania 1997)
Solution: Since
bc
a2 + 2bc+
ca
b2 + 2ca+
ab
c2 + 2ab=
3
2− 1
2
(a2
a2 + 2bc+
b2
b2 + 2ca+
c2
c2 + 2ab
),
we can see that the right hand side inequality follows immediately from theleft hand side inequality, so it suffices to prove the left inequality. However,this inequality is obviously true since by the Cauchy Schwarz Inequality,
a2
a2 + 2bc+
b2
b2 + 2ca+
c2
c2 + 2ab≥ (a+ b+ c)2
a2 + 2bc+ b2 + 2ca+ c2 + 2ab= 1.
The proof is completed. Note that the equality holds if and only if a = b = c.?F?
97.12. Show that if 1 < a < b < c, then
loga(loga b) + logb(logb c) + logc(logc a) > 0.
(Russia 1997)
Solution: Since loga b > 1, we have loga(loga b) > logb(loga b). Also, sincelogc a < 1, we find that logc(logc a) > logb(logc a). Thus the left side of thegiven inequality exceeds
logb(loga b logb c logc a) = 0.
?F?
97.13. Prove that for x, y, z ≥ 2,
(y3 + x)(z3 + y)(x3 + z) ≥ 125xyz.
(Saint Petersburg 1997)
First solution: The left side is at least (4y + x)(4z + y)(4x + y). By theWeighted AM-GM Inequality, 4y+x ≥ 5y4/5x1/5. This and the two analogousinequalities imply the claim. Note that the equality holds if and only if x =y = z = 2.
Second solution: Dividing both sides by xyz > 0, we can rewrite the originalinequality as (
x2 +y
x
)(y2 +
z
y
)(z3 +
x
z
)≥ 125.
Now, applying the AM-GM Inequality, we have
x2 +y
x=x2
4+x2
4+x2
4+x2
4+y
x≥ 5
5
√x7y
44.
274
Multiplying this to the two analogous inequalities, we obtain(x2 +
y
x
)(y2 +
z
y
)(z3 +
x
z
)≥ 125
5
√x8y8z8
412≥ 125
5
√224
224= 125,
as desired.
Third solution: We have to prove that(x3
y+ 1
)(y3
z+ 1
)(z3
x+ 1
)≥ 125.
This is true, since by the Holder’s Inequality, we have(x3
y+ 1
)(y3
z+ 1
)(z3
x+ 1
)≥
(1 + 3
√x3
y· y
3
z· z
3
x
)3
≥ 125.
?F?
97.14. Given an integer n ≥ 2, find the minimal value of
x51x2 + x3 + · · ·+ xn
+x52
x3 + · · ·+ xn + x1+ · · ·+ x5n
x1 + x2 + · · ·+ xn−1,
for positive real numbers x1, . . . , xn subject to the condition x21 + · · ·+x2n = 1.(Turkey 1997)
First solution: Let S = x1 + · · · + xn. By the Chebyshev’s Inequality, the
average ofx5i
S − xiis at least the average of x4i times the average of
xiS − xi
(since both are increasing functions of xi). The latter function is convex, so
its average is at least1
n− 1. We apply the Power Mean Inequality to the
former, which gives
n∑i=1
x4i
n
1/2
≥
n∑i=1
x2i
n=
1
n. We conclude
n∑i=1
x5iS − xi
≥ n · 1
n2· 1
n− 1=
1
n(n− 1),
with equality iff x1 = · · · = xn =1√n.
Second solution: By the Cauchy Schwarz Inequality, we have
P =
n∑i=1
x5in∑i=1
xi − xi≥
(n∑i=1
x3i
)2
n∑i=1
xi
(n∑i=1
xi − xi
) =
(n∑i=1
x3i
)2
(n∑i=1
xi
)2
− 1
.
275
In addition, the Chebyshev’s Inequality implies that
n∑i=1
x3i ≥1
n
(n∑i=1
xi
)(n∑i=1
x2i
)=
1
n
(n∑i=1
xi
).
From this and the above inequality, we deduce that
P ≥ 1
n2·
(n∑i=1
xi
)2
(n∑i=1
xi
)2
− 1
=1
n2
1 +1(
n∑i=1
xi
)2
− 1
≥ 1
n2
1 +1
n
n∑i=1
x2i − 1
=1
n2
(1 +
1
n− 1
)=
1
n(n− 1).
Note that the equality holds iff x1 = x2 = · · · = xn =1√n.
?F?
97.15. Let x, y and z be positive real numbers.
(a) If x+ y + z ≥ 3, is it necessarily true that1
x+
1
y+
1
z≤ 3?
(b) If x+ y + z ≤ 3, is it necessarily true that1
x+
1
y+
1
z≥ 3?
(United Kingdom 1997)
Solution: (a) The answer is ”No”. For example, take x = 2, y = 1, z =1
3.
(b) The answer is ”Yes”, since by the Cauchy Schwarz Inequality, we have
1
x+
1
y+
1
z≥ 9
x+ y + z≥ 9
3= 3.
?F?
97.16. Prove that, for all positive real numbers a, b, c,
1
a3 + b3 + abc+
1
b3 + c3 + abc+
1
c3 + a3 + abc≤ 1
abc.
(USA 1997)
Solution: The inequality (a− b)(a2 − b2) ≥ 0 implies a3 + b3 ≥ ab(a+ b), so
1
a3 + b3 + abc≤ 1
ab(a+ b) + abc=
c
abc(a+ b+ c).
Similarly, we have
1
b3 + c3 + abc≤ a
abc(a+ b+ c), and
1
c3 + a3 + abc≤ b
abc(a+ b+ c).
276
Thus
1
a3 + b3 + abc+
1
b3 + c3 + abc+
1
c3 + a3 + abc≤ a+ b+ c
abc(a+ b+ c)=
1
abc.
?F?
98.1. Let a, b, c be positive real numbers. Prove that(1 +
a
b
)(1 +
b
c
)(1 +
c
a
)≥ 2
(1 +
a+ b+ c3√abc
).
(APMO 1998)
First solution: After expanding, we find that the original inequality is equiv-alent to (
a
b+b
c+c
a
)+
(a
c+b
a+c
b
)≥ 2 · a+ b+ c
3√abc
.
Now, we apply the AM-GM Inequality to get
3
(a
b+b
c+c
a
)=
(a
b+a
b+b
c
)+
(b
c+b
c+c
a
)+( ca
+c
a+a
b
)≥ 3
3
√a
b· ab· bc
+ 33
√b
c· bc· ca
+ 3 3
√c
a· ca· ab
=3 (a+ b+ c)
3√abc
,
and
3
(a
c+b
a+c
b
)=(ac
+a
c+c
b
)+
(b
a+b
a+a
c
)+
(c
b+c
b+b
a
)≥ 3 3
√a
c· ac· cb
+ 33
√b
a· ba· ac
+ 33
√c
b· cb· ba
=3(a+ b+ c)
3√abc
.
These two inequalities imply that
a
b+b
c+c
a≥ a+ b+ c
3√abc
, andb
a+c
b+a
c≥ a+ b+ c
3√abc
,
and hence the result follows. Note that the equality holds if and only if a =b = c = 1.
Second solution: According to the well-known inequalities (a+ b)(b+ c)(c+
a) ≥ 8
9(a+ b+ c)(ab+ bc+ ca) and ab+ bc+ ca ≥
√3abc(a+ b+ c), we infer
that (1 +
a
b
)(1 +
b
c
)(1 +
c
a
)=
(a+ b)(b+ c)(c+ a)
abc
≥ 8(a+ b+ c)(ab+ bc+ ca)
9abc
≥8(a+ b+ c)
√3(a+ b+ c)
9√abc
.
277
On the other hand, it is clear that
2
(1 +
a+ b+ c3√abc
)≤ 8(a+ b+ c)
3 3√abc
.
Therefore, it suffices to prove that
8(a+ b+ c)√
3(a+ b+ c)
9√abc
≥ 8(a+ b+ c)
3 3√abc
,
which is equivalently to √a+ b+ c
3≥ 6√abc.
Of course, this is clearly true from the AM-GM Inequality and hence, ourproof is completed.
?F?
98.2. Let x1, x2, y1, y2 be real numbers such that x21 + x22 ≤ 1. Prove theinequality
(x1y1 + x2y2 − 1)2 ≥ (x21 + x22 − 1)(y21 + y22 − 1).
(Austrian-Polish Competition 1998)
Solution: Since (x1y1 +x2y2− 1)2 = (x1y1 +x2y2)2− 2(x1y1 +x2y2) + 1, and
(x21 + x22 − 1)(y21 + y22 − 1) = (x21 + x22)(y21 + y22)− (x21 + x22 + y21 + y22) + 1,
the original inequality can be rewritten as
x21 + x22 + y21 + y22 − 2(x1y1 + x2y2) ≥ (x21 + x22)(y21 + y22)− (x1y1 + x2y2)
2,
or equivalently,
(x1 − y1)2 + (x2 − y2)2 ≥ (x1y2 − x2y1)2.
Now, since x21 + x22 ≤ 1, it suffices to prove that
(x21 + x22)[(x1 − y1)2 + (x2 − y2)2
]≥ (x1y2 − x2y1)2.
Of course, this is true since by the Cauchy Schwarz Inequality, we have
(x21 + x22)[(x1 − y1)2 + (x2 − y2)2
]≥ [x1(x2 − y2) + x2(y1 − x1)]2
= (x1y2 − x2y1)2.
?F?
98.3. If n ≥ 2 is an integer and 0 < a1 < a2 < . . . < a2n+1 are real numbers,prove the inequality
n√a1− n√a2+ n√a3−· · ·− n
√a2n+ n
√a2n+1 <
n√a1 − a2 + a3 − . . .− a2n + a2n+1.
(Balkan 1998)
278
Solution: Denote bi = n√ai, then 0 < b1 < b2 < · · · < b2n+1 and our
inequality becomes
n
√bn1 − bn2 + bn3 − · · · − bn2n + bn2n+1 > b1 − b2 + b3 − · · · − b2n + b2n+1.
Consider the function
f(t) = n
√tn − bn2 + bn3 − · · · − bn2n + bn2n+1 − (t− b2 + b3 − · · · − b2n + b2n+1),
where 0 < t < b2. We have
f ′(t) =tn−1(
tn − bn2 + bn3 − · · · − bn2n + bn2n+1
)n−1n
− 1
=bn−11[
bn1 + (bn2n+1 − bn2n) + · · ·+ (bn3 − bn2 )]n−1
n
− 1
<bn−11
(bn1 + 0 + · · ·+ 0)n−1n
− 1 = 0.
This meanst that f(t) is strictly decreasing, and hence
f(t) > f(b2) = n
√bn3 − · · · − bn2n + bn2n+1 − (b3 − · · · − b2n + b2n+1),
for any 0 < t < b2. Since 0 < b1 < b2, we infer that f(b1) > f(b2), or
n
√bn1 − bn2 + bn3 − · · · − bn2n + bn2n+1 − (b1 − b2 + b3 − · · · − b2n + b2n+1) >
> n
√bn3 − · · · − bn2n + bn2n+1 − (b3 − · · · − b2n + b2n+1).
In the same manner, we can establish the chain inequalities
n
√bn1 − bn2 + bn3 − · · · − bn2n + bn2n+1 − (b1 − b2 + b3 − · · · − b2n + b2n+1) >
> n
√bn3 − · · · − bn2n + bn2n+1 − (b3 − · · · − b2n + b2n+1)
> n
√bn5 − · · · − bn2n + bn2n+1 − (b5 − · · · − b2n + b2n+1)
> · · · > n
√bn2n−1 − bn2n + bn2n+1 − (b2n−1 − b2n + b2n+1)
> n
√bn2n − bn2n + bn2n+1 − (b2n − b2n + b2n+1) = 0,
and so, our proof is completed.?F?
98.4. Let a, b, c be positive real numbers. Prove that
a
b+b
c+c
a≥ a+ b
b+ c+b+ c
a+ b+ 1.
(Belarus 1998)
279
First solution: Note that the inequality is equivalent to
(a+ b+ c)
(a
b+b
c+c
a− 3
)≥ (a+ b+ c)
(a+ b
b+ c+b+ c
a+ b− 2
),
and thus it can be rewritten as
a2
b+b2
c+c2
a+ab
c+bc
a+ca
b− 2(a+ b+ c) ≥ (a+ b+ c)(a− c)2
(a+ b)(b+ c).
Now, since
a2
b+b2
c+c2
a− (a+ b+ c) =
(a− b)2
b+
(b− c)2
c+
(c− a)2
a,
and
ab
c+bc
a+ca
b− (a+ b+ c) =
a2(b− c)2 + b2(c− a)2 + c2(a− b)2
2abc≥ 0,
it suffices to prove that
(a− b)2
b+
(b− c)2
c+
(c− a)2
a≥ (a+ b+ c)(a− c)2
(a+ b)(b+ c).
By the Cauchy Schwarz Inequality, we get
(a− b)2
b+
(b− c)2
c≥ (a− c)2
b+ c,
and thus, it remains to show that
1
b+ c+
1
a≥ a+ b+ c
(a+ b)(b+ c).
Of course, this is obvious since it is equivalent with
b(a+ b+ c)
a(a+ b)(b+ c)≥ 0.
The equality holds if and only if a = b = c.
Second solution: After writing x =a
band y =
c
b, we get
c
a=y
x,
a+ b
b+ c=x+ 1
1 + y,
b+ c
b+ a=
1 + y
x+ 1.
One may rewrite the inequality as
x3y2 + x2 + x+ y3 + y2 ≥ x2y + 2xy + 2xy2.
Now, we apply the AM-GM Inequality to obtain
x3y2 + x
2≥ x2y, x3y2 + x+ y3 + y3
2≥ 2xy2, x2 + y2 ≥ 2xy.
280
Adding up these three inequalities, we get the result.?F?
98.5. Let n be a natural number such that n ≥ 2. Prove that
1
n+ 1
(1 +
1
3+ . . .+
1
2n− 1
)>
1
n
(1
2+
1
4+ . . .+
1
2n
).
(Canada 1998)
Solution: The inequality is equivalent to
1
n+ 1
n∑k=1
1
2k − 1>
1
n
n∑k=1
1
2k, or
n∑k=1
n+ 1− 2k
2k(2k − 1)> 0.
Now, we see that
n+ 1− 2 · 1 > n+ 1− 2 · 2 > · · · > n+ 1− 2 · n,
and1
1 · 2>
1
3 · 4> · · · > 1
2n(2n− 1).
Therefore, applying the Chebyshev’s Inequality, we have
n∑k=1
n+ 1− 2k
2k(2k − 1)≥ 1
n
[n∑k=1
(n+ 1− 2k)
][n∑k=1
1
2k(2k − 1)
]
=1
n
[n(n+ 1)− 2
n∑k=1
k
][n∑k=1
1
2k(2k − 1)
]= 0.
On the other hand, it is clear that the equality cannot occur, so we must have
n∑k=1
n+ 1− 2k
2k(2k − 1)> 0,
and the proof is completed.?F?
98.6. Let n ≥ 2 be a positive integer and let x1, x2, . . . , xn be real numberssuch that
n∑i=1
x2i +
n−1∑i=1
xixi+1 = 1.
For every positive integer k, 1 ≤ k ≤ n, determine the maximum value of |xk| .(China 1998)
Solution: For convenience, let us denote x0 = xn+1 = 0. Now, we will provefirst that the following inequality holds for every n ≥ 1,
x20 + x21 + · · ·+ x2n + x0x1 + · · ·+ xn−1xn + xnx0 ≥n+ 1
2nx2m,
281
wherem = 1 orm = n, and the equality holds when x1 = − n
n− 1x2, . . . , xn−1 =
−2xn for n ≥ 2,m = 1 and xn = − n
n− 1xn−1, . . . , x2 = −2x1 for n ≥ 2,m =
n. We will prove this statement by induction on n. And with noting that theproofs for m = 1 and m = n are quite similar, it suffices to prove the state-ment for m = 1 (by induction, of course). For n = 1 and n = 2, it is clear.Suppose that the statement holds for n = k ≥ 2 and let us prove that it holdsfor n = k + 1. Indeed, from the inductive hypothesis, we see that for anyy0, . . . , yk,
y20 + y21 + · · ·+ y2k + y0y1 + · · ·+ yk−1yk + yky0 ≥k + 1
2ky21,
Choosing (y0, y1, . . . , yk) = (0, x2, . . . , xk+1), we get
x22 + · · ·+ x2k+1 + x2x3 + · · ·+ xkxk+1 ≥k + 1
2kx22.
On the other hand, we have
x21 +k + 1
2kx22 + x1x2 −
k + 2
2(k + 1)x21 =
[kx1 + (k + 1)x2]2
2k (k + 1)≥ 0.
Using this in combination with the above inequality, it is easy to deduce that
x20 + x21 + · · ·+ x2k+1 + x0x1 + · · ·+ xkxk+1 + xk+1x0 ≥k + 2
2(k + 1)x21.
This proves our statement. Now, turning back to our problem, applying theabove inequality, we can see that
1 =n∑i=1
x2i +n−1∑i=1
xixi+1 =n∑i=0
x2i +n∑i=0
xixi+1 ≥n+ 1
2nx21.
Therefore max |x1| =√
2n
n+ 1, with equality if x1 = ±
√2n
n+ 1, x2 = −n− 1
nx1,
. . . , xn = −1
2xn−1. Similarly, we can find that max |xn| =
√2n
n+ 1. These are
also all values we need to find for the case n = 2. In case n ≥ 3, for all2 ≤ k ≤ n− 1, we have
1 =
n∑i=1
x2i +
n−1∑i=1
xixi+1
=(x20 + x21 + · · ·+ x2k−1 + x0x1 + · · ·+ xk−2xk−1 + xk−1x0
)+ (x20 + x2k+1+
+ · · ·+ x2n + x0xk+1 + · · ·+ xn−1xn + xnx0) + x2k + xk−1xk + xkxk+1
≥ k
2(k − 1)x2k−1 +
n− k + 1
2(n− k)x2k+1 + x2k + xk−1xk + xkxk+1
=[kxk−1 + (k − 1)xk]
2
2k(k − 1)+
[(n− k)xk + (n− k + 1)xk+1]2
2(n− k)(n− k + 1)+
n+ 1
2k(n+ 1− k)x2k
≥ n+ 1
2k(n+ 1− k)x2k.
282
Hence |xk| ≤√
2k(n+ 1− k)
n+ 1, and it is easy to check that the equality can be
attained (according to the equality of the statement we have proved above),thus
max |xk| =√
2k(n+ 1− k)
n+ 1.
?F?
98.7. Let a, b, c ≥ 1. Prove that
√a− 1 +
√b− 1 +
√c− 1 ≤
√c (ab+ 1).
(Hong Kong 1998)
Solution: The given inequality is a consequence of the Cauchy Schwarz In-equality. Indeed, from the Cauchy Schwarz Inequality, we get
√a− 1 +
√b− 1 +
√c− 1 ≤
√(1 + b− 1) (a− 1 + 1) +
√c− 1 =
√ab+
√c− 1
≤√
(1 + c− 1) (ab+ 1) =√c (ab+ 1),
as desired.?F?
98.8. Let a1, a2, . . . , an > 0 such that a1 + a2 + · · ·+ an < 1. Prove that
a1a2 · · · an (1− a1 − a2 − · · · − an)
(a1 + a2 + · · ·+ an) (1− a1) (1− a2) · · · (1− an)≤ 1
nn+1.
(IMO Shortlist 1998)
Solution: Put an+1 = 1−a1−a2−· · ·−an. The original inequality becomes
a1a2 · · · anan+1
(1− a1) (1− a2) · · · (1− an) (1− an+1)≤ 1
nn+1.
Now, for each i (1 ≤ i ≤ n+ 1), we apply the AM-GM Inequality as follows
1− ai =∑j 6=i
aj ≥ n n
√∏j 6=i
aj .
Setting i = 1, 2, . . . , n+1, we get the n+1 similar inequalities and multiplyingthem up, we obtain
(1− a2) (1− a2) · · · (1− an) (1− an+1) ≥ nn+1 n
√an1a
n2 · · · annann+1
= nn+1a1a2 · · · anan+1.
From this, it follows that
a1a2 · · · anan+1
(1− a1) (1− a2) · · · (1− an) (1− an+1)≤ 1
nn+1,
283
as desired. The equality holds if and only if a1 = a2 = · · · = an =1
n+ 1.
?F?
98.9. Let a, b, c be positive real numbers such that abc = 1. Prove that
a3
(1 + b) (1 + c)+
b3
(1 + a) (1 + c)+
c3
(1 + a) (1 + b)≥ 3
4.
(IMO Shortlist 1998)
Solution: Applying the AM-GM Inequality, we have
a3
(1 + b) (1 + c)+
1 + b
8+
1 + c
8≥ 3 3
√a3 (1 + b) (1 + c)
82 (1 + b) (1 + c)=
3a
4.
Adding this to the two analogous inequalities, we deduce that∑ a3
(1 + b)(1 + c)+a+ b+ c+ 3
4≥ 3(a+ b+ c)
4,
and hence, it follows that
∑ a3
(1 + b)(1 + c)≥ a+ b+ c
2− 3
4≥ 3 3√abc
2− 3
4,
as desired. The equality holds if and only if a = b = c = 1.?F?
98.10. Let x, y, z > 1 such that1
x+
1
y+
1
z= 2. Prove that
√x+ y + z ≥
√x− 1 +
√y − 1 +
√z − 1.
(Iran 1998)
First solution: By applying the Cauchy Schwarz Inequality, we have that
(√x− 1 +
√y − 1 +
√z − 1
)2=
(√x− 1
x·√x+
√y − 1
y· √y +
√z − 1
z·√z
)2
≤ (x+ y + z)
(x− 1
x+y − 1
y+z − 1
z
)= (x+ y + z)
(3− 1
x− 1
y− 1
z
)= (x+ y + z) (3− 2) = x+ y + z,
and then, we deduce that
√x− 1 +
√y − 1 +
√z − 1 ≤
√x+ y + z,
as desired. Note that the equality holds if and only if x = y = z =3
2.
284
Second solution: After the algebraic substitution a =1
x, b =
1
y, c =
1
z, we
are required to prove that√1
a+
1
b+
1
c≥√
1− aa
+
√1− bb
+
√1− cc
,
where a, b, c ∈ (0, 1) and a+ b+ c = 2. Using the constraint a+ b+ c = 2, weobtain a homogeneous inequality
√1
2(a+ b+ c)
(1
a+
1
b+
1
c
)≥∑√√√√ a+ b+ c
2− a
a,
or √(a+ b+ c)
(1
a+
1
b+
1
c
)≥√b+ c− a
a+
√c+ a− b
b+
√a+ b− c
c,
which immediately follows from the Cauchy Schwarz Inequality√[(b+ c− a) + (c+ a− b) + (a+ b− c)]
(1
a+
1
b+
1
c
)≥
≥√b+ c− a
a+
√c+ a− b
b+
√a+ b− c
c.
?F?
98.11. Suppose that a1 < a2 < · · · < an are real numbers. Prove that
a1a42 + a2a
43 + · · ·+ an−1a
4n + ana
41 ≥ a2a41 + a3a
42 + · · ·+ ana
4n−1 + a1a
4n.
(Iran 1998)
Solution: We prove the result by induction on n. For n = 2, we have equality.The case n = 3 will be needed below. For n = 3, we have to show that
a1a42 + a2a
43 + a3a
41 ≥ a2a41 + a3a
42 + a1a
43.
This is true, since
a1a42 + a2a
43 + a3a
41 − (a2a
41 + a3a
42 + a1a
43) =
=1
2(a2 − a1)(a3 − a2)(a3 − a1)[(a1 + a2)
2 + (a2 + a3)2 + (a3 + a1)
2] ≥ 0.
Assume that the claim is true for n− 1, and let us prove it for n. By applyingthe induction hypothesis, we find that it is sufficient to prove that
an−1a4n + ana
41 − an−1a41 ≥ ana4n−1 + a1a
4n − a1a4n−1,
which is the case n = 3.?F?
285
98.12. Show that if x is a nonzero real number, then
x8 − x5 − 1
x+
1
x4≥ 0.
(Ireland 1998)
Solution: We have
x8 − x5 − 1
x+
1
x4=
(x8 − 1
x
)−(x5 − 1
x4
)=x9 − 1
x− x9 − 1
x4
=(x9 − 1)(x3 − 1)
x4=
(x3 − 1)2(x6 + x3 + 1)
x4≥ 0.
The equality holds if and only if x = 1.?F?
98.13. Prove that if a, b, c are positive real numbers, then
9
a+ b+ c≤ 2
(1
a+ b+
1
b+ c+
1
c+ a
),
and1
a+ b+
1
b+ c+
1
c+ a≤ 1
2
(1
a+
1
b+
1
c
).
(Ireland 1998)
Solution: Both inequalities are the consequences of the Cauchy Schwarz In-equality. Indeed, from the Cauchy Schwarz Inequality, we have that
2
(1
a+ b+
1
b+ c+
1
c+ a
)≥ 2 · 9
(a+ b) + (b+ c) + (c+ a)=
9
a+ b+ c,
and
1
2
(1
a+
1
b+
1
c
)=
(1
4a+
1
4b
)+
(1
4b+
1
4c
)+
(1
4c+
1
4a
)≥ 4
4a+ 4b+
4
4b+ 4c+
4
4c+ 4a
=1
a+ b+
1
b+ c+
1
c+ a.
The equality (in both inequalities) occurs if and only if a = b = c.?F?
98.14. If x, y, z are positive real numbers such that x+ y + z = xyz, then
1√1 + x2
+1√
1 + y2+
1√1 + z2
≤ 3
2.
(Korea 1998)
286
Solution: Applying the AM-GM Inequality, we find that
1√x2 + 1
=1√
x2 + xyzx+y+z
=
√x+ y + z
x(x+ y)(x+ z)
=
√yz
(x+ y)(x+ z)≤ y
2(x+ y)+
z
2(x+ z).
Adding this to the two analogous inequalities, we conclude that
1√1 + x2
+1√
1 + y2+
1√1 + z2
≤ x+ y
2(x+ y)+
y + z
2(y + z)+
z + x
2(z + x)=
3
2,
as desired. Note that the equality holds if and only if x = y = z =√
3.?F?
98.15. Let a, b, c, d, e, f be positive real numbers such that
a+ b+ c+ d+ e+ f = 1, ace+ bdf ≥ 1
108.
Prove that
abc+ bcd+ cde+ def + efa+ fab ≤ 1
36.
(Poland 1998)
Solution: Put M = ace + bdf and N = abc + bcd + cde + def + efa + fab.Applying the AM-GM Inequality, we find that
M +N = (a+ d) (b+ e) (c+ f) ≤(a+ d+ b+ e+ c+ f
3
)3
=1
27,
and hence, we deduce that
N ≤ 1
27−M ≤ 1
27− 1
108=
1
36,
as desired.?F?
98.16. Let a be a positive real numbers and let x1, x2, . . . , xn be positive realnumbers such that x1 + x2 + · · ·+ xn = 1. Prove that
ax1−x2
x1 + x2+
ax2−x3
x2 + x3+ · · ·+ axn−x1
xn + x1≥ n2
2.
(Serbia 1998)
Solution: The original inequality is a direct consequence of the AM-GMInequality. Indeed, denote with P the left hand side of the original inequality,
287
then by applying the AM-GM Inequality, we have that
P ≥ n n
√ax1−x2
x1 + x2· a
x2−x3
x2 + x3· · · a
xn−x1
xn + x1
= n n
√a(x1−x2)+(x2−x3)+···+(xn−x1)
(x1 + x2) (x2 + x3) · · · (xn + x1)
≥ n · n
(x1 + x2) + (x2 + x3) + · · ·+ (xn + x1)=n2
2.
The equality holds if and only if x1 = x2 = · · · = xn =1
n.
?F?
98.17. Find the minimum of the expression
F (x, y) =√
(x+ 1)2 + (y − 1)2+√
(x− 1)2 + (y + 1)2+√
(x+ 2)2 + (y + 2)2,
where x, y are real numbers.(Vietnam 1998)
First solution: According to the Cauchy Schwarz Inequality, we have
√(x+ 1)2 + (y − 1)2 =
√[(√3 + 1
)2+(√
3− 1)2]
[(1− y)2 + (x+ 1)2]
2√
2
≥(1− y)
(√3 + 1
)+ (1 + x)
(√3− 1
)2√
2.
Similarly, we find that
√(x− 1)2 + (y + 1)2 ≥
(1− x)(√
3 + 1)
+ (1 + y)(√
3− 1)
2√
2.
On the other hand, the Cauchy Schwarz Inequality also implies√(x+ 2)2 + (y + 2)2 ≥ (x+ 2) + (y + 2)√
2.
Adding up these three inequalities, we get F (x, y) ≥√
6+2√
2. As we can see,
this upper bound is attained for x = y = − 1√3, and so the minimum value we
are looking for is√
6 + 2√
2.
Second solution: In the coordinate plane, consider four points A(−1, 1),B(1,−1), C(−2,−2) and M(x, y). Then we can reformulate the problem in ageometric ones as follows: Find a point M such that the sum of its distancesto the vertices of the triangle ABC is minimum. Note that this is an acutetriangle. The well-known solution is that all angles ∠AMB = ∠BMC =∠CMA = 120◦. From this, one can easily deduce that the minimum of F (x, y)is√
6 + 2√
2.
288
?F?
98.18. Let n ≥ 2 and x1, x2, . . . , xn be positive real numbers satisfying
1
x1 + 1998+
1
x2 + 1998+ · · ·+ 1
xn + 1998=
1
1998.
Prove thatn√x1x2 · · ·xnn− 1
≥ 1998.
(Vietnam 1998)
Solution: For each i (1 ≤ i ≤ n), the AM-GM Inequality gives us
xi1998(xi + 1998)
=1
1998− 1
xi + 1998=∑j 6=i
1
xj + 1998≥ n− 1
n−1
√∏j 6=i
(xj + 1998)
.
From this, it follows that
xi ≥ 1998(n− 1)n−1
√√√√√(xi + 1998)n−1∏j 6=i
(xj + 1998).
Setting i = 1, 2, . . . , n, we get the n similar inequalities and multiplying themup, the result follows immediately. Note that the equality holds if and only ifx1 = x2 = · · · = xn = 1998(n− 1).
?F?
99.1. Let {an} be a sequence of real numbers such that ai+j ≤ ai + aj for alli, j. Prove that the following inequality holds
a1 +a22
+ · · ·+ ann≥ an.
(APMO 1999)
Solution: We will prove the required inequality by induction on n. For n = 1and n = 2, it is clear. Suppose that the inequality holds for n = k ≥ 2 and letus prove it for n = k + 1. Indeed, from the inductive hypothesis, we have
a1 ≥ a1, a1 +a22≥ a2, . . . , a1 +
a22
+ · · ·+ akk≥ ak.
Adding up these k inequalities, we get
ka1 +(k − 1)a2
2+ · · ·+ ak
k≥ a1 + a2 + · · ·+ ak.
This inequality is equivalent to
(k + 1)(a1 +
a22
+ · · ·+ akk
)≥ 2(a1 + a2 + · · ·+ ak).
289
On the other hand, from the given condition, we have
2k∑i=1
ai =k∑i=1
ai +k∑i=1
ai =k∑i=1
ai +k∑i=1
ak+1−i
=
k∑i=1
(ai + ak+1−i) ≥k∑i=1
ai+k+1−i = kak+1.
From this and the above inequality, we deduce that
a1 +a22
+ · · ·+ akk≥ k
k + 1ak+1,
or
a1 +a22
+ · · ·+ akk
+ak+1
k + 1≥ ak+1,
as desired.?F?
99.2. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Provethat
1
1 + ab+
1
1 + bc+
1
1 + ca≥ 3
2.
(Belarus 1999)
Solution: From the Cauchy Schwarz Inequality and the AM-GM Inequality,we deduce that
1
1 + ab+
1
1 + bc+
1
1 + ca≥ (1 + 1 + 1)2
1 + ab+ 1 + bc+ 1 + ca
≥ 9
3 + a2 + b2 + c2=
3
2,
as desired. The equality holds if and only if a = b = c = 1.?F?
99.3. For any nonnegative real numbers x, y, z such that x+ y+ z = 1, provethe following inequality
x2y + y2z + z2x ≤ 4
27.
(Canada 1999)
Solution: Assume without loss of generality that x = max{x, y, z}. We havetwo cases
+ If x ≥ y ≥ z, then
x2y + y2z + z2x ≤ x2y + y2z + z2x+ z[xy + (x− y)(y − z)]
= (x+ z)2y = 4
(1
2− y
2
)(1
2− y
2
)y ≤ 4
27,
290
where the last inequality follows from the AM-GM Inequality. The equality
occurs if and only if z = 0 (from the first inequality) and y =1
3, in which case
(x, y, z) =
(2
3,1
3, 0
).
+ If x ≥ z ≥ y, then
x2y + y2z + z2x = x2z + z2y + y2x− (x− z)(z − y)(x− y)
≤ x2z + z2y + y2x ≤ 4
27,
where the second inequality is true from the result we proved for x ≥ y ≥z (except with y and z reversed). The quality holds in the first inequalityonly when two of x, y, z are equal, and in the second only when (x, z, y) =(
2
3,1
3, 0
). Since these conditions cant both be true, the inequality is actually
strict in this case.
Therefore the inequality is indeed true, and the equality holds when (x, y, z)
equals
(2
3,1
3, 0
), or
(1
3, 0,
2
3
), or
(0,
2
3,1
3
).
?F?
99.4. Let a, b, c be positive real numbers. Prove that
a
b+ 2c+
b
c+ 2a+
c
a+ 2b≥ 1.
(Czech-Slovak 1999)
Solution: By applying Cauchy Schwarz Inequality in combination with thewell-known inequality (a+ b+ c)2 ≥ 3(ab+ bc+ ca), we get
a
b+ 2c+
b
c+ 2a+
c
a+ 2b≥ (a+ b+ c)2
a (b+ 2c) + b (c+ 2a) + c (a+ 2b)
≥ 3 (ab+ bc+ ca)
3 (ab+ bc+ ca)= 1,
as desired. The equality holds if and only if a = b = c.?F?
99.5. Let n ≥ 2 be a fixed integer. Find the least constant C such that theinequality ∑
1≤i<j≤nxixj(x
2i + x2j ) ≤ C(x1 + x2 + · · ·+ xn)4
holds for every x1, . . . , xn ≥ 0. For this constant C, characterize the instancesof equality.
(IMO 1999)
291
First solution: Applying the AM-GM Inequality, we have
∑1≤i<j≤n
xixj(x2i + x2j ) ≤
∑1≤i<j≤n
xixj
x2i + x2j +∑
k 6=i,k 6=jx2k
=
∑1≤i<j≤n
xixj
( n∑i=1
x2i
)
=1
2·
2∑
1≤i<j≤nxixj
·( n∑i=1
x2i
)
≤ 1
2
2∑
1≤i<j≤nxixj +
n∑i=1
x2i
2
2
=1
8
(n∑i=1
xi
)4
.
The equality holds if and only if xixj(x2i + x2j ) = xixj(x
21 + x22 + · · ·+ x2n) for
any i < j and x21 + x22 + · · · + x2n = 2∑
1≤i<j≤nxixj , which holds if and only
if n − 2 of the xi are zero and the remaining two are equal. And since the
equality can occur, it follows that1
8is the smallest possible value of C.
Second solution: For x1 = x2 = · · · = xn = 0, it holds for any C ≥ 0.Hence, we consider the case when x1 + x2 + · · ·+ xn > 0. Since the inequalityis homogeneous, we may normalize to x1 + x2 + · · ·+ xn = 1. We denote
F (x1, x2, . . . , xn) =∑
1≤i<j≤nxixj(x
2i + x2j ).
From the assumption x1 + x2 + · · ·+ xn = 1, we have
F (x1, x2, . . . , xn) =∑
1≤i<j≤nx3ixj +
∑1≤i<j≤n
xix3j =
∑1≤i≤n
x3i∑j 6=i
xi
=
n∑i=1
x3i (1− xi) =
n∑i=1
xi(x2i − x3i ).
We claim that C =1
8. It suffices to show that
F (x1, x2, . . . , xn) ≤ 1
8= F
(1
2,1
2, 0, . . . , 0
).
Now, note that for any 0 ≤ x ≤ y ≤ 1
2, we have
x2 − x3 ≤ y2 − y3.
Indeed, since x+y ≤ 1, we get x+y ≥ (x+y)2 ≥ x2+xy+y2. Since y−x ≥ 0,this implies that y2 − x2 ≥ y3 − x3 or y2 − y3 ≥ x2 − x3, as desired.
292
To prove our claim, we need to consider two cases
The first one is when1
2≥ x1 ≥ x2 ≥ · · · ≥ xn. In this case, we apply the
above note and get
n∑i=1
xi(x2i − x3i ) ≤
n∑i=1
xi
[(1
2
)2
−(
1
2
)3]
=1
8
n∑i=1
xi =1
8.
The second case is when x1 ≥1
2≥ x2 ≥ . . . ≥ xn. Let x1 = x and y = 1−x =
x2 + . . . + xn. Since1
2≥ y ≥ x2, . . . , xn, we can apply the above note again
to get
F (x1, . . . , xn) = x3y+n∑i=2
xi(x2i−x3i ) ≤ x3y+
n∑i=2
xi(y2−y3) = x3y+y(y2−y3).
Since x3y + y(y2 − y3) = x3y + y3(1 − y) = xy(x2 + y2), it remains to showthat
xy(x2 + y2) ≤ 1
8.
Using x+ y = 1, we homogenize the above inequality as following
xy(x2 + y2) ≤ 1
8(x+ y)4.
However, we immediately find that (x+ y)4 − 8xy(x2 + y2) = (x− y)4 ≥ 0.?F?
99.6. For real numbers x1, x2, . . . , x6 ∈ [0, 1], prove the inequality
x31x52 + x53 + x54 + x55 + x56 + 5
+ · · ·+ x36x51 + x52 + x53 + x54 + x55 + 5
≤ 3
5.
(Ukraine 1999)
Solution: Since x1, . . . , x6 are in the interval [0, 1],
x52 + x53 + x54 + x55 + x56 + 5 ≥ x51 + x52 + x53 + x54 + x55 + x56 + 4.
By permuting the subscripts, we see that the left side of the inequality in theproblem is at most
6∑i=1
x3ix51 + x52 + · · ·+ x56 + 4
=
6∑i=1
x3i
6∑i=1
x5i + 4
.
For y ≥ 0, the AM-GM Inequality gives us 3y5 + 2 ≥ 5y3. Thus
56∑i=1
x3i ≤6∑i=1
(3x5i + 2) = 3
(6∑i=1
x5i + 4
),
293
that is6∑i=1
x3i
6∑i=1
x5i + 4
≤ 3
5.
This, together with our initial observations, completes the proof.?F?
99.7. Nonnegative real numbers p, q and r satisfy p+ q + r = 1. Prove that
7(pq + qr + rp) ≤ 2 + 9pqr.
(United Kingdom 1999)
Solution: According to the Schur’s Inequality (applied for third degree), wehave
4(p+ q + r)(pq + qr + rp) ≤ (p+ q + r)3 + 9pqr.
Using the constraint p+ q + r = 1, we deduce that
pq + qr + rp ≤ 1 + 9pqr
4.
Therefore, it suffices to prove that
7 · 1 + 9pqr
4≤ 2 + 9pqr, or equivalently, 27pqr ≤ 1.
Applying the AM-GM Inequality, we see that
27pqr ≤ (p+ q + r)3 = 1.
Therefore, the last inequality is obviously true and our proof is completed.
The equality holds if and only if p = q = r =1
3.
?F?
99.8. Let n > 3 and a1, a2, . . . , an be real numbers such that a1+a2+· · ·+an ≥n and a21 + a22 + · · ·+ a2n ≥ n2. Prove that max {a1, a2, . . . , an} ≥ 2.
(USA 1999)
First solution: The most natural idea is to suppose that ai < 2 for all i and
to substitute xi = 2 − ai > 0. Then we have
n∑i=1
(2 − xi) ≥ n or
n∑i=1
xi ≤ n,
and also
n2 ≤n∑i=1
a2i =
n∑i=1
(2− xi)2 = 4n− 4
n∑i=1
xi +
n∑i=1
x2i .
Now, using the fact that xi > 0, we obtain
n∑i=1
x2i <
(n∑i=1
xi
)2
, which com-
bined with the above inequality yields
n2 < 4n− 4n∑i=1
xi +
(n∑i=1
xi
)2
≤ 4n+ (n− 4)
n∑i=1
xi.
294
(we have used the fact thatn∑i=1
xi ≤ n). Thus, we have (n−4)
(n∑i=1
xi − n
)>
0, which is clearly impossible since n ≥ 4 and
n∑i=1
xi ≤ n. So, our assumption
was wrong and consequently max{a1, a2, . . . , an} ≥ 2.
Second solution: Assume that ai < 2 for all i = 1, 2, . . . , n, then we have
a1 + a2 + · · ·+ ai − 2(i− 1) < 2 ∀i = 1, 2, . . . , n.
Now, notice that for any m,n < 2, we have
(m+ n− 2)2 + 22 −m2 − n2 = 2(2−m)(2− n) > 0.
Using this inequality in combination with the above observation, we have
(a1 + a2 − 2)2 + 22 > a21 + a22,
[a1 + a2 + a3 − 2(3− 1)]2 + 22 > (a1 + a2 − 2)2 + a23,· · · · · · · · · · · · · · · · · · · · ·
[a1 + a2 + . . .+ an − 2(n− 1)]2 + 22 > [a1 + a2 + . . .+ an−1 − 2(n− 2)]2 + a2n.
Summing up these inequalities, we get
[a1 + a2 + · · ·+ an − 2(n− 1)]2 + 4(n− 1) > a21 + a22 + · · ·+ a2n.
Since ai < 2, n ≥ 4 and a1 + a2 + . . .+ an ≥ n, we have
n− 2 ≥ 2 > a1 + a2 + · · ·+ an − 2(n− 1) ≥ 2− n.
From here, we deduce that
[a1 + a2 + · · ·+ an − 2(n− 1)]2 ≤ (n− 2)2,
and hencea21 + a22 + · · ·+ a2n < (n− 2)2 + 4(n− 1) = n2,
which contradicts with the given hypothesis. So, our assumption was wrong,or in the other words, we must have max{a1, a2, . . . , an} ≥ 2.
?F?
99.9. Let a0, a1, . . . , an be numbers from the interval(
0,π
2
)such that
tan(a0 −
π
4
)+ tan
(a1 −
π
4
)+ · · ·+ tan
(an −
π
4
)≥ n− 1.
Prove thattan a0 tan a1 · · · tan an ≥ nn+1.
(USA 1999)
Solution: Let tan ai = xi for all i = 0, 1, . . . , n. It follows from the hypothesisthat for each i, xi > 0, and
x0 − 1
x0 + 1+x1 − 1
x1 + 1+ · · ·+ xn − 1
xn + 1≥ n− 1,
295
or equivalently,
1 ≥ 1
x0 + 1+
1
x1 + 1+ · · ·+ 1
xn + 1.
Now, for each i (0 ≤ i ≤ n), the AM-GM Inequality gives us
xixi + 1
= 1− 1
xi + 1≥∑j 6=i
1
xj + 1≥ n
n
√∏j 6=i
(xj + 1)
,
and hence, it follows that
xi ≥ nn
√√√√√ (xi + 1)n∏j 6=i
(xj + 1).
Setting i = 0, 1, . . . , n, we get the n + 1 similar inequalities and multiplyingthem up, we deduce that
x0x1 · · ·xn ≥ nn+1, or equivalently, tan a0 tan a1 · · · tan an ≥ nn+1,
as desired. The equality holds if and only if a0 = a1 = · · · = an = arctann.?F?
99.10. Let x, y, z > 1. Prove that
xx2+2yzyy
2+2zxzz2+2xy ≥ (xyz)xy+yz+zx .
(USA (Shortlist) 1999)
Solution: The original inequality is equivalent to(x2 + 2yz
)lnx+
(y2 + 2zx
)ln y +
(z2 + 2xy
)ln z ≥ (xy + yz + zx) ln (xyz) ,
or
(x− y) (x− z) lnx+ (y − x) (y − z) ln y + (z − x) (z − y) ln z ≥ 0.
Without loss of generality, we may assume that x ≥ y ≥ z. By this assumption,we have (x−y)(x−z) ≥ 0, (z−x)(z−y) ≥ 0. Therefore, with noting x, y, z > 1,we infer that
(x− y) (x− z) lnx ≥ (x− y)(x− z) ln y, and (z − x)(z − y) ln z ≥ 0.
From this, we can see that the left hand side of the above inequality is notless than
(x− y)(x− z) ln y + (y − z)(y − x) ln y = (x− y)2 ln y ≥ 0.
Our proof is completed. Note that the equality holds if and only if x = y = z.?F?
296
00.1. Let a, b be real numbers and a 6= 0. Prove that
a2 + b2 +1
a2+b
a≥√
3.
(Austria 2000)
Solution: Applying the AM-GM Inequality, we get
a2 + b2 +1
a2+b
a=
(b+
1
2a
)2
+ a2 +3
4a2≥ a2 +
3
4a2≥ 2
√a2 · 3
4a2=√
3,
as desired. The equality holds if and only if b = − 1
2aand a2 =
3
4a2, i.e. when
(a, b) equals
(4
√3
4,−1
24
√4
3
), or
(− 4
√3
4,1
24
√4
3
).
?F?
00.2. For all real numbers a, b, c ≥ 0 such that a+ b+ c = 1, prove that
2 ≤ (1− a2)2 + (1− b2)2 + (1− c2)2 ≤ (1 + a)(1 + b)(1 + c).
(Austrian-Polish Competition 2000)
Solution: Let us first prove the left inequality. Without loss of generality,
assume that a = max {a, b, c} . By this assumption, we have a ≥ 1
3. Now, with
noting that b2 + c2 ≤ (b + c)2 = (1 − a)2 ≤ 1, we apply the Cauchy SchwarzInequality and get
(1− b2)2 + (1− c2)2 ≥ (2− b2 − c2)2
2≥[2− (1− a)2
]22
=(1 + 2a− a2)2
2.
Therefore, it suffices to prove that
(1− a2)2 +(1 + 2a− a2)2
2≥ 2.
After some simple computations, we see that
(1− a2)2 +(1 + 2a− a2)2
2− 2 =
(1− a)2(1 + a)(3a− 1)
2,
which is clearly nonnegative since a ≥ 1
3. This ends the proof for the left
inequality. Note that the equality holds if and only if (a, b, c) equals (1, 0, 0),or (0, 1, 0), or (0, 0, 1).
For the right inequality, to prove it, let us notice that for every nonnegativenumbers x,
9(1− x)(1 + x)2 + 15x2 − 10x− 9 = −x(1− 3x)2 ≤ 0.
297
This implies that
(1− x)(1 + x)2 ≤ 9 + 10x− 15x2
9.
Using this inequality, we deduce that∑(1− a2)2 =
∑(1− a)(1− a)(1 + a)2 ≤ 1
9
∑(1− a)(9 + 10a− 15a2),
and hence, it suffices to prove that∑(1− a)(9 + 10a− 15a2) ≤ 9(1 + a)(1 + b)(1 + c).
Now, expanding with noting that a+ b+ c = 1, we can rewrite this inequalityas
15∑
a3 − 25∑
a2 + 10 ≤ 9∑
ab+ 9abc,
or
15∑
a3 − 25(∑
a2)(∑
a)
+ 10(∑
a)3≤ 9
(∑a)(∑
ab)
+ 9abc.
The last one can be simplified to
4∑
ab(a+ b) ≥ 24abc,
which is obviously true by the AM-GM Inequality. Therefore, the right handside inequality is proved. It is easy to see that the equality holds if and only
if (a, b, c) is one of the triples (1, 0, 0), (0, 1, 0), (0, 0, 1),
(1
3,1
3,1
3
).
?F?
00.3. Let a, b, c, x, y, z be positive real numbers. Prove that
a3
x+b3
y+c3
z≥ (a+ b+ c)3
3 (x+ y + z).
(Belarus 2000)
First solution: From the Cauchy Schwarz Inequality, we have
a3
x+b3
y+c3
z≥
(a√a+ b
√b+ c
√c)2
x+ y + z.
Using this inequality, it suffices to prove that
3(a√a+ b
√b+ c
√c)2≥ (a+ b+ c)3 .
Because of the homogeneity of this inequality, we may homogenize a+b+c = 3,and hence, it becomes
a√a+ b
√b+ c
√c ≥ 3.
298
Now, applying the Bernoulli’s Inequality, we have
a√a = [1 + (a− 1)]
32 ≥ 1 +
3
2(a− 1).
Adding this to the two analogous inequalities, we conclude that
a√a+ b
√b+ c
√c ≥ 3 +
3
2(a+ b+ c− 3) = 3,
as desired.
Second solution: By the Holder’s Inequality,(a3
x+b3
y+c3
z
)1/3
(1 + 1 + 1)1/3(x+ y + z)1/3 ≥ a+ b+ c.
Cubing both sides and then dividing both sides by 3(x+y+z) gives the desiredresult.
?F?
00.4. Suppose that the real numbers a1, a2, . . . , a100 satisfy(i) a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0;(ii) a1 + a2 ≤ 100;(iii) a3 + a4 + · · ·+ a100 ≤ 100.Determine the maximum possible value of a21 + a22 + · · · + a2100, and find allpossible sequences a1, a2, . . . , a100 which achieve this maximum.
(Canada 2000)
Solution: We have
a21 + a22 − a23 − (a1 + a2 − a3)2 = −2(a1 − a3)(a2 − a3) ≤ 0,
and hence
a21 + a22 ≤ a23 + (a1 + a2 − a3)2 ≤ a23 + (100− a3)2.
Note that the equality in this inequality holds iff a2 = a3 and a1 + a2 = 100.Now, since a3 ≥ a4 ≥ · · · ≥ a100, we have
a24 + · · ·+ a2100 ≤ a3a4 + · · ·+ a3a100 = a3(a4 + · · ·+ a100) ≤ a3(100− a3),
with equality iff a24 = a3a4, . . . , a2100 = a3a100 and a3(a3+a4+· · ·+a100−100) =
0. In addition, since a1 ≥ a2 ≥ a3 and a1 + a2 ≤ 100, we deduce that a3 ≤ 50,and hence a3(a3 − 50) ≤ 0 with equality iff a3(a3 − 50) = 0. From theseinequalities, we infer that
a21 + a22 + · · ·+ a2100 ≤ a23 + (100− a3)2 + a23 + a3(100− a3)= 10000 + 2a3(a3 − 50) ≤ 10000.
The equality holds if and only if the equality must hold in each inequalityfound above - that is, we must have:(a) a1 + a2 = 100, a2 = a3,
299
(b) a24 = a3a4, . . . , a2100 = a3a100, a3(a3 + a4 + · · ·+ a100 − 100) = 0,
(c) a3(a3 − 50) = 0.Solving the system of equations (a), (b) and (c) (with noting at the givenhypothesis), we can find that the equality holds only when the sequencea1, a2, . . . , a100 equals
100, 0, 0, . . . , 0, or 50, 50, 50, 50, 0, 0, . . . , 0.
?F?
00.5. Show that
3
√2 (a+ b)
(1
a+
1
b
)≥ 3
√a
b+
3
√b
a
for all positive real numbers a and b, and determine when the equality occurs.(Czech-Slovak 2000)
First solution: Multiplying both sides of the desired inequality by 3√ab gives
the equivalent inequality
3√a2 +
3√b2 ≤ 3
√2 (a+ b)2.
Setting 3√a = x and 3
√b = y, we see that it suffices to prove that
x2 + y2 ≤ 3
√2 (x3 + y3)2, or equivalently,
(x2 + y2
)3 ≤ 2(x3 + y3
)2.
Now, by applying the Cauchy Schwarz Inequality and the Chebyshev’s In-equality, we get(x2 + y2
)3=(x2 + y2
) (√x ·√x3 +
√y ·√y3)2≤(x2 + y2
)(x+ y)
(x3 + y3
)≤ 2
(x3 + y3
) (x3 + y3
)= 2
(x3 + y3
)2,
as desired. It is easy to see that the equality holds if and only if a = b.
Second solution: By the Power Mean Inequality, we have3
√a
b+ 3
√b
a
2
3
≤
√a
b+
√b
a
2
2
,
with equality if and only ifa
b=b
a, or equivalently a = b. It is easy to see that
the desired result follows from this inequality and the identity(√a
b+
√b
a
)2
= (a+ b)
(1
a+
1
b
).
?F?
300
00.6. Let a, b, c be positive real numbers such that abc = 1. Prove that
1 + ab2
c3+
1 + bc2
a3+
1 + ca2
b3≥ 18
a3 + b3 + c3.
(Hong Kong 2000)
Solution: Multiplying both sides of the desired inequality by a3 + b3 + c3 > 0gives the equivalent inequality
(a3 + b3 + c3
)( 1
a3+
1
b3+
1
c3
)+(a3 + b3 + c3
)(ab2c3
+bc2
a3+ca2
b3
)≥ 18.
Now, applying the AM-GM Inequality, we have
(a3 + b3 + c3
)( 1
a3+
1
b3+
1
c3
)≥ 3
3√a3b3c3 · 3 3
√1
a3b3c3= 9,
and
(a3 + b3 + c3
)(ab2c3
+bc2
a3+ca2
b3
)≥ 3
3√a3b3c3 · 3 3
√a3b3c3
a3b3c3= 9.
Adding up these two inequalities, we get the result. It is easy to see that theequality holds if and only if a = b = c = 1.
?F?
00.7. Let x, y, and z denote positive real numbers, each less than 4. Prove
that at least one of the numbers1
x+
1
4− y,
1
y+
1
4− z, and
1
z+
1
4− xis greater
than or equal to 1.(Hungary 2000)
Solution: From the Cauchy Schwarz Inequality, we have
1
x+
1
4− x≥ 4
x+ (4− x)= 1.
Now, if the three numbers1
x+
1
4− y,
1
y+
1
4− z, and
1
z+
1
4− xwere all less
than 1, their sum S would be less than 3. However,
S =
(1
x+
1
4− x
)+
(1
y+
1
4− y
)+
(1
z+
1
4− z
)≥ 3.
This contradiction proves the requested result.?F?
00.8. Let a, b, c be positive real numbers such that abc = 1. Prove the in-equality (
a+1
b− 1
)(b+
1
c− 1
)(c+
1
a− 1
)≤ 1.
(IMO 2000)
301
First solution: Without loss of generality, we can suppose that b = min{a, b, c},then it is clear that b ≤ 1. Now, replacing c =
1
ab, the inequality becomes
(a+
1
b− 1
)(b+ ab− 1)
(1
ab+
1
a− 1
)≤ 1,
or
(ab+ 1− b)(ab+ b− 1)(b+ 1− ab) ≤ ab2.
Because (ab+b−1)+(b+1−ab) = 2b > 0, we can see that max{ab+b−1, b+1 − ab} > 0. On the other hand, it is obvious that ab + 1 − b > 0, thereforeif min{ab + b − 1, b + 1 − ab} ≤ 0, the left hand side of the above inequalityis not greater than 0 while the right hand side is, so the inequality is trivial.Otherwise, one can see that
0 < (ab+ 1− b)(ab+ b− 1) = a2b2 − (b− 1)2 ≤ a2b2,
0 < (ab+ b− 1)(b+ 1− ab) = b2 − (ab− 1)2 ≤ b2,
0 < (b+ 1− ab)(ab+ 1− b) = 1− b2(a− 1)2 ≤ 1.
Multiplying these three inequalities and taking the square root, we can getthe desired result. The equality holds if and only if a = b = c = 1.
Second solution: Using the condition abc = 1, it is straightforward to verifythe equalities
2 =1
a
(a− 1 +
1
b
)+ c
(b− 1 +
1
c
),
2 =1
b
(b− 1 +
1
c
)+ a
(c− 1 +
1
a
),
2 =1
c
(c− 1 +
1
a
)+ b
(a− 1 +
1
c
).
In particular, they show that at most one of the numbers u = a − 1 +1
b,
v = b− 1 +1
c, w = c− 1 +
1
ais negative. If there is such a number, we have
(a− 1 +
1
b
)(b− 1 +
1
c
)(c− 1 +
1
a
)= uvw < 0 < 1.
And if u, v, w ≥ 0, the AM-GM Inequality yields
2 =1
au+ cv ≥ 2
√c
auv, 2 =
1
bv+aw ≥ 2
√a
bvw, 2 =
1
cw+aw ≥ 2
√b
cwu.
Thus, uv ≤ a
c, vw ≤ b
a, wu ≤ c
b, so (uvw)2 ≤ a
c· ba· cb
= 1. Since u, v, w ≥ 0,
this completes the proof.
302
Third solution: From the given hypothesis, we may substitute a =x
y, b =
y
z,
c =z
x, where x, y, z > 0. By this substitution, our inequality becomes(
x
y− 1 +
z
y
)(yz− 1 +
x
z
)( zx− 1 +
y
x
)≤ 1,
orxyz ≥ (y + z − x) (z + x− y) (x+ y − z) .
By expanding, we find that the last inequality is equivalent to
x3 + y3 + z3 + 3xyz ≥ xy(x+ y) + yz(y + z) + zx(z + x),
which is just the Schur’s Inequality (in the special case third degree).?F?
00.9. Let x, y ≥ 0 with x+ y = 2. Prove that
x2y2(x2 + y2) ≤ 2.
(Ireland 2000)
First solution: After homogenizing it, we need to prove
2
(x+ y
2
)6
≥ x2y2(x2 + y2), or (x+ y)6 ≥ 32x2y2(x2 + y2).
(Now, forget the constraint x + y = 2!) In case xy = 0, it clearly holds. Wenow assume that xy 6= 0. Because of the homogeneity of the inequality, thismeans that we may normalize to xy = 1. Then, it becomes(
x+1
x
)6
≥ 32
(x2 +
1
x2
), or p3 ≥ 32(p− 2),
where p =
(x+
1
x
)2
≥ 4. Our job is now to minimize F (p) = p3 − 32(p− 2)
on [4,∞). Since F ′(p) = 3p2 − 32 ≥ 0, where p ≥√
32
3, F is (monotone)
increasing on [4,∞). So, F (p) ≥ F (4) = 0 for all p ≥ 4.
Second solution: As in the first solution, we prove that (x+ y)6 ≥ 32(x2 +y2)(xy)2 for all x, y ≥ 0. In case x = y = 0, it’s clear. Now, if x2+y2 > 0, then
we may normalize to x2+y2 = 2. Setting p = xy, we have 0 ≤ p ≤ x2 + y2
2= 1
and (x+ y)2 = x2 + y2 + 2xy = 2 + 2p. It now becomes
(2 + 2p)3 ≥ 64p2, or p3 − 5p2 + 3p+ 1 ≥ 0.
We want to minimize F (p) = p3− 5p2 + 3p+ 1 on [0, 1]. We compute F ′(p) =
3
(p− 1
3
)(p − 3). We find that F is monotone increasing on
[0,
1
3
]and
303
monotone decreasing on
[1
3, 1
]. Since F (0) = 1 and F (1) = 0, we conclude
that F (p) ≥ F (1) = 0 for all p ∈ [0, 1].
Third solution: We show that (x+ y)6 ≥ 32(x2 + y2)(xy)2 where x ≥ y ≥ 0.We make the substitution u = x+ y and v = x− y. Then, we have u ≥ v ≥ 0.It becomes
u6 ≥ 32
(u2 + v2
2
)(u2 − v2
4
)2
, or u6 ≥ (u2 + v2)(u2 − v2)2.
Note that u4 ≥ u4 − v4 ≥ 0 and that u2 ≥ u2 − v2 ≥ 0. So,
u6 ≥ (u4 − v4)(u2 − v2) = (u2 + v2)(u2 − v2)2.
Fourth solution: According to the AM-GM Inequality, we find that
xy ≤(x+ y
2
)2
= 1,
and
xy(x2 + y2) =1
2· 2xy · (x2 + y2) ≤ 1
2
(2xy + x2 + y2
2
)2
= 2.
Therefore, by combining these two results, we conclude that
x2y2(x2 + y2) = xy · xy(x2 + y2) ≤ 1 · 2 = 2,
as desired.?F?
00.10. The real numbers a, b, c, x, y, z satisfy a ≥ b ≥ c > 0 and x ≥ y ≥ z >0. Prove that
a2x2
(by + cz)(bz + cy)+
b2y2
(cz + ax)(cx+ az)+
c2z2
(ax+ by)(ay + bx)≥ 3
4.
(Korea 2000)
First solution: Denote the left hand side of the given inequality by S. Be-cause a ≥ b ≥ c and x ≥ y ≥ z, by the Rearrangement Inequality, we havebz+ cy ≤ by+ cz so (by+ cz)(bz+ cy) ≤ (by+ cz)2 ≤ 2[(by)2 + (cz)2]. Settingm = (ax)2, n = (by)2, p = (cz)2, we obtain
a2x2
(by + cz)(bz + cy)≥ a2x2
2[(by)2 + (cz)2]=
m
2(n+ p).
Adding this to the two analogous inequalities, we find that
S ≥ 1
2
(m
n+ p+
n
p+m+
p
m+ n
).
304
On the other hand, according to the Nesbitt’s Inequality, it is easy to see that
m
n+ p+
n
p+m+
p
m+ n≥ 3
2,
and therefore S ≥ 3
4, as desired.
Second solution: According to the Cauchy Schwarz Inequality, we haveby+ cz ≤
√(b2 + c2)(y2 + z2) and bz+ cy ≤
√(b2 + c2)(z2 + y2). Multiplying
these inequalities, we get (by + cz)(bz + cy) ≤ (b2 + c2)(y2 + z2), and hence
a2x2
(by + cz)(bz + cy)≥ a2
b2 + c2· x2
y2 + z2.
Adding this to the two analogous inequalities, we find that
a2x2
(by + cz)(bz + cy)+
b2y2
(cz + ax)(cx+ az)+
c2z2
(ax+ by)(ay + bx)≥
≥ a2
b2 + c2· x2
y2 + z2+
b2
c2 + a2· y2
z2 + x2+
c2
a2 + b2· z2
x2 + y2.
On the other hand, from the given hypothesis, it is easy to verify that
a2
b2 + c2≥ b2
c2 + a2≥ c2
a2 + b2,
x2
y2 + z2≥ y2
z2 + x2≥ z2
x2 + y2.
Therefore, by the Chebyshev’s Inequality, we have
a2
b2 + c2· x2
y2 + z2+
b2
c2 + a2· y2
z2 + x2+
c2
a2 + b2· z2
x2 + y2≥
≥ 1
3
(a2
b2 + c2+
b2
c2 + a2+
c2
a2 + b2
)(x2
y2 + z2+
y2
z2 + x2+
z2
x2 + y2
)≥ 3
4,
where the last inequality holds according to the Nesbitt’s Inequalitym
n+ p+
n
p+m+
p
m+ n≥ 3
2. From this and the above estimation, we can get the
result.?F?
00.11. Let x, y, z be real numbers. Prove that
x2 + y2 + z2 ≥√
2 (xy + yz) .
(Macedonia 2000)
Solution: By the AM-GM Inequality, we find that
x2 + y2 + z2 =
(x2 +
y2
2
)+
(y2
2+ z2
)≥ 2
√x2 · y
2
2+ 2
√y2
2· z2
=√
2 (|xy|+ |yz|) ≥√
2(xy + yz),
305
as desired. It is easy to see that the equality holds if and only if x = z =y
2.
?F?
00.12. Let a, b, x, y, z be positive real numbers. Prove that
x
ay + bz+
y
az + bx+
z
ax+ by≥ 3
a+ b.
(MOSP 2000)
Solution: Using the Cauchy Schwarz Inequality in combination with the well-known inequality (x+ y + z)2 ≥ 3(xy + yz + zx), we deduce that
x
ay + bz+
y
az + bx+
z
ax+ by≥ (x+ y + z)2
x(ay + bz) + y(az + bx) + z(ax+ by)
=1
a+ b· (x+ y + z)2
xy + yz + zx≥ 3
a+ b,
as desired. It is easy to see that the equality holds if and only if x = y = z = 1.?F?
00.13. Let ABC be an acute triangle. Prove that(cosA
cosB
)2
+
(cosB
cosC
)2
+
(cosC
cosA
)2
+ 8 cosA cosB cosC ≥ 4.
(MOSP 2000)
Solution: Since 4−8 cosA cosB cosC = 4(cos2A+cos2B+cos2C), it sufficesto prove that(
cosA
cosB
)2
+
(cosB
cosC
)2
+
(cosC
cosA
)2
≥ 4(cos2A+ cos2B + cos2C).
Now, applying the AM-GM Inequality in combination with the well-known
inequality cosA cosB cosC ≤ 1
8, we have
(cosA
cosB
)2
+
(cosA
cosB
)2
+
(cosB
cosC
)2
≥ 33
√cos4A
cos2B cos2C
=3 cos2A
3√
cos2A cos2 b cos2C
≥ 3 cos2A
3
√1
82
= 12 cos2A.
Adding this inequality with its analogous forms and dividing both sides ofthe resulting inequality by 3, we obtain desired result. It is easy to see that
equality holds if and only if A = B = C =π
3.
?F?
306
00.14. Let a, b, c be positive real numbers such that min {a, b} ≥ c. Provethat √
c(a− c) +√c(b− c) ≤
√ab.
(MOSP 2000)
Solution: By the Cauchy Schwarz Inequality, we have[√c(a− c) +
√c(b− c)
]2≤ [c+ (b− c)] [(a− c) + c] = ab.
Taking square root of each side of this inequality yields the desired result.?F?
00.15. Let a1, a2, . . . , an be nonnegative real numbers. Prove that
a1 + a2 + · · ·+ ann
− n√a1a2 · · · an ≥
≥ 1
2min
{(√a1 −
√a2)
2 , (√a2 −
√a3)
2 , . . . , (√an −
√a1)
2}.
(MOSP 2000)
Solution: Denote an+1 = a1, because
nmin{
(√a1 −
√a2)
2 , (√a2 −
√a3)
2 , . . . , (√an −
√a1)
2}≤
n∑i=1
(√ai −
√ai+1)
2 ,
one can see that the original inequality is deduced from the following inequality
a1 + a2 + · · ·+ an − n n√a1a2 · · · an ≥
1
2
n∑i=1
(√ai −
√ai+1)
2 .
This is equivalent to
a1+a2+ · · ·+an−n n√a1a2 · · · an ≥ a1+a2+ . . .+an−(
√a1a2 + · · ·+
√ana1) ,
or √a1a2 + · · ·+
√ana1 ≥ n n
√a1a2 · · · an.
The last one is clearly true by the AM-GM Inequality, so our proof is com-pleted. Note that the equality holds if and only if a1 = a2 = · · · = an.
?F?
00.16. Let (an) be an infinite sequence of positive numbers such that
a11
+a42
+ · · ·+ ak2
k≤ 1
for all k. Prove thata11
+a22
+ · · ·+ ann< 2,
for all n.(MOSP 2000)
307
Solution: One can see that we can put n = k2+p where k ≥ 1 and 0 ≤ p ≤ 2k.Then
a11
+a22
+ · · ·+ ann
=a11
+a22
+ . . .+ak2+pk2 + p
≤ a11
+a22
+ · · ·+ak2+pk2 + p
+ · · ·+ ak2+2k
k2 + 2k
=k∑i=1
(ai2
i2+ · · ·+ ai2+2i
i2 + 2i
).
Since (an) is an decreasing sequence, it is clear that
ai2
i2+ · · ·+ ai2+2i
i2 + 2i≤(
1
i2+
1
i2 + 1+ · · ·+ 1
i2 + 2i
)ai2 .
On the other hand, we have
1
i2+ · · ·+ 1
i2 + 2i=
(1
i2+ · · ·+ 1
i2 + i− 1
)+
(1
i2 + i+ . . .+
1
i2 + 2i
)<
i
i2+i+ 1
i2 + i=
2
i.
Therefore, for any i = 1, 2, . . . , k, we have
ai2
i2+ · · ·+ ai2+2i
i2 + 2i≤ 2ai2
i,
and from this, it follows that
a11
+a22
+ · · ·+ ann< 2
k∑i=1
ai2
i≤ 2,
as desired.?F?
00.17. Let a, b, c be nonnegative real numbers such that ab + bc + ca = 1.Prove that
1
b+ c+
1
c+ a+
1
a+ b≥ 5
2.
(MOSP 2000)
First solution: If a+ b+ c ≥ 2, then we have
1
b+ c+
1
c+ a+
1
a+ b=ab+ bc+ ca
b+ c+ab+ bc+ ca
c+ a+ab+ bc+ ca
a+ b
= a+ b+ c+bc
b+ c+
ca
c+ a+
ab
a+ b
≥ a+ b+ c+bc
b+ c+ a+
ca
c+ a+ b+
ab
a+ b+ c
= a+ b+ c+1
a+ b+ c≥ 2 +
1
2=
5
2.
308
So, let us consider now the case a + b + c ≤ 2. Without loss of generality,
we can assume that a = max {a, b, c} , then a ≥ 1√3>
1
2. Accordingly, this
assumption allows us to proceed the following estimation
1
b+ c+
1
c+ a+
1
a+ b=ab+ bc+ ca
a+ b+ab+ bc+ ca
a+ c+
1
b+ c
=
(b+ c+
1
b+ c
)+
(ab
a+ b+
ac
a+ c
)≥ 2 +
ab
a+ b+
ac
a+ c= 2 +
a(1 + bc)
a(a+ b+ c) + bc
≥ 2 +a(1 + bc)
2a+ bc≥ 2 +
a(1 + bc)
2a+ 2abc
= 2 +1
2=
5
2.
Therefore, in any cases, we always have
1
b+ c+
1
c+ a+
1
a+ b≥ 5
2.
The equality holds if and only if (a, b, c) is a permutation of (1, 1, 0).?F?
00.18. Let a1, a2, . . . , an be positive real numbers such that
1
a1+
1
a2+ · · ·+ 1
an≤ 1.
Prove that for any positive integer k,
(ak1 − 1)(ak2 − 1) · · · (akn − 1) ≥ (nk − 1)n.
(MOSP 2000)
Solution: In order to prove the given inequality, we put bi =1
ai, and then
b1 + b2 + · · ·+ bn ≤ 1. By this substitution, we need to prove that(1
bk1− 1
)(1
bk2− 1
)· · ·(
1
bkn− 1
)≥ (nk − 1)n.
Firstly, we will show that(1
b1− 1
)(1
b2− 1
). . .
(1
bn− 1
)≥ (n− 1)n.
Indeed, by the AM-GM Inequality, we have
1
bi− 1 ≥ b1 + b2 + · · ·+ bn
bi− 1 =
∑j 6=i
bj
bi≥
(n− 1) n−1
√∏j 6=ibj
bi.
309
Setting i = 1, 2, . . . , n, we get the n similar inequalities and multiplying themup, we can get the above inequality. Now, since k is an positive integer, theHolder’s Inequality implies
n∏i=1
(1
bki− 1
)=
n∏i=1
(1
bi− 1
)·n∏i=1
(1 +
1
bi+ · · ·+ 1
bk−1i
)
≥ (n− 1)nn∏i=1
(1 +
1
bi+ · · ·+ 1
bk−1i
)
≥ (n− 1)n
[1 +
1n√b1b2 . . . bn
+ · · ·+ 1(n√b1b2 . . . bn
)k−1]n.
On the other hand, from the condition b1+b2+· · ·+bn ≤ 1, we can easily deduce
that n√b1b2 · · · bn ≤
1
n. Using this in combination with the above inequality,
we get
n∏i=1
(1
bki− 1
)≥ (n− 1)n(1 + n+ · · ·+ nk−1)n = (nk − 1)n,
as desired. Note that the inequality becomes equality if and only if a1 = a2 =· · · = an = n.
?F?
00.19. Let a, b, c be positive real numbers. Prove that
1
a+ b+
1
b+ c+
1
c+ a+
1
2 3√abc≥
(a+ b+ c+ 3
√abc)2
(a+ b)(b+ c)(c+ a).
(MOSP 2000)
First solution: Applying the Cauchy Schwarz Inequality, we have
1
a+ b+
1
b+ c+
1
c+ a+
1
2 3√abc
=a2
a2(b+ c)+
b2
b2(c+ a)+
c2
c2(a+ b)+
(3√abc)2
2abc
≥
(a+ b+ c+ 3
√abc)2
a2(b+ c) + b2(c+ a) + c2(a+ b) + 2abc.
On the other hand, it is easy to verify that a2(b+c)+b2(c+a)+c2(a+b)+2abc =(a+ b)(b+ c)(c+ a). Therefore, the above inequality implies
1
a+ b+
1
b+ c+
1
c+ a+
1
2 3√abc≥
(a+ b+ c+ 3
√abc)2
(a+ b)(b+ c)(c+ a).
The equality holds if and only if a = b = c = 1.
Second solution: The original inequality is equivalent to∑(a+ b)(a+ c) +
(a+ b)(b+ c)(c+ a)
2 3√abc
≥(a+ b+ c+
3√abc)2,
310
or
ab+ bc+ ca+(a+ b)(b+ c)(c+ a)
2 3√abc
≥ 23√abc(a+ b+ c) +
3√a2b2c2.
Since ab+ bc+ ca ≥ 33√a2b2c2, the above inequality follows from
(a+ b)(b+ c)(c+ a)
2 3√abc
+ 23√a2b2c2 ≥ 2
3√abc(a+ b+ c).
Multiplying both sides for 2 3√abc > 0, we can rewrite this inequality as
(a+ b)(b+ c)(c+ a) + 4abc ≥ 43√a2b2c2(a+ b+ c).
Without loss of generality, we can assume that a ≥ b ≥ c. Now, we write theinequality in form
(b+ c)[(a+ b)(a+ c)− 4
3√a2b2c2
]≥ 4
3√a2b2c2
(a− 3√abc).
Because (a+ b)(a+ c) = a2 + ab+ ac+ bc, it is equivalent to
(b+ c)(a2 + ab+ bc+ ca− 4
3√a2b2c2
)≥ 4
3√a2b2c2
(a− 3√abc).
Using again the estimation ab+ bc+ ca ≥ 33√a2b2c2, it suffices to prove that
(b+ c)(a2 − 3
√a2b2c2
)≥ 4
3√a2b2c2
(a− 3√abc),
or equivalently,
(b+ c)(a+
3√abc)≥ 4
3√a2b2c2.
Of course, this is obvious since by the AM-GM Inequality, we have
(b+ c)(a+
3√abc)≥ 2√bc · 2
√a
3√abc = 4
3√a2b2c2.
?F?
00.20. For any integer n ≥ 2, consider n−1 positive real numbers a1, a2, . . . , an−1having sum 1, and n real numbers b1, b2, . . . , bn. Prove that
b21 +b22a1
+b23a2
+ . . .+b2nan−1
≥ 2b1(b2 + b3 + . . .+ bn).
(Romania 2000)
Solution: By the Cauchy Schwarz Inequality, we have
b22a1
+b23a2
+ · · ·+ b2nan−1
≥ (b2 + b3 + · · ·+ bn)2
a1 + a2 + · · ·+ an−1= (b2 + b3 + · · ·+ bn)2.
From this inequality, we deduce that
b21 +b22a1
+b23a2
+ · · ·+ b2nan−1
≥ b21 + (b2 + b3 + · · ·+ bn)2
≥ 2 |b1(b2 + b3 + · · ·+ bn)|≥ 2b1(b2 + b3 + · · ·+ bn),
311
as desired.?F?
00.21. Positive real numbers x, y, z satisfy xyz = 1. Prove that the followinginequality holds
x2 + y2 + z2 + x+ y + z ≥ 2 (xy + yz + zx) .
(Russia 2000)
Solution: From the AM-GM Inequality and the given hypothesis, we havex+ y + z ≥ 3 3
√xyz = 3. Therefore, it suffices to prove that
x2 + y2 + z2 + 3 ≥ 2 (xy + yz + zx) .
Multiplying both sides of this inequality by x+ y+ z > 0, we can rewrite it as
(x2 + y2 + z2)(x+ y + z) + 3(x+ y + z) ≥ 2(x+ y + z)(xy + yz + zx).
Because(∑
x2)(∑
x)
=∑
x3 +∑
xy(x + y) and(∑
x)(∑
xy)
=∑xy(x+ y) + 3xyz =
∑xy(x+ y) + 3, the latter inequality is equivalent to
x3 + y3 + z3 + 3 (x+ y + z) ≥ xy (x+ y) + yz (y + z) + zx (z + x) + 6.
Now, we use the AM-GM Inequality and the Schur’s Inequality (applied forthird degree), and get
x3 + y3 + z3 + 3 (x+ y + z) =[x3 + y3 + z3 + (x+ y + z)
]+ 2 (x+ y + z)
≥(x3 + y3 + z3 + 3 3
√xyz
)+ 2 · 3 3
√xyz
=(x3 + y3 + z3 + 3xyz
)+ 2 · 3 3
√xyz
≥ xy (x+ y) + yz (y + z) + zx (z + x) + 6.
Therefore, the last inequality is valid and so, our proof is completed. Notethat the equality holds if and only if x = y = z = 1.
?F?
00.22. Let x1, x2, . . . , xn be real numbers (n ≥ 2), satisfying the conditions−1 < x1 < x2 < · · · < xn < 1 and
x131 + x132 + · · ·+ x13n = x1 + x2 + · · ·+ xn.
Prove that
x131 y1 + x132 y2 + · · ·+ x13n yn < x1y1 + x2y2 + · · ·+ xnyn
for any real numbers y1 < y2 < · · · < yn.
(Russia 2000)
312
Solution: Using the Abel’s summation formula, we have
n∑i=1
xiyi −n∑i=1
x13i yi =n∑i=1
yi(xi − x13i )
= y1
n∑i=1
(xi − x13i ) +n∑i=2
(yi − yi−1)n∑j=i
(xj − x13j )
=n∑i=2
(yi − yi−1)n∑j=i
(xj − x13j ).
Therefore, in order to prove the desired inequality, it suffices to prove that eachterm of the above sum is positive, and since yi− yi−1 > 0 for any i = 2, . . . , n,it suffices to prove that
n∑j=i
(xj − x13j ) > 0
for any i = 2, . . . , n. Indeed, if xi > 0 then we have 1 > xn > · · · > xi >0, from which it follows that xj > x13j for any j = i, . . . , n, and thereforen∑j=i
(xj−x13j ) > 0.Alternatively, if xi ≤ 0 then we have−1 < x1 < . . . < xi ≤ 0,
from which it follows that xj < x13j for any j = 1, . . . , i− 1, so it is clear that
i−1∑j=1
(xj − x13j ) < 0,
and since
n∑j=i
(xj − x13j ) = −i−1∑j=1
(xj − x13j ), we conclude that
n∑j=i
(xj − x13j ) > 0.
This completes our proof.?F?
00.23. Show that for all n ∈ N and x ∈ R,
sinn 2x+ (sinn x− cosn x)2 ≤ 1.
(Russia 2000)
Solution: For n = 0 and n = 1, the inequality becomes equality. In casen ≥ 2, denote a = sinx, b = cosx, then we have a2 + b2 = 1. The left handside of the desired inequality equals
(2ab)n + (an − bn)2 = a2n + b2n + (2n − 2)anbn,
while the right hand side equals
1 = (a2 + b2)n = a2n + b2n +
n−1∑j=1
(n
j
)a2(n−j)b2j .
313
It thus suffices to prove that
n−1∑j=1
(n
j
)a2(n−j)b2j ≥ (2n − 2)anbn. We can do so
by viewing
n−1∑j=1
(n
j
)a2(n−j)b2j as a sum of
n−1∑j=1
(n
j
)= 2n−2 terms of the form
a2(n−j)b2j , and then applying the AM-GM Inequality to these terms.?F?
00.24. Let n ≥ 3 be a positive integer. Prove that for all positive real numbersa1, a2, . . . , an, we have
a1 + a22
· a2 + a32
· · · an + a12
≤ a1 + a2 + a3
2√
2· · · an + a1 + a2
2√
2.
(Saint Petersburg 2000)
Solution: We take the indices of the ai modulo n. Observe that
4(ai−1 + ai + ai+1)2 = [(2ai−1 + ai) + (ai + 2ai+1)]
2
≥ 4(2ai−1 + ai)(ai + 2ai+1)
by the AM-GM Inequality. Equivalently,
(ai−1 + ai + ai+1)2 ≥ (2ai−1 + ai)(ai + 2ai+1).
In addition to this inequality, note that
(2ai−1 + ai)(ai + 2ai−1) ≥ 2(ai−1 + ai)2
since ai is positive number for each i. Multiplying these two inequalitiestogether for i = 1, 2, . . . , n, and then taking the square root of each side of theresulting inequality, gives the desired result.
?F?
00.25. Let n ≥ 3 be an integer. Prove that for positive numbers x1 ≤ x2 ≤· · · ≤ xn,
xnx1x2
+x1x2x3
+ · · ·+ xn−1xnx1
≥ x1 + x2 + · · ·+ xn.
(Saint Petersburg 2000)
First solution: Suppose that 0 ≤ x ≤ y and 0 < a ≤ 1. We have 1 ≥ a andy ≥ x ≥ ax, implying that (1− a)(y− ax) ≥ 0 or ax+ ay ≤ a2x+ y. Dividingboth sides of this final inequality by a, we find that
x+ y ≤ ax+y
a. (1)
We may set (x, y, a) =
(xn, xn ·
xn−1x2
,x1x2
)in (1) to find that
xn +xn−1xnx2
≤ xnx1x2
+xn−1xnx1
. (2)
314
Furthermore, for i = 1, 2, . . . , n−2, we may set (x, y, a) =
(x1, xn−1 ·
xi+1
x2,xi+1
xi+2
)in (1) to find that
xi + xn−1 ·xi+1
x2≤ xi ·
xi+1
xi+2+ xn−1 ·
xi+1
x2· xi+2
xi+1= xi ·
xi+1
xi+2+ xn−1 ·
xi+2
xi.
Summing this inequality for i = 1, 2, . . . , n− 2 yields
(x1 + · · ·+ xn−2) + xn−1 ≤(x1x2x3
+ · · ·+ xn−2xn−1xn
)+xn−1xnx2
.
Summing this last inequality with (2) yields the desired inequality.
Second solution: We will prove the inequality by induction on n. For n = 3,it becomes
x1x2x3
+x3x1x2
+x2x3x1≥ x1 + x2 + x3,
which is true since by the AM-GM Inequality, we have
2
(x1x2x3
+x3x1x2
+x2x3x1
)=∑(
x1x2x3
+x3x1x2
)≥ 2(x1 + x2 + x3).
Now, suppose that the equality holds for n ≥ 3 and let us prove it for n + 1,that is to prove
x1x2x3
+· · ·+xn−2xn−1xn
+xn−1xnxn+1
+xnxn+1
x1+xn+1x1x2
≥ x1+x2+· · ·+xn+xn+1.
Since x1 ≤ x2 ≤ · · · ≤ xn, we can apply the inductive hypothesis to get
x1x2x3
+ · · ·+ xn−2xn−1xn
+xn−1xnx1
+xnx1x2≥ x1 + x2 + · · ·+ xn.
Therefore, in order to prove that the inequality holds for n + 1, it suffices toprove that
xn−1xnxn+1
+xnxn+1
x1+xn+1x1x2
− xn+1 ≥xn−1xnx1
+xnx1x2
.
We have
xn−1xnxn+1
+xnxn+1
x1+xn+1x1x2
− xn−1xnx1
− xnx1x2− xn+1 =
=xn(xn+1 − xn−1)
x1+x1(xn+1 − xn)
x2+xn−1xn − x2n+1
xn+1
=xn(xn+1 − xn + xn − xn−1)
x1+x1(xn+1 − xn)
x2+xn−1xn − x2n + x2n − x2n+1
xn+1
= (xn+1 − xn)
(xnx1
+x1x2− xnxn+1
− 1
)+ (xn − xn−1)
(xnx1− xnxn+1
).
Because xn+1 ≥ xn ≥ xn−1, we find thatxnx1≥ 1 ≥ xn
xn+1, and
xnx1
+x1x2≥ 2
√xnx2≥ 2 ≥ xn
xn+1+ 1.
315
Therefore, the last quantity is obviously nonnegative and it implies that
xn−1xnxn+1
+xnxn+1
x1+xn+1x1x2
− xn+1 ≥xn−1xnx1
+xnx1x2
,
as desired. Our proof is completed.?F?
00.26. Let a, b, c, d be positive real numbers such that c2 + d2 =(a2 + b2
)3.
Prove thata3
c+b3
d≥ 1.
(Singapore 2000)
Solution: From the given hypothesis and the Cauchy Schwarz Inequality, wehave(
a3
c+b3
d
)(ac+ bd) =
(√a3
c
)2
+
(√b3
d
)2[(√ac)2 +
(√bd)2]
≥
(√a3
c·√ac+
√b3
d·√bd
)2
=(a2 + b2
)2=√
(a2 + b2) (c2 + d2) ≥ ac+ bd.
This implies thata3
c+b3
d≥ 1,
as desired.?F?
00.27. Given that x, y, z are positive real numbers satisfying xyz = 32, findthe minimum value of
x2 + 4xy + 4y2 + 2z2.
(United Kingdom 2000)
Solution: By the AM-GM Inequality, we have that
x2 + 4xy + 4y2 + 2z2 = (x− 2y)2 + 4xy + 4xy + 2z2 ≥ 4xy + 4xy + 2z2
≥ 3 3√
4xy · 4xy · 2z2 = 3 3√
32(xyz)2 = 96.
The equality holds when x = 2y and 4xy = 2z2, i.e. when x = 4, y = 2, z = 4.?F?
00.28. Prove that for any nonnegative real numbers a, b, c, the followinginequality holds
a+ b+ c
3− 3√abc ≤ max
{(√a−√b)2,(√
b−√c)2,(√c−√a)2}
.
(USA 2000)
316
First solution: We prove the stronger inequality
a+ b+ c− 33√abc ≤
(√a−√b)2
+(√
b−√c)2
+(√c−√a)2. (1)
The conclusion is immediate if abc = 0, so we assume that a, b, c > 0. Bymultiplying a, b, c by a suitable factor, we may reduce to the case abc = 1.Without loss of generality, assume that a and b are both greater than or equalto 1, or both less than or equal to 1. The desired inequality now becomes
a+ b+ c− 2√ab− 2
√bc− 2
√ca+ 3 ≥ 0.
By replacing c =1
ab,√bc =
1√a, and
√ca =
1√b, we find that it is equivalent
to
P = a+ b+1
ab− 2√ab− 2√
a− 2√
b+ 3 ≥ 0.
We have
P =(√
a−√b)2
+1
ab− 2√
a− 2√
b+ 3
=(√
a−√b)2
+
(1√a− 1
)2
+
(1√b− 1
)2
+1
ab− 1
a− 1
b+ 1
=(√
a−√b)2
+
(1√a− 1
)2
+
(1√b− 1
)2
+
(1
a− 1
)(1
b− 1
)≥ 0.
Therefore the inequality (1) is proved and from it, the result follows immedi-ately. Note that the equality holds if and only if a = b = c.
Second solution: We again prove the stronger inequality (1), which can bewritten ∑
sym
[a− 2(ab)1/2 + (abc)1/3] ≥ 0.
But this inequality follows from adding the two inequalities∑sym
[a− 2a2/3b1/3 + (abc)1/3] ≥ 0,
and ∑sym
(a2/3b1/3 + a1/3b2/3 − 2a1/2b1/2) ≥ 0.
The first of these is the Schur’s Inequality with x = a1/3, y = b1/3, z = c1/3,while the second follows from the AM-GM Inequality.
Third solution: Without loss of generality, assume that b is between a andc. The desired inequality reads
a+ b+ c− 33√abc ≤ 3
(a+ c− 2
√ac).
As a function of b, the right side minus the left side is concave (its second
derivative is −2(ac)1/3
3b5/3), so its minimum value in the range [a, c] occurs at
317
one of the endpoints. Thus, without loss of generality, we may assume a = b.Moreover, we may rescale the variables to get a = b = 1. Now the claim reads
2c+ 3c1/3 + 1
6≥ c1/2.
This is an instance of weighted AM-GM Inequality.?F?
01.1. Let a, b, c ≥ 0 such that a + b + c ≥ abc. Prove that the followinginequality holds
a2 + b2 + c2 ≥√
3abc.
(Balkan 2001)
Solution: From the AM-GM Inequality, we
(a+ b+ c)3 ≥ 27abc.
Multiplying both sides of this inequality by a+ b+ c > 0 and using the givenhypothesis, we deduce that
(a+ b+ c)4 = (a+ b+ c)3 (a+ b+ c) ≥ 27a2b2c2.
Taking the square root of each side of this inequality, we get
(a+ b+ c)2 ≥ 3√
3abc.
On the other hand, it is clear that 3(a2 + b2 + c2) ≥ (a + b + c)2 from theCauchy Schwarz Inequality. Combining this and the above inequality, we getthe result. It is easy to that the equality holds if and only if a = b = c =
√3.
?F?
01.2. Let x1, x2, x3 be real numbers in [−1, 1], and let y1, y2, y3 be real num-bers in [0, 1). Find the maximum possible value of the expression
1− x11− x2y3
· 1− x21− x3y1
· 1− x31− x1y2
.
(Belarus 2001)
Solution: Since 1−x1y2 = (1−x1)y2+(1−y2) > 0, 1+x1 ≥ 0 and 1−y2 > 0,we have
1− x11− x1y2
− 2
1 + y2= − (1 + x1)(1− y2)
(1 + y2)(1− x1y2)≤ 0.
From this, we deduce that
1− x11− x1y2
≤ 2
1 + y2≤ 2.
Similarly, we have1− x2
1− x2y3≤ 2,
1− x31− x3y1
≤ 2.
318
Now, note that from the given hypothesis, three expressions1− x1
1− x1y2,
1− x21− x2y3
,
1− x31− x3y1
must be nonnegative. From this note and the three inequalities
above, we get
1− x11− x2y3
· 1− x21− x3y1
· 1− x31− x1y2
=1− x1
1− x1y2· 1− x2
1− x2y3· 1− x3
1− x3y1≤ 8.
On the other hand, it is easy to see that the equality can be attained (forexample, take x1 = x2 = x3 = −1, y1 = y2 = y3 = 0), therefore the searchedmaximum is 8.
?F?
01.3. Let x and y be any two real numbers. Prove that
3(x+ y + 1)2 + 1 ≥ 3xy.
Under what conditions does equality hold?(Colombia 2001)
First solution: Denote t =x+ y
2, then we have xy ≤ t2 (according to the
AM-GM Inequality). And hence, it suffices to prove that
3(2t+ 1)2 + 1 ≥ 3t2.
We have 3(2t+1)2+1−3t2 = (3t+2)2 which is clearly nonnegative. Therefore,the above inequality holds and our proof is completed. Note that the equality
holds if and only if x = y = −2
3.
Second solution: For any real numbers X and Y , we have
X2 + Y 2 +XY =
(X +
Y
2
)2
+3Y 2
4≥ 0,
with equality if and only if X = Y = 0. Let x and y be any two real numbers.
Letting X = x+2
3and Y = y +
2
3, we obtain
(x+
2
3
)2
+
(y +
2
3
)2
+
(x+
2
3
)(y +
2
3
)≥ 0.
Expanding and multiplying by 3 gives
3x2 + 3y2 + 3xy + 6x+ 6y + 4 ≥ 0.
This may be written as
3(x+ y + 1)2 − 3xy + 1 ≥ 0,
from which we arrive at the desired inequality.?F?
319
01.4. Let n (n ≥ 2) be an integer and let a1, a2, . . . , an be positive realnumbers. Prove the inequality
(a31 + 1)(a32 + 1) · · · (a3n + 1) ≥ (a21a2 + 1) · · · (a2na1 + 1).
(Czech-Slovak-Polish 2001)
First solution: We try to apply the Cauchy Schwarz Inequality for eachfactor of the product in the right hand side. It is natural to write (1+a21a2)
2 ≤(1 + a31)(1 + a1a
22), since we need 1 + a31, which appears in the left hand side.
Similarly, we can write
(1 + a22a3)2 ≤ (1 + a32)(1 + a2a
23), . . . , (1 + a2na1)
2 ≤ (1 + a3n)(1 + ana21).
Multiplying we obtain
[(a21a2+1) · · · (a2na1+1)]2 ≤ [(a31+1) · · · (a3n+1)][(1+a1a22) · · · (1+ana
21)]. (∗)
Now, we use again the same argument to find that
[(1+a1a22) · · · (1+ana
21)]
2 ≤ [(a31+1) · · · (a3n+1)][(a21a2+1) · · · (a2na1+1)]. (∗∗)
Thus, if (a21a2 + 1) · · · (a2na1 + 1) ≥ (1 + a1a22) · · · (1 + ana
21), then (∗) will give
the answer, otherwise (∗∗) will. It is easy to see that the equality holds if andonly if a1 = a2 = · · · = an.
Second solution: From the Holder’s Inequality,
(a31 + 1)1/3(a31 + 1)1/3(a32 + 1)1/3 ≥ a21a2 + 1,
we get
(a31 + 1)2(a32 + 1) ≥ (a21a2 + 1)3.
In the same manner, we can establish the folllowing inequalities(a31 + 1
)2 (a32 + 1
)≥(a21a2 + 1
)3, . . . ,
(a3n + 1
)2 (a31 + 1
)≥(a2na1 + 1
)3.
Multiplying them up and taking the cube root of each side of the resultinginequality, we get the desired result.
?F?
01.5. Prove that for any positive real numbers a, b, c, the following inequalityholds
a√a2 + 8bc
+b√
b2 + 8ca+
c√c2 + 8ab
≥ 1.
(IMO 2001)
First solution: Applying the AM-GM Inequality, we have
a√a2 + 8bc
=2a(a+ b+ c)
2(a+ b+ c)√a2 + 8bc
≥ 2a(a+ b+ c)
(a+ b+ c)2 + a2 + 8bc.
320
By establishing the similar inequalities for the two other expressions, we get∑ a√a2 + 8bc
≥ 2(a+ b+ c)∑ a
(a+ b+ c)2 + a2 + 8bc.
On the other hand, the Cauchy Schwarz Inequality implies that
∑ a
(a+ b+ c)2 + a2 + 8bc=∑ a2
a(a+ b+ c)2 + a3 + 8abc
≥ (a+ b+ c)2∑[a(a+ b+ c)2 + a3 + 8abc]
=(a+ b+ c)2
(a+ b+ c)3 + a3 + b3 + c3 + 24abc.
It coud be said that∑ a√a2 + 8bc
≥ 2(a+ b+ c)3
(a+ b+ c)3 + a3 + b3 + c3 + 24abc.
Therefore, in order to prove the original inequality, it suffices to prove that
(a+ b+ c)3 ≥ a3 + b3 + c3 + 24abc.
Of course, this is trivial since by the AM-GM Inequality, we have
(a+ b+ c)3 = a3 + b3 + c3 + 3(a+ b)(b+ c)(c+ a) ≥ a3 + b3 + c3 + 24abc.
The equality occurs iff a = b = c.
Second solution: First we shall prove that
a√a2 + 8bc
≥ a43
a43 + b
43 + c
43
,
or equivalently, that (a
43 + b
43 + c
43
)2≥ a
23 (a2 + 8bc).
The AM-GM inequality yields(a
43 + b
43 + c
43
)2− a
83 =
(b43 + c
43
)(a
43 + a
43 + b
43 + c
43
)≥ 2b
23 c
23 · 8a
23 b
13 c
13 = 8a
23 bc.
Thus (a
43 + b
43 + c
43
)2≥ a
83 + 8a
23 bc = a
23 (a2 + 8bc),
so our statement is proved. Similarly, we have
b√b2 + 8ca
≥ b43
a43 + b
43 + c
43
, andc√
c2 + 8ab≥ c
43
a43 + b
43 + c
43
.
321
Adding these three inequalities yields
a√a2 + 8bc
+b√
b2 + 8ca+
c√c2 + 8ab
≥ 1.
Third solution: To remove the square roots, we make the following substi-tution
x =a√
a2 + 8bc, y =
b√b2 + 8ca
, z =c√
c2 + 8ab.
Clearly, x, y, z ∈ (0, 1). Our aim is to show that x+ y+ z ≥ 1. We notice that
a2
8bc=
x2
1− x2,
b2
8ac=
y2
1− y2,
c2
8ab=
z2
1− z2.
Hence1
512=
(x2
1− x2
)(y2
1− y2
)(z2
1− z2
).
Thus, we need to show that x+y+z ≥ 1 where 0 < x, y, z < 1 and (1−x2)(1−y2)(1 − z2) = 512(xyz)2. We use contradiction method. Assume that thereexist x, y, z satisfying both conditions 0 < x, y, z < 1, (1−x2)(1−y2)(1−z2) =512(xyz)2 and also x+y+z < 1. We will prove that this is impossible. Indeed,from 1 > x+ y + z, it follows that
(1− x2)(1− y2)(1− z2) >> [(x+ y + z)2 − x2][(x+ y + z)2 − y2][(x+ y + z)2 − z2]= (x+ x+ y + z)(y + z)(x+ y + y + z)(z + x)(x+ y + z + z)(x+ y)
≥ 4(x2yz)14 · 2(yz)
12 · 4(y2zx)
14 · 2(zx)
12 · 4(z2xy)
14 · 2(xy)
12
= 512(xyz)2.
This is a contradiction.?F?
01.6. Let a, b, c be positive real numbers such that abc = 1. Prove that
ab+cbc+aca+b ≤ 1.
(India 2001)
Solution: The desired inequality is equivalent to
(b+ c) ln a+ (c+ a) ln b+ (a+ b) ln c ≤ 0.
Without loss of generality, we may assume that a ≥ b ≥ c. By this assumption,we have b + c ≤ c + a ≤ a + b, and ln a ≥ ln b ≥ ln c. Therefore, by theChebyshev’s Inequality, we get∑
(b+ c) ln a ≤ 1
3[(b+ c) + (c+ a) + (a+ b)] (ln a+ ln b+ ln c)
=2
3(a+ b+ c) ln(abc) = 0,
322
as claimed. It is easy to see that the equality holds if and only if a = b = c = 1.?F?
01.7. Let x, y, z be positive real numbers such that xyz ≥ xy+yz+zx. Provethat
xyz ≥ 3 (x+ y + z) .
(India 2001)
Solution: Applying the well-known inequality (a+ b+ c)2 ≥ 3(ab+ bc+ ca)for the triple (a, b, c) = (xy, yz, zx), we deduce that
(xy + yz + zx)2 ≥ 3xyz(x+ y + z).
On the other hand, from the given hypothesis, we have x2y2z2 ≥ (xy + yz +zx)2. It follows that
x2y2z2 ≥ 3xyz(x+ y + z).
Dividing both sides of the last inequality by xyz > 0, we get the desired result.Note that the equality holds if and only if x = y = z = 3.
?F?
01.8. Prove that for any real numbers a, b, c the following inequality holds
(b+ c− a)2(c+ a− b)2(a+ b− c)2 ≥ (b2 + c2 − a2)(c2 + a2 − b2)(a2 + b2 − c2).
(Japan 2001)
First solution: The inequality is equivalent to
3(a2 + b2 + c2)(b+ c− a)2(c+ a− b)2(a+ b− c)2 ≥≥ 3(a2 + b2 + c2)(b2 + c2 − a2)(c2 + a2 − b2)(a2 + b2 − c2).
By the Cauchy Schwarz Inequality, we have
3(a2 + b2 + c2) ≥ (a+ b+ c)2,
and thus it is suffice to prove that
(a+ b+ c)2(b+ c− a)2(c+ a− b)2(a+ b− c)2 ≥≥ 3(a2 + b2 + c2)(b2 + c2 − a2)(c2 + a2 − b2)(a2 + b2 − c2).
Since
(a+ b+ c)2(b+ c− a)2(c+ a− b)2(a+ b− c)2 =(
2∑
a2b2 −∑
a4)2
and
3(a2+b2+c2)(b2+c2−a2)(c2+a2−b2)(a2+b2−c2) = 3(
2∑
a4b4 −∑
a8),
the last inequality is equivalent to(2∑
xy −∑
x2)2≥ 3
(2∑
x2y2 −∑
x4),
323
where x = a2, y = b2, z = c2. This last one rewrites as
4∑
x2(x− y)(x− z) ≥ 0,
which is a particular case of the Schur’s Inequality. Note that the equalityholds iff a = b = c, or c = 0 and a = b, or b = 0 and c = a, or a = 0 and b = c.
Second solution: It is clear that we only need to consider the case a2, b2, c2
are the sidelengths of a triangle (because in the conversed case, the giveninequality is trivial). Then, we will prove that
[a2 − (b− c)2]2 ≥ (a2 + b2 − c2)(a2 + c2 − b2).
Indeed, it is equivalent to
a4 − 2a2(b− c)2 + (b− c)4 ≥ a4 − (b− c)2(b+ c)2 ≥ 0,
or
(b− c)2(b2 + c2 − a2) ≥ 0.
Of course, this is true and so the claim is proved. Proceeding by the sameway, we can establish the similar inequalities as follows
[b2 − (c− a)2]2 ≥ (b2 + c2 − a2)(b2 + a2 − c2),
[c2 − (a− b)2]2 ≥ (c2 + a2 − b2)(c2 + b2 − a2).
Multiplying these three inequalities, we get
[a2−(b−c)2]2[b2−(c−a)2]2[c2−(a−b)2]2 ≥ (a2+b2−c2)2(b2+c2−a2)2(c2+a2−b2)2,
which is equivalent to
(a+b−c)4(b+c−a)4(c+a−b)4 ≥≥ (a2+b2−c2)2(b2+c2−a2)2(c2+a2−b2)2.
Now, taking square root for both sides, we can get the result.?F?
01.9. Let a, b, c be the sidelengths of an acute-angled triangle. Prove that
(a+ b+ c)(a2 + b2 + c2)(a3 + b3 + c3) ≥ 4(a6 + b6 + c6).
(Japan 2001)
Solution: By the Cauchy Schwarz Inequality, we have
(a+ b+ c)(a3 + b3 + c3) ≥ (a2 + b2 + c2)2,
and therefore it is enough to prove that
(a2 + b2 + c2)3 ≥ 4(a6 + b6 + c6).
324
Set x = a2, y = b2, z = c2. Since a, b, c are the sidelengths of an acute-angledtriangle, (it is well-known and easy to prove that) the numbers x, y, z aretheirselves the sidelengths of a triangle. The inequality to prove now becomes
(x+ y + z)3 ≥ 4(x3 + y3 + z3).
According to the triangle Inequality, we proceed as follows
(x+ y + z)3 = x3 + y3 + z3 + 3x2(y + z) + 3y2(z + x) + 3z2(x+ y) + 6xyz
≥ x3 + y3 + z3 + 3x2(y + z) + 3y2(z + x) + 3z2(x+ y)
≥ x3 + y3 + z3 + 3x2 · x+ 3y2 · y + 3z2 · z= 4(x3 + y3 + z3).
Note that the equality holds if and only if the triangle is degenerated, havingsidelengths of the type (0, t, t), where t is a positive real number.
?F?
01.10. Prove that for any real numbers x1, x2, . . . , xn, y1, y2, . . . , yn such thatx21 + x22 + · · ·+ x2n = y21 + y22 + · · ·+ y2n = 1,
(x1y2 − x2y1)2 ≤ 2
(1−
n∑k=1
xkyk
).
(Korea 2001)
First solution: We clearly have the inequality
(x1y2 − x2y1)2 ≤∑
1≤i<j≤n(xiyj − xjyi)2 =
(n∑i=1
x2i
)(n∑i=1
y2i
)−
(n∑i=1
xiyi
)2
=
(1−
n∑i=1
xiyi
)(1 +
n∑i=1
xiyi
).
Because we also have
∣∣∣∣∣n∑i=1
xiyi
∣∣∣∣∣ ≤ 1, we find immediately that
(1−
n∑i=1
xiyi
)(1 +
n∑i=1
xiyi
)≤ 2
(1−
n∑i=1
xiyi
),
and the problem is solved.
Second solution: Observe that
2
(1−
n∑k=1
xkyk
)=
n∑k=1
x2k +n∑k=1
y2k − 2n∑k=1
xkyk =n∑k=1
(xk − yk)2
≥ (x1 − y1)2 + (x2 − y2)2,
therefore, it suffices to prove that
(x1 − y1)2 + (x2 − y2)2 ≥ (x1y2 − x2y1)2.
325
However, this is true since by the Cauchy Schwarz Inequality, we have
(x1 − y1)2 + (x2 − y2)2 ≥ (x22 + x21)[(x1 − y1)2 + (x2 − y2)2]≥ [x2(x1 − y1) + x1(y2 − x2)]2 = (x1y2 − x2y1)2.
The proof is completed.?F?
01.11. Prove that if a, b, c are positive real numbers, then√a4 + b4 + c4 +
√a2b2 + b2c2 + c2a2 ≥
√a3b+ b3c+ c3a+
√ab3 + bc3 + ca3.
(Korea 2001)
Solution: Without loss of generality, we may assume that
max{a3b+ b3c+ c3a, ab3 + bc3 + ca3
}= a3b+ b3c+ c3a.
Then, we have√a3b+ b3c+ c3a+
√ab3 + bc3 + ca3 ≤ 2
√a3b+ b3c+ c3a.
On the other hand, applying the AM-GM Inequality in combination with theCauchy Schwarz Inequality, we find that√a4 + b4 + c4 +
√a2b2 + b2c2 + c2a2 ≥ 2 4
√(a4 + b4 + c4)(a2b2 + b2c2 + c2a2)
≥ 24
√(a2 · ab+ b2 · bc+ c2 · ca)2
= 2√a3b+ b3c+ c3a.
From this and the above inequality, it follows immediately that√a4 + b4 + c4 +
√a2b2 + b2c2 + c2a2 ≥
√a3b+ b3c+ c3a+
√ab3 + bc3 + ca3.
Note that the equality holds if and only if a = b = c.?F?
01.12. Prove that for all a, b, c > 0,√(a2b+ b2c+ c2a) (ab2 + bc2 + ca2) ≥ abc+ 3
√(a3 + abc) (b3 + abc) (c3 + abc).
(Korea 2001)
Solution: Dividing by abc, it becomes√(a
c+b
a+c
b
)(c
a+a
b+b
c
)≥ abc+ 3
√(a2
bc+ 1
)(b2
ca+ 1
)(c2
ab+ 1
).
After the substitution x =a
b, y =
b
c, z =
c
a, we obtain the constraint xyz = 1.
It takes the form
√(x+ y + z) (xy + yz + zx) ≥ 1 + 3
√(xz
+ 1)(y
x+ 1)(z
y+ 1
).
326
From the constraint xyz = 1, we find two identities(xz
+ 1)(y
x+ 1)(z
y+ 1
)= (z + x)(x+ y)(y + z),
and
(x+ y + z) (xy + yz + zx) = (x+y)(y+z)(z+x)+xyz = (x+y)(y+z)(z+x)+1.
Letting p = 3√
(x+ y)(y + z)(z + x), the inequality now becomes√p3 + 1 ≥
1+p. Applying the AM-GM Inequality, we have p ≥ 3
√2√xy · 2√yz · 2
√zx =
2. It follows that
(p3 + 1)− (1 + p)2 = p(p+ 1)(p− 2) ≥ 0.
The proof is completed. Note that the equality holds if and only if a = b = c.?F?
01.13. Prove that if a, b, c > 0 have product 1, then
(a+ b)(b+ c)(c+ a) ≥ 4(a+ b+ c− 1).
(MOSP 2001)
First solution: Using the identity (a+ b)(b+ c)(c+a) = (a+ b+ c)(ab+ bc+ca)− 1, we reduce the problem to the following one
ab+ bc+ ca+3
a+ b+ c≥ 4.
Now, we can apply the AM-GM Inequality in the following form
ab+ bc+ ca+3
a+ b+ c= 3 · ab+ bc+ ca
3+
3
a+ b+ c≥ 4 4
√(ab+ bc+ ca)3
9(a+ b+ c).
And so, it is enough to prove that
(ab+ bc+ ca)3 ≥ 9(a+ b+ c).
But this is easy, because we clearly have ab+ bc+ ca ≥ 3 and (ab+ bc+ ca)2 ≥3abc(a+ b+ c) = 3(a+ b+ c). It is easy to see that the equality holds if andonly if a = b = c = 1.
Second solution: We will use the fact that (a+ b)(b+ c)(c+ a) ≥ 8
9(a+ b+
c)(ab+ bc+ ca). So, it is enough to prove that
2
9(ab+ bc+ ca) +
1
a+ b+ c≥ 1.
Using the AM-GM Inequality, we can write
2
9(ab+ bc+ ca) +
1
a+ b+ c≥ 3 3
√(ab+ bc+ ca)2
81(a+ b+ c)≥ 1,
327
because (ab+ bc+ ca)2 ≥ 3abc(a+ b+ c) = 3(a+ b+ c).
Third solution: By homogenizing, we can write the inequality as
(a+ b)(b+ c)(c+ a) + 4abc ≥ 43√a2b2c2(a+ b+ c).
Without loss of generality, we can assume that a ≥ b ≥ c. Now, we write theinequality in form
(b+ c)[(a+ b)(a+ c)− 4
3√a2b2c2
]≥ 4
3√a2b2c2
(a− 3√abc).
Because (a+ b)(a+ c) = a2 + ab+ ac+ bc, it is equivalent to
(b+ c)(a2 + ab+ bc+ ca− 4
3√a2b2c2
)≥ 4
3√a2b2c2
(a− 3√abc).
Using again the estimation ab+ bc+ ca ≥ 33√a2b2c2, it suffices to prove that
(b+ c)(a2 − 3
√a2b2c2
)≥ 4
3√a2b2c2
(a− 3√abc),
or equivalently,
(b+ c)(a+
3√abc)≥ 4
3√a2b2c2.
Of course, this is obvious since by the AM-GM Inequality, we have
(b+ c)(a+
3√abc)≥ 2√bc · 2
√a
3√abc = 4
3√a2b2c2.
?F?
01.14. Show that the inequality
n∑i=1
ixi ≤(n
2
)+
n∑i=1
xii
holds for every integer n ≥ 2 and all real numbers x1, x2, . . . , xN ≥ 0.(Poland 2001)
Solution: From the Bernoulli’s Inequality, for each nonnegative real numberx and for each positive integer k, we have xk = [1 + (x− 1)]k ≥ 1 + k(x− 1),and from this, we deduce that kx ≤ xk +k− 1, with equality, for k ≥ 2, if andonly if x = 1. Using this for each xi and summing leads to
n∑i=1
ixi ≤n∑i=1
[xii + (i− 1)] =
(n
2
)+
n∑i=1
xii,
as desired. Note that the equality holds if and only if x2 = · · · = xn = 1 (andx1 ≥ 0 is arbitrary).
?F?
01.15. Let a and b be positive real numbers in the interval (0, 1]. Prove that
1√1 + a2
+1√
1 + b2≤ 2√
1 + ab.
328
(Russia 2001)
Solution: Because a and b are positive real numbers, there are angles x and y,with 0◦ < x, y < 90◦, such that tanx = a and tan y = b. The desired inequalityis clearly true when a = b. Hence we assume that a 6= b, or equivalently, x 6= y.
Then 1 + a2 = sec2 x and1√
1 + a2= cosx. Note that
1 + ab =cosx cos y + sinx sin y
cosx cos y=
cos(x− y)
cosx cos y
by the addition and subtraction formulas. The desired inequality reduces to
cosx+ cos y ≤ 2
√cosx cos y
cos(x− y).
Squaring both sides, we can rewrite it as
cos2 x+ cos2 y + 2 cosx cos y ≤ 4 cosx cos y
cos(x− y).
Because 0◦ < |x − y| < 90◦, it follows that 0 < cos(x − y) < 1. Hence
2 cosx cos y ≤ 2 cosx cos y
cos(x− y). It suffices to show that
cos(x− y)(cos2 x+ cos2 y) ≤ 2 cosx cos y,
orcos(x− y)(cos 2x+ cos 2y + 2) ≤ 4 cosx cos y
by the double-angle formulas. By the sum-to-product formulas, the last in-equality is equivalent to
cos(x− y)[2 cos(x− y) cos(x+ y) + 2] ≤ 2[cos(x− y) + cos(x+ y)],
or cos 2(x − y) cos(x + y) ≤ cos(x + y), which is clearly true, because for0 < a, b ≤ 1, we have 0◦ < x, y ≤ 45◦, and so 0◦ < x + y ≤ 90◦ andcos(x+ y) > 0. This completes our proof.
?F?
01.16. Let a, b, c, x, y, z be positive real numbers such that x + y + z = 1.Prove that
ax+ by + cz + 2√
(xy + yz + zx)(ab+ bc+ ca) ≤ a+ b+ c.
(Ukraine 2001)
First solution: We will use the Cauchy Schwarz Inequality twice. First, wecan write ax + by + cz ≤
√a2 + b2 + c2 ·
√x2 + y2 + z2 and then we apply
again the Cauchy Schwarz Inequality to obtain
ax+ by + cz + 2√
(xy + yz + zx)(ab+ bc+ ca) ≤
≤√∑
a2 ·√∑
x2 +
√2∑cyc
xy ·√
2∑
ab
≤√∑
a2 + 2∑
ab ·√∑
x2 + 2∑
xy =∑
a.
329
Second solution: The inequality being homogeneous in a, b, c, so we canassume that a+ b+ c = 1. We apply this time the AM-GM Inequality and wefind that the left hand side of the original inequality is not greater than
ax+ by + cz + xy + yz + zx+ ab+ bc+ ca.
Consequently,
xy + yz + zx+ ab+ bc+ ca =1− x2 − y2 − z2
2+
1− a2 − b2 − c2
2≤ 1− ax− by − cz,
the last one being equivalent to (x− a)2 + (x− b)2 + (x− c)2 ≥ 0.
Third solution: Applying the AM-GM Inequality, we have
2√
(ab+ bc+ ca)(xy + yz + zx) =
= 2
√[a(y + z) +
bc(y + z)
b+ c
] [x(b+ c) +
yz(b+ c)
y + z
]≤ a(y + z) +
bc(y + z)
b+ c+ x(b+ c) +
yz(b+ c)
y + z.
Therefore, if we denote with P the left hand side of the original inequality,then we find that
P ≤ ax+ by + cz + a(y + z) +bc(y + z)
b+ c+ x(b+ c) +
yz(b+ c)
y + z.
And so, it suffices to prove that
ax+by+cz+a(y+z)+bc(y + z)
b+ c+x(b+c)+
yz(b+ c)
y + z≤ (a+b+c)(x+y+z),
or
bz + cy ≥ bc(y + z)
b+ c+yz(b+ c)
y + z.
The last one is true since by the Cauchy Schwarz Inequality, we have
bz + cy − bc(y + z)
b+ c=b2z + c2y
b+ c=
(b2z + c2y)(y + z)
(b+ c)(y + z)
≥(b√zy + c
√yz)2
(b+ c)(y + z)=yz(b+ c)
y + z.
?F?
01.17. Let a, b, c be positive real numbers such that a + b + c ≥ abc. Provethat at least two of the inequalities
2
a+
3
b+
6
c≥ 6,
2
b+
3
c+
6
a≥ 6,
2
c+
3
a+
6
b≥ 6
330
are true.(USA 2001)
Solution: Denote x =1
a, y =
1
b, z =
1
c, then we have xy + yz + zx ≥ 1, and
we are required to prove that: There are at least two of the inequalities
2x+ 3y + 6z ≥ 6, 2y + 3z + 6x ≥ 6, 2z + 3x+ 6y ≥ 6
are true. We have that
(2x+ 3y + 6z) + (2y + 3z + 6x) = 8x+ 5y + 9z,
and
(8x+ 5y + 9z)2 − 144(xy + yz + zx) = 64x2 − 64xy + 25y2 − 54yz + 81z2
= 16(2x− y)2 + 9(y − 3z)2 ≥ 0,
therefore
(2x+ 3y + 6z) + (2y + 3z + 6x) ≥ 12√xy + yz + zx ≥ 12,
and it follows that
max {2x+ 3y + 6z, 2y + 3z + 6x} ≥ 6.
Without loss of generality, we may assume that max {2x+ 3y + 6z, 2y + 3z + 6x} =2x+ 3y + 6z, then from the above inequality, we have 2x+ 3y + 6z ≥ 6.On the other hand, proceeding by the same way as above, we can also findthat
(2y + 3z + 6x) + (2z + 3x+ 6y) ≥ 12,
from which it follows
max {2y + 3z + 6x, 2z + 3x+ 6y} ≥ 6.
These arguments show that there are at least 2 inequalities among the giveninequalities holds.
?F?
01.18. Prove that for any nonnegative real numbers a, b, c such that a2 + b2 +c2 + abc = 4, we have
0 ≤ ab+ bc+ ca− abc ≤ 2.
(USA 2001)
First solution: From the condition, at least one of a, b and c does not exceed1, say a ≤ 1. Then
ab+ bc+ ca− abc = a(b+ c) + bc(1− a) ≥ 0.
To obtain the equality, we have a(b+c) = bc(1−a) = 0. If a = 1, then b+c = 0or b = c = 0, which contradicts the given condition a2 + b2 + c2 + abc = 4.
331
Hence 1− a 6= 0 and only one of b and c is 0. Without loss of generality, sayb = 0. Therefore b+ c > 0 and a = 0. Plugging a = b = 0 back into the givencondition gives c = 2. By permutation, the lower bound holds if and only if(a, b, c) is one of the triples (2, 0, 0), (0, 2, 0), and (0, 0, 2).
Now we prove the upper bound. Let us note that at least two of the threenumbers a, b, and c are both greater than or equal to 1 or less than or equal to1. Without loss of generality, we assume that the numbers with this propertyare b and c. Then we have
(1− b)(1− c) ≥ 0. (1)
The given equality a2 + b2 + c2 + abc = 4 and the inequality b2 + c2 ≥ 2bcimply
a2 + 2bc+ abc ≤ 4, or bc(2 + a) ≤ 4− a2.
Dividing both sides of the last inequality by 2 + a yields
bc ≤ 2− a. (2)
Combining (1) and (2) gives
ab+ bc+ ca− abc ≤ a(b+ c) + (2− a)− abc= 2− a(1 + bc− b− c)= 2− a(1− b)(1− c) ≤ 2,
as desired. The last equality holds if and only if b = c and a(1− b)(1− c) = 0.Hence, the equality for the upper bound holds if and only if (a, b, c) is one ofthe triples (1, 1, 1),
(0,√
2,√
2),(√
2, 0,√
2), and
(√2,√
2, 0).
Second solution: We won’t prove again the lower part, since this is an easyproblem. Let us concentrate on the upper bound. Let a ≥ b ≥ c and leta = x+ y, b = x− y (x ≥ y ≥ 0). The hypothesis becomes x2(2 + c) + y2(2−c) = 4 − c2, and we have to prove that (x2 − y2)(1 − c) ≤ 2(1 − xc). Since
y2 = 2 + c− 2 + c
2− cx2, the problem asks to prove the inequality
4x2 − (4− c2)2− c
(1− c) ≤ 2(1− xc).
Of course, we have c ≤ 1 and 0 ≤ y2 = 2 + c − 2 + c
2− cx2, therefore x2 ≤ 2 − c
and hence x ≤√
2− c. Now, consider the function f :[0,√
2− c]→ R,
f(x) = 2(1− cx)− 4x2 − (4− c2)2− c
(1− c).
We have f ′(x) = −2c − 8x(1− c)2− c
≤ 0 and thus f is decreasing and f(x) ≥
f(√
2− c). So we have to prove that f
(√2− c
)≥ 0, or equivalently
2(1− c
√2− c
)≥ (2− c)(1− c).
332
By some simple calculations, we find that it is equivalent to c(1−√
2− c)2 ≥
0, clearly true. Thus, the problem is solved.
Third solution: We give another way to prove the upper part
ab+ bc+ ca− abc ≤ 2.
Letting a = 2p, b = 2q, c = 2r, we get p2 + q2 + r2 + 2pqr = 1. Therefore, we
may put a = 2 cosA, b = 2 cosB, c = 2 cosC for some A,B,C ∈[0,π
2
]with
A+B + C = π. We are required to prove
cosA cosB + cosB cosC + cosC cosA− 2 cosA cosB cosC ≤ 1
2.
Assume that A ≥ π
3or 1− 2 cosA ≥ 0. Note that
cosA cosB + cosB cosC + cosC cosA− 2 cosA cosB cosC =
= cosA(cosB + cosC) + cosB cosC(1− 2 cosA).
We apply the Jensen’s Inequality to deduce cosB + cosC ≤ 32 − cosA. Note
that 2 cosB cosC = cos(B − C) + cos(B + C) ≤ 1− cosA. These imply that
cosA(cosB + cosC) + cosB cosC(1− 2 cosA) ≤
≤ cosA
(3
2− cosA
)+
(1− cosA
2
)(1− 2 cosA).
However, it’s easy to verify that cosA
(3
2− cosA
)+
(1− cosA
2
)(1−2 cosA) =
1
2.
?F?
01.19. Let x, y, z be positive real numbers satisfying
(i)1√2≤ z ≤ 1
2min
{x√
2, y√
3}
;
(ii) x+ z√
3 ≥√
6;(iii) y
√3 + z
√10 ≥ 2
√5.
Find the maximum of P (x, y, z) =1
x2+
2
y2+
3
z2.
(Vietnam 2001)
First solution: From the first condition, we have√
2 ≥ 1
z≥√
2
x, and it
follows that1
z2≤ 2,
z2
x2≤ 1
2. The second condition leads us to x2 + 3z2 ≥ 3,
and from it, we deduce that2
3+
2z2
x2≥ 2
x2. Thus
1
x2+
1
z2=
2
x2+
1
z2− 1
x2≤ 2
3+
2z2
x2+
1
z2
(1− z2
x2
)≤ 2
3+
2z2
x2+ 2
(1− z2
x2
)=
8
3.
333
Similarly, using the first and third conditions, we get
1
y2+
1
z2≤ 13
5.
From these two inequalities, we obtain
P =1
x2+
1
z2+ 2
(1
y2+
1
z2
)≤ 8
3+
26
3=
118
15.
In addition, from above proofs, it is easy to see that P =118
15if and only if
x =
√3
2, y =
√5
3, z =
1√2. Obviously, these values of x, y, z satisfy (i), (ii)
and (iii). So maxP =118
15.
Second solution: From the given conditions, we have x, y > z and
x ≥√
3(√
2− z), y ≥
√10
3
(√2− z
).
To find the maximum of P, we need to consider two cases:
Case 1.√
3(√
2− z)≥ z. In this case, we find that
1√2≤ z ≤
√6√
3 + 1, and
1
x2+
2
y2+
3
z2≤ 1
3(√
2− z)2 +
210
3
(√2− z
)2 +3
z2
=14
15(√
2− z)2 +
3
z2= f(z).
Now, we find that f ′′(z) =18
z4+
28
5(√
2− z)4 > 0, therefore f(z) is convex,
and since1√2≤ z ≤
√6√
3 + 1, it follows that
f(z) ≤ max
{f
(1√2
), f
( √6√
3 + 1
)}= max
{118
15,58 + 29
√3
15
}=
118
15.
Case 2.√
3(√
2− z)≤ z. In this case, we find that z ≥
√6√
3 + 1. Therefore,
by using the inequalities x > z and y > z, we get
1
x2+
2
y2+
3
z2≤ 6
z2≤ 6( √
6√3 + 1
)2 = 4 + 2√
3 <118
15.
Actually, we have proved in both cases the following inequality
P =1
x2+
2
y2+
3
z2≤ 118
15.
334
Moreover, let x =
√3
2, y =
√5
3, z =
1√2, we can easily see that the conditions
(i), (ii) and (iii) are still satisfied and also P =118
15. Therefore maxP =
118
15.
?F?
01.20. Let x, y, z be positive real numbers such that
(i)2
5≤ z ≤ min {x, y} ;
(ii) xz ≥ 4
15;
(iii) yz ≥ 1
5.
Determine the maximum possible value of
P (x, y, z) =1
x+
2
y+
3
z.
(Vietnam 2001)
First solution: From the given condition, we have x ≥ 4
15z, y ≥ 1
5z. To find
the maximum value of P, we will consider two cases
The first case is when4
15z≥ z. In this case, we have
2
5≤ z ≤ 2√
15. Now,
observe that
1
x+
2
y+
3
z≤ 15z
4+ 2 · 5z +
3
z=
55
4z +
3
z= f(z).
It is easy to check that f(z) is convex, therefore
f(z) ≤ max
{f
(2
5
), f
(2√15
)}= max
{13,
10√
5
3
}= 13,
and hence, we deduce that P ≤ 13 with equality iff z =2
5, x =
4
15z=
2
3, y =
1
5z=
1
2.
The second case is when4
15z≤ z. In this case, we have z ≥ 2√
15, and hence
1
x+
2
y+
3
z≤ 6
z≤ 3√
15 < 13.
So, in both cases, we always have P ≤ 13 with equality iff x =2
3, y =
1
2, z =
2
5.
Therefore, the searched maximum of P is 13.
Second solution: Applying the AM-GM Inequality in combination with thegiven hypothesis, we deduce that
3x+ 5z ≥ 2√
15xz ≥ 4, 4y + 5z ≥ 4√
5yz ≥ 4.
335
From this, it follows that
2
x≤ 5z
2x+
3
2,
2
y≤ 5z
2y+ 2.
Using these two inequalities with noting that 1− z
x≥ 0, 1− z
y≥ 0 and z ≥ 2
5,
we have
1
x+
1
z=
2
x+
1
z
(1− z
x
)≤ 5z
2x+
3
2+
1
z
(1− z
x
)≤ 5z
2x+
3
2+
5
2
(1− z
x
)= 4,
and
1
y+
1
z=
2
y+
1
z
(1− z
y
)≤ 5z
2y+ 2 +
1
z
(1− z
y
)≤ 5z
2y+ 2 +
5
2
(1− z
y
)=
9
2.
Therefore, we conclude that
P =1
x+
2
y+
3
z=
(1
x+
1
z
)+ 2
(1
y+
1
z
)≤ 4 + 2 · 9
2= 13,
with equality if and only if 3x = 4y = 5z, xz =4
15, yz =
1
5, z =
2
5, i.e. when
x =2
3, y =
1
2, z =
2
5.
?F?
01.21. Find the minimum value of the expression1
a+
2
b+
3
cwhere a, b, c are
positive real numbers such that 21ab+ 2bc+ 8ca ≤ 12.(Vietnam 2001)
First solution: Let x =1
a, y =
2
b, z =
3
c. Then it is easy to check that
the condition of the problem becomes 2xyz ≥ 2x + 4y + 7z. And we need tominimize x+y+z. But from the hypothesis, we find that z(2xy−7) ≥ 2x+4y
and hence 2xy > 7, z ≥ 2x+ 4y
2xy − 7. Now, we transform the expression so that
after one application of the AM-GM Inequality, the numerator 2xy− 7 shouldvanish
x+y+z ≥ x+y+2x+ 4y
2xy − 7= x+
11
2x+
(y − 7
2x
)+
2x+14
x2xy − 7
≥ x+11
2x+2
√1 +
7
x2.
But, it is immediate to prove that 2
√1 +
7
x2≥
3 +7
x2
and so x + y + z ≥3
2+ x +
9
x≥ 15
2. We have equality for x = 3, y =
5
2, z = 2. Therefore, in the
initial problem, the answer is15
2, achieved for a =
1
3, b =
4
5, c =
3
2.
336
Second solution: We use the same substitution and reduce the problem tofinding the minimum value of x+ y + z when 2xyz ≥ 2x+ 4y + 7z. Applyingthe weighted AM-GM Inequality, we find that
x+ y + z ≥(
5x
2
) 25
(3y)13
(15z
4
) 415
.
And also 2x + 4y + 7z ≥ 1015 · 12
13 · 5
715 · x
15 · y
13 · z
715 . This means that
(x + y + z)2(2x + 4y + 7z) ≥ 225
2xyz. Because 2xyz ≥ 2x + 4y + 7z, we will
have
(x+ y + z)2 ≥ 225
4, or x+ y + z ≥ 15
2,
with equality for x = 3, y =5
2, z = 2.
?F?
02.1. Let x, y, z be positive real numbers such that
1
x+
1
y+
1
z= 1.
Prove that
√x+ yz +
√y + zx+
√z + xy ≥ √xyz +
√x+√y +√z.
(APMO 2002)
Solution: Let a =1
x, b =
1
y, c =
1
z. Then from the given hypothesis, we have
a+ b+ c = 1, and the original inequality becomes√1
a+
1
bc+
√1
b+
1
ca+
√1
c+
1
ab≥ 1√
abc+
1√a
+1√b
+1√c,
which is equivalent to
√a+ bc+
√b+ ca+
√c+ ab ≥
√ab+
√bc+
√ca+ 1,
or ∑√a(a+ b+ c) + bc ≥
√ab+
√bc+
√ca+ a+ b+ c.
By the Cauchy Schwarz Inequality, we have∑√a(a+ b+ c) + bc =
∑√(a+ b)(a+ c) ≥
∑(a+√bc)
=∑
a+∑√
ab.
Therefore, the last inequality is true and thus, our problem is solved. Notethat the equality holds if and only if x = y = z = 3.
?F?
02.2. Let a, b, c be positive real numbers. Prove that
a3
b2+b3
c2+c3
a2≥ a2
b+b2
c+c2
a.
337
(Balkan (Shortslit) 2002)
Solution: Using the AM-GM Inequality, we have
a3
b2+ a ≥ 2a2
b,
b3
c2+ b ≥ 2b2
c,
c3
a2+ c ≥ 2c2
a.
Adding up these three inequalities, we deduce that
a3
b2+b3
c2+c3
a2≥ 2
(a2
b+b2
c+c2
a
)− (a+ b+ c).
From this inequality, we see that it suffices to prove that
a2
b+b2
c+c2
a≥ a+ b+ c,
which is true since the AM-GM Inequality yields
a2
b+ b ≥ 2a,
b2
c+ c ≥ 2b,
c2
a+ a ≥ 2c.
The equality occurs if and only if a = b = c.?F?
02.3. If a, b, c are positive real numbers such that abc = 2, then
a3 + b3 + c3 ≥ a√b+ c+ b
√c+ a+ c
√a+ b.
(Balkan (Shortlist) 2002)
Solution: Because
2(a3 + b3 + c3)− a2(b+ c)− b2(c+ a)− c2(a+ b) =
=[a3 + b3 − ab(a+ b)
]+[b3 + c3 − bc(b+ c)
]+[c3 + a3 − ca(a+ b)
]= (a− b)2(a+ b) + (b− c)2(b+ c) + (c− a)2(c+ a) ≥ 0,
it suffices to prove that
a2(b+ c) + b2(c+ a) + c2(a+ b) ≥ 2(a√b+ c+ b
√c+ a+ c
√a+ b
).
On the other hand, the Cauchy Schwarz Inequality yields
a2(b+ c) + b2(c+ a) + c2(a+ b) ≥ 1
3
(a√b+ c+ b
√c+ a+ c
√a+ b
)2,
so it is enough to prove that
a√b+ c+ b
√c+ a+ c
√a+ b ≥ 6,
which is true since from the AM-GM Inequality, we have
a√b+ c+ b
√c+ a+ c
√a+ b ≥ 3
3
√abc√
(a+ b)(b+ c)(c+ a)
≥ 33
√abc√
8abc = 6.
338
The equality holds if and only if a = b = c = 3√
2.?F?
02.4. Let a, b, c be real numbers such that a2+b2+c2 = 1. Prove the inequality
a2
1 + 2bc+
b2
1 + 2ca+
c2
1 + 2ab≥ 3
5.
(Bosnia and Herzegovina 2002)
Solution: In view of the inequality 2xy ≤ x2 + y2 and the observation that1 + 2bc = a2 + (b+ c)2 > 0, etc., we see that
a2
1 + 2bc+
b2
1 + 2ca+
c2
1 + 2ab≥ a2
1 + b2 + c2+
b2
1 + c2 + a2+
c2
1 + a2 + b2
=a2
2− a2+
b2
2− b2+
c2
2− c2
= −3 + 2
(1
2− a2+
1
2− b2+
1
2− c2
).
On the other hand, according to the Cauchy Schwarz Inequality, one can findthat
1
2− a2+
1
2− b2+
1
2− c2≥ 9
6− a2 − b2 − c2=
9
5,
from which it follows
a2
1 + 2bc+
b2
1 + 2ca+
c2
1 + 2ab≥ −3 + 2 · 9
5=
3
5,
as desired. Note that the equality holds if and only if a = b = c = ± 1√3.
?F?
02.5. Show that for any positive real numbers a, b, c, we have
a3
bc+b3
ca+c3
ab≥ a+ b+ c.
(Canada 2002)
First solution: From the AM-GM Inequality, we get
a3
bc+ b+ c ≥ 3
3
√a3
bc· b · c = 3a.
Adding this to the two analogous inequalities, we get the desired result. Notethat the equality holds if and only if a = b = c.
Second solution: Multiplying both sides of the desired inequality by abc > 0,we can rewrite it as
a4 + b4 + c4 ≥ abc (a+ b+ c) .
339
Now, we apply the well-known inequality x2 + y2 + z2 ≥ xy + yz + zx with(x, y, z) = (a2, b2, c2) and (x, y, z) = (ab, bc, ca) to get
a4 + b4 + c4 ≥ a2b2 + b2c2 + c2a2, and a2b2 + b2c2 + c2a2 ≥ abc(a+ b+ c).
Combining these two inequalities, we get the desired result.?F?
02.6. Assume (P1, P2, . . . , Pn) (n ≥ 2) is an arbitrary permutation of (1, 2, . . . , n).Prove that
1
P1 + P2+
1
P2 + P3+ . . .+
1
Pn−1 + Pn>n− 1
n+ 2.
(China 2002)
Solution: Denote with A the left hand side of the desired inequality. By theCauchy Schwarz Inequality, we find that
A ≥ (n− 1)2
(P1 + P2) + (P2 + P3) + · · ·+ (Pn−1 + Pn)
=(n− 1)2
2(P1 + P2 + · · ·+ Pn)− P1 − Pn=
(n− 1)2
2(1 + 2 + · · ·+ n)− P1 − Pn
=(n− 1)2
n(n+ 1)− P1 − Pn≥ (n− 1)2
n(n+ 1)− 1− 2=
(n− 1)2
(n− 1)(n+ 2)− 1
>(n− 1)2
(n− 1)(n+ 2)=n− 1
n+ 2,
as desired.?F?
02.7. Let x, y be positive real numbers such that x+ y = 2. Prove that
x3y3(x3 + y3) ≤ 2.
(India 2002)
Solution: According to the AM-GM Inequality and the given hypothesis, wefind that
x3y3(x3 + y3) = 2x3y3(x2 − xy + y2)
= 2 · xy · xy · xy · (x2 − xy + y2)
≤ 2
(xy + xy + xy + x2 − xy + y2
4
)4
= 2
[(x+ y)2
4
]4= 2,
as desired. The equality holds if and only if x = y = 1.?F?
340
02.8. For any positive real numbers a, b, c, show that the following inequalityholds
a
b+b
c+c
a≥ c+ a
c+ b+a+ b
a+ c+b+ c
b+ a.
(India 2002)
First solution: Rewrite the inequality in the form(a
b− a+ c
b+ c+ 1
)+
(b
c− a+ b
a+ c+ 1
)+
(c
a− b+ c
b+ a+ 1
)≥ 3,
orb2 + ca
b(b+ c)+c2 + ab
c(c+ a)+
a2 + bc
a(a+ b)≥ 3.
From this, using the AM-GM Inequality, we find that this inequality is deducedfrom
(a2 + bc)(b2 + ca)(c2 + ab) ≥ abc(a+ b)(b+ c)(c+ a).
Since (a2 + bc)(b2 + ca)− ab(a+ c)(b+ c) = c(a+ b)(a− b)2, we get
(a2 + bc)(b2 + ca) ≥ ab(a+ c)(b+ c),
and similarly, (b2 + ca)(c2 + ab) ≥ bc(a + b)(c + a) and (a2 + bc)(c2 + ab) ≥ac(a+ b)(b+ c). Thus
(a2 + bc)2(b2 + ca)2(c2 + ab)2 ≥ a2b2c2(a+ b)2(b+ c)2(c+ a)2,
and so(a2 + bc)(b2 + ca)(c2 + ab) ≥ abc(a+ b)(b+ c)(c+ a).
It is easy to see that the equality holds if and only if a = b = c.
Second solution: Let us take x =a
b, y =
b
c, z =
c
a. Observe that
a+ c
b+ c=
1 + xy
1 + y= x+
1− x1 + y
.
Using similar relations, the problem reduces to proving that if xyz = 1, then
x− 1
y + 1+y − 1
z + 1+z − 1
x+ 1≥ 0.
By expanding, we find that it is equivalent to
(x2 − 1)(z + 1) + (y2 − 1)(x+ 1) + (z2 − 1)(y + 1) ≥ 0,
or(xy2 + yz2 + zx2) + (x2 + y2 + z2) ≥ (x+ y + z) + 3.
But this inequality is very easy. Indeed, using the AM-GM Inequality, we havexy2 + yz2 + zx2 ≥ 3 and so it remains to prove that x2 + y2 + z2 ≥ x+ y + z,which follows from the inequalities
x2 + y2 + z2 ≥ (x+ y + z)2
3≥ x+ y + z.
341
?F?
02.9. Let x1, x2, . . . , xn be positive real numbers. Prove that
x11 + x21
+x2
1 + x21 + x22+ · · ·+ xn
1 + x21 + x22 + · · ·+ x2n<√n.
(India 2002)
Solution: Applying the Cauchy Schwarz Inequality, we can see that the lefthand side of the original inequality is not greater than
√n
√x21
(1 + x21)2
+x22
(1 + x21 + x22)2
+ · · ·+ x2n(1 + x21 + x22 + · · ·+ x2n)2
.
Now, we havex21
(1 + x21)2≤ x21
1 + x21= 1− 1
1 + x21,
and for all 2 ≤ i ≤ n, we can assert that
x2i(1 + x21 + · · ·+ x2i )
2≤ x2i
(1 + x21 + · · ·+ x2i−1)(1 + x21 + · · ·+ x2i )
=1
1 + x21 + · · ·+ x2i−1− 1
1 + x21 + · · ·+ x2i.
Adding the preceding expressions, we obtain
n∑i=1
x2i(1 + x21 + · · ·+ x2i )
2≤ 1− 1
1 + x21 + · · ·+ x2n< 1.
From this, it follows that the left hand side of the original inequality is lessthan
√n. This is what we want to prove.
?F?
02.10. Let a, b, c, d be the positive real numbers such that
1
1 + a4+
1
1 + b4+
1
1 + c4+
1
1 + d4= 1.
Prove that abcd ≥ 3.(Latvia 2002)
First solution: We need to prove the inequality a4b4c4d4 ≥ 81. After makingthe substitution
A =1
1 + a4, B =
1
1 + b4, C =
1
1 + c4, D =
1
1 + d4,
we obtain
a4 =1−AA
, b4 =1−BB
, c4 =1− CC
, d4 =1−DD
.
342
The constraint becomes A+B+C+D = 1 and the inequality can be writtenas
1−AA· 1−B
B· 1− C
C· 1−D
D≥ 81,
orB + C +D
A· C +D +A
B· D +A+B
C· A+B + C
D≥ 81,
or
(B + C +D)(C +D +A)(D +A+B)(A+B + C) ≥ 81ABCD.
However, this is an immediate consequence of the AM-GM Inequality
(B + C +D)(C +D +A)(D +A+B)(A+B + C) ≥
≥ 3 (BCD)13 · 3 (CDA)
13 · 3 (DAB)
13 · 3 (ABC)
13 .
Note that the equality holds if and only if a = b = c = d = 4√
3.
Second solution: We can write a2 = tanA, b2 = tanB, c2 = tanC, d2 =
tanD, where A,B,C,D ∈(
0,π
2
). Then, the algebraic identity becomes the
following trigonometric identity
cos2A+ cos2B + cos2C + cos2D = 1.
Applying the AM-GM Inequality, we obtain
sin2A = 1− cos2A = cos2B + cos2C + cos2D ≥ 3 (cosB cosC cosD)23 .
Similarly, we obtain
sin2B ≥ 3 (cosC cosD cosA)23 ,
sin2C ≥ 3 (cosD cosA cosB)23 ,
sin2D ≥ 3 (cosA cosB cosC)23 .
Multiplying these four inequalities, we get the result.?F?
02.11. Prove that for any positive real numbers a, b, c, we have
a
2a+ b+
b
2b+ c+
c
2c+ a≤ 1.
(Moldova 2002)
Solution: The original inequality is equivalent to(1− 2a
2a+ b
)+
(1− 2b
2b+ c
)+
(1− 2c
2c+ a
)≥ 1,
orb
2a+ b+
c
2b+ c+
a
2c+ a≥ 1,
343
which is true since by Cauchy Schwarz Inequality, we have
b
2a+ b+
c
2b+ c+
a
2c+ a≥ (b+ c+ a)2
b (2a+ b) + c (2b+ c) + a (2c+ a)= 1.
It is easy to see that the equality holds if and only if a = b = c.?F?
02.12. Positive numbers α, β, x1, x2, . . . , xn (n ≥ 1) satisfy the conditionx1 + x2 + · · ·+ xn = 1. Prove that
x31αx1 + βx2
+x32
αx2 + βx3+ · · ·+ x3n
αxn + βx1≥ 1
n(α+ β).
(Moldova 2002)
First solution: By the AM-GM Inequality, we have
x3iαxi + βxi+1
+αxi + βxi+1
n(α+ β)2+
1
n2(α+ β)≥ 3xin(α+ β)
,
for all i = 1, 2, . . . , n. Therefore
n∑i=1
x3iαxi + βxi+1
+n∑i=1
αxi + βxi+1
n(α+ β)2+
1
n(α+ β)≥
n∑i=1
3xin(α+ β)
.
Sincen∑i=1
αxi + βxi+1
n(α+ β)2=
1
n(α+ β)and
n∑i=1
3xin(α+ β)
=3
n(α+ β), we thus
have thatn∑i=1
x3iαxi + βxi+1
+2
n(α+ β)≥ 3
n(α+ β),
which yieldsn∑i=1
x3iαxi + βxi+1
≥ 1
n(α+ β).
The equality holds if and only if xi =1
nfor all i = 1, 2, . . . , n.
Second solution: By the Holder’s Inequality, we have that
n
(n∑i=1
x3iαxi + βxi+1
)[n∑i=1
(αxi + βxi+1)
]≥
(n∑i=1
xi
)3
= 1,
and sincen∑i=1
(αxi + βxi+1) = α+ β, we conclude that
n∑i=1
x3iαxi + βxi+1
≥ 1
n(α+ β).
?F?
344
02.13. Let a, b, c be positive real numbers. Prove that(2a
b+ c
) 23
+
(2b
c+ a
) 23
+
(2c
a+ b
) 23
≥ 3.
(MOSP 2002)
Solution: By the AM-GM Inequality, we have that(2a
b+ c
) 23
=2a
3√
2a · (b+ c) · (b+ c)≥ 6a
2a+ (b+ c) + (b+ c)=
3a
a+ b+ c.
Similarly, we have(2b
c+ a
) 23
≥ 3b
a+ b+ c,
(2c
a+ b
) 23
≥ 3c
a+ b+ c.
Adding up these three inequalities, we get the result. It is easy to see that theequality holds if and only if a = b = c.
?F?
02.14. If a, b, c ∈ (0, 1), prove that
√abc+
√(1− a)(1− b)(1− c) < 1.
(Romania 2002)
First solution: Observe that√x < 3
√x for x ∈ (0, 1). Thus
√abc < 3
√abc
and√
(1− a)(1− b)(1− c) < 3√
(1− a)(1− b)(1− c). On the other hand, theAM-GM Inequality implies
3√abc ≤ a+ b+ c
3, and 3
√(1− a)(1− b)(1− c) ≤ (1− a) + (1− b) + (1− c)
3.
Therefore
√abc+
√(1− a)(1− b)(1− c) < a+ b+ c
3+
(1− a) + (1− b) + (1− c)3
= 1,
as desired.
Second solution: We have
√abc+
√(1− a)(1− b)(1− c) <
√b ·√c+√
1− b ·√
1− c < 1,
by the Cauchy Schwarz Inequality.
Third solution: Let a = sin2 x, b = sin2 y, c = sin2 z, where x, y, z ∈(
0,π
2
).
The inequality becomes
sinx · sin y · sin z + cosx · cos y · cos z < 1,
345
and it follows from the inequalities
sinx ·sin y ·sin z+cosx ·cos y ·cos z < sinx ·sin y+cosx ·cos y < cos(x−y) ≤ 1.
?F?
02.15. Given positive real numbers a, b, c and x, y, z, for which a+x = b+y =c+ z = 1. Prove that
(abc+ xyz)
(1
ay+
1
bz+
1
cx
)≥ 3.
(Russia 2002)
First solution: Our inequality is equivalent to
[abc+ (1− a)(1− b)(1− c)][
1
a(1− b)+
1
b(1− c)+
1
c(1− a)
]≥ 3.
Since a + x = b + y = c + z = 1 and x, y, z are positive numbers, we have0 < a, b, c < 1. And thus, there exist positive numbers m,n, p such that
a =1
m+ 1, b =
1
n+ 1, c =
1
p+ 1. By this substitution, we have
abc+ (1− a)(1− b)(1− c) =mnp+ 1
(m+ 1)(n+ 1)(p+ 1),
and
1
a(1− b)+
1
b(1− c)+
1
c(1− a)=
(m+ 1)(n+ 1)
n+
(n+ 1)(p+ 1)
p+
(p+ 1)(m+ 1)
m.
Therefore, the above inequality can be written as
mnp+ 1
m(n+ 1)+mnp+ 1
n(p+ 1)+mnp+ 1
p(m+ 1)≥ 3.
By the AM-GM Inequality, we have
1 +mnp
m(1 + n)+
1 +mnp
n(1 + p)+
1 +mnp
p(1 +m)+ 3 =
=1 +m+mn+mnp
m(1 + n)+
1 + n+ np+mnp
n(1 + p)+
1 + p+ pm+mnp
p(1 +m)
=(1 +m) +mn(1 + p)
m(1 + n)+
(1 + n) + np(1 +m)
n(1 + p)+
(1 + p) + pm(1 + n)
p(1 +m)
=
[1 +m
m(1 + n)+
1 + n
n(1 + p)+
1 + p
p(1 +m)
]+
[n(1 + p)
1 + n+p(1 +m)
1 + p+m(1 + n)
1 +m
]≥ 3
3√mnp
+ 3 3√mnp ≥ 6.
From this, we deduce that
mnp+ 1
m(n+ 1)+mnp+ 1
n(p+ 1)+mnp+ 1
p(m+ 1)≥ 3,
346
as desired. It is easy to see that the equality holds if and only if m = n = p = 1,
i.e. if and only if a = b = c = x = y = z =1
2.
Second solution: Proceeding similar as above, we see that it suffices to provethat the inequality
(1 + abc)
[1
a(1 + b)+
1
b(1 + c)+
1
c(1 + a)
]≥ 3
holds for any positive real numbers a, b, c. Note that in general we have abc 6= 1,
so we cannot apply the substitutions a =v
u, b =
w
v, c =
u
wwith positive
reals u, v, w. However, we can substitute a = k · vu, b = k · w
v, c = k · u
w,
where k, u, v, w are positive real numbers. Then abc = k3, and our inequalitysimplifies to(
1 + k3) [ u
k (v + kw)+
v
k (w + ku)+
w
k (u+ kv)
]≥ 3.
This can be rewritten as
1 + k3
k·(
u
v + kw+
v
w + ku+
w
u+ kv
)≥ 3.
Now, using the Cauchy Schwarz Inequality and the well-known (u+v+w)2 ≥3(uv + vw + wu), we have
u
v + kw+
v
w + ku+
w
u+ kv≥ (u+ v + v)2
(k + 1)(uv + vw + wu)≥ 3
k + 1.
So, it is sufficient to prove thatk3 + 1
k· 3
k + 1≥ 3, which is true.
Third solution: We will give another proof for the inequality
(1 + abc)
[1
a(1 + b)+
1
b(1 + c)+
1
c(1 + a)
]≥ 3.
We have∑ 1 + abc
a(1 + b)− 3 =
∑ 1− a− ab+ abc
a(1 + b)=∑ (1− ab)− a(1− bc)
a(1 + b)
=∑ 1− ab
a(1 + b)−∑ 1− bc
1 + b=∑ 1− ab
a(1 + b)−∑ 1− ab
1 + a
=∑
(1− ab)[
1
a(1 + b)− 1
1 + a
]=∑ (1− ab)2
a(1 + a)(1 + b).
The last quantity is obviously nonnegative, so we must have
(1 + abc)
[1
a(1 + b)+
1
b(1 + c)+
1
c(1 + a)
]≥ 3,
as desired.
347
Fourth solution: We present another approach to the inequality
(1 + abc)
[1
a(1 + b)+
1
b(1 + c)+
1
c(1 + a)
]≥ 3.
Multiplying each side of this inequality by(1 + a)(1 + b)(1 + c)
1 + abc> 0, we can
rewrite it as ∑ (1 + a)(1 + c)
a≥ 3(1 + a)(1 + b)(1 + c)
1 + abc,
or ∑ 1
a+∑
a+∑ a
b+ 3 ≥ 3(1 + a)(1 + b)(1 + c)
1 + abc.
Because (1 + a)(1 + b)(1 + c) =∑
a+∑
ab+ abc+ 1, it is equivalent to
∑a+
∑ 1
a+∑ a
b≥
3∑
a+ 3∑
ab
1 + abc,
or
abc∑
a+∑ 1
a+∑
a2c+∑ a
b≥ 2
∑a+ 2
∑ab.
But this follows from the inequalities
a2bc+b
c≥ 2ab, b2ca+
c
a≥ 2bc, c2ab+
a
b≥ 2ca,
and
a2c+1
c≥ 2a, b2a+
1
a≥ 2b, c2b+
1
b≥ 2c.
Fifth solution: In the same manner with all above solutions, we shall provethat
(abc+ 1)
[1
a(1 + b)+
1
b(1 + c)+
1
c(1 + a)
]≥ 3
for positive real numbers a, b, c. Squaring both sides and using the well-knowninequality (x+ y + z)2 ≥ 3(xy + yz + zx), we see that it suffices to prove
(abc+ 1)2 · 3∑ 1
ab(1 + b)(1 + c)≥ 9,
or equivalently,
(abc+ 1)2 ·
∑a+
∑ab
abc(1 + a)(1 + b)(1 + c)≥ 3.
Now, using the AM-GM Inequality, we have∑a+
∑ab
(1 + a)(1 + b)(1 + c)=
1
1 + abc∑a+
∑ab
+ 1
≥ 1
1 + abc
3 3√abc+ 3
3√a2b2c2
+ 1
=3 3√abc+ 3
3√a2b2c2(
3√abc+ 1
)3 =3 3√abc(
3√abc+ 1
)2 .348
So, it is enough to prove that
(abc+ 1)2 · 3 3√abc
abc(
3√abc+ 1
)2 ≥ 3,
which is not hard to prove.?F?
02.16. Let x, y, z be positive real numbers with sum 3. Prove that
√x+√y +√z ≥ xy + yz + zx.
(Russia 2002)
First solution: By applying the Holder’s Inequality, we have(√x+√y +√z)2 (
x2 + y2 + z2)≥ (x+ y + z)3 = 27,
and hence, it suffices to prove that(x2 + y2 + z2
)(xy + yz + zx)2 ≤ 27.
This is true since by the AM-GM Inequality, we have
(x2 + y2 + z2
)(xy + yz + zx)2 ≤
[x2 + y2 + z2 + 2(xy + yz + zx)
3
]3=
[(x+ y + z)2
3
]3= 27.
Note that the equality holds if and only if x = y = z = 1.
Second solution: Rewrite the inequality in the form
x2 + 2√x+ y2 + 2
√y + z2 + 2
√z ≥ x2 + y2 + z2 + 2(xy + yz + zx).
Since x2 + y2 + z2 + 2(xy + yz + zx) = (x+ y + z)2 = 9, it is equivalent to
x2 + 2√x+ y2 + 2
√y + z2 + 2
√z ≥ 9.
Now, from the AM-GM Inequality, we have
x2 + 2√x = x2 +
√x+√x ≥ 3
3
√x2 ·√x ·√x = 3x.
Therefore
x2 + 2√x+ y2 + 2
√y + z2 + 2
√z ≥ 3(x+ y + z) = 9,
as desired.?F?
349
02.17. Let a1, a2, . . . , an and b1, b2, . . . , bn be real numbers between 1001 and2002 inclusive. Suppose a21 + · · ·+ a2n = b21 + · · ·+ b2n. Prove that
n∑i=1
a3ibi≤ 17
10
n∑i=1
a2i .
Determine when equality holds.(Singapore 2002)
Solution: For each i, we have1
2=
1001
2002≤ ai
bi≤ 2002
1001= 2, and therefore
(2ai − bi)(2bi − ai) ≥ 0, that is
5aibi ≥ 2a2i + 2b2i . (1)
Multiplying this inequality byaibi> 0, we get
5a2i ≥ 2a3ibi
+ 2aibi. (2)
From (1), we have 2aibi ≥4
5(a2i + b2i ). Using this inequality in (2), we obtain
5a2i ≥ 2a3ibi
+4
5(a2i + b2i ),
which may be rewritten as
a3ibi≤ 21
10a2i −
2
5b2i . (3)
Note that the equality in (3) holds if and only if bi = 2ai or ai = 2bi, that is,if and ony if (ai, bi) = (1001, 2002) or (ai, bi) = (2002, 1001). Now, summing
over i in (3) and recallingn∑i=1
b2i =n∑i=1
a2i , we get
n∑i=1
a3ibi≤ 21
10
n∑i=1
a2i −2
5
n∑i=1
b2i =17
10
n∑i=1
a2i ,
as desired. The equality holds if and only if, for each i, either (ai, bi) =
(1001, 2002) or (ai, bi) = (2002, 1001). The condition
n∑i=1
b2i =
n∑i=1
a2i can be
rewritten as 10012p+ (n− p)20022 = 20022p+ (n− p)10012, which is p =n
2.
Thus, the equality holds if and only if n is even and (ai, bi) = (1001, 2002) forhalf of the subscripts i while (ai, bi) = (2002, 1001) for the other half.
?F?
02.18. Let a, b, c, d be real numbers contained in the interval
(0,
1
2
). Prove
that
a4 + b4 + c4 + d4
abcd≥ (1− a)4 + (1− b)4 + (1− c)4 + (1− d)4
(1− a)(1− b)(1− c)(1− d).
350
(Taiwan 2002)
First solution: Without loss of generality, we may assume that a ≥ b ≥ c ≥d. Then, by noting at the identity x4 + y4 + z4 + t4 = (x2− y2)2 + (z2− t2)2 +2(x2y2 + z2t2), we can see that the original inequality follows from adding thefollowing three inequalities
(a2 − b2)2
abcd≥ (a− b)2(2− a− b)2
(1− a)(1− b)(1− c)(1− d),
(c2 − d2)2
abcd≥ (c− d)2(2− c− d)2
(1− a)(1− b)(1− c)(1− d),
and2(a2b2 + c2d2)
abcd≥ 2[(1− a)2(1− b)2 + (1− c)2(1− d)2]
(1− a)(1− b)(1− c)(1− d).
Hence, in order to prove it, it suffices to show that these three inequalitiesholds. Indeed, the first one is equivalent to
(1− c)(1− d)
cd
(1
a− 1
)(1
b− 1
)≥(
2
a+ b− 1
)2
,
and since(1− c)(1− d)
cd≥ 1, it suffices to prove that
(1
a− 1
)(1
b− 1
)≥(
2
a+ b− 1
)2
.
This is true, because by the AM-GM Inequality, we have(1
a− 1
)(1
b− 1
)=
1− a− bab
+ 1 ≥ 4(1− a− b)(a+ b)2
+ 1 =
(2
a+ b− 1
)2
.
Since the proof for
(c2 − d2)2
abcd≥ (c− d)2(2− c− d)2
(1− a)(1− b)(1− c)(1− d)
is similar to the previous one, we are left to show that
a2b2 + c2d2
abcd≥ (1− a)2(1− b)2 + (1− c)2(1− d)2
(1− a)(1− b)(1− c)(1− d),
or equivalently,
f
(ab
cd
)≥ f
((1− c)(1− d)
(1− a)(1− b)
),
where f is the real-valued function such that f(x) = x+1
x. Now, since f(x)
is monotonically increasing for all x ≥ 1, and moreover since
ab
cd≥ 1,
(1− c)(1− d)
(1− a)(1− b)≥ 1,
351
we only need to show that
ab
cd≥ (1− c)(1− d)
(1− a)(1− b), or
a(1− a)
c(1− c)· b(1− b)d(1− d)
≥ 1.
But this is obviously true, because
a(1− a)
c(1− c)− 1 =
(a− c)(1− a− c)c(1− c)
≥ 0,
andb(1− b)d(1− d)
− 1 =(b− d)(1− b− d)
d(1− d)≥ 0.
Note that the equality holds iff a = b = c = d.
Second solution: First, we claim that for any x, y, z, t ∈(
0,1
2
), the in-
equality holds
(1− x)4 + (1− y)4
(1− x)(1− y)+
(x2 − y2)2[zt(1− x)(1− y)− xy(1− z)(1− t)]x2y2zt
≤
≤ (1− x)(1− y)(x4 + y4)
x2y2.
Indeed, since z, t ∈(
0,1
2
), we have
zt(1− x)(1− y)− xy(1− z)(1− t)zt
= (1− x)(1− y)− xy · (1− z)(1− t)zt
< (1− x)(1− y)− xy = 1− x− y,
and thus, it suffices to prove that
(1− x)(1− y)(x4 + y4)
x2y2− (1− x)4 + (1− y)4
(1− x)(1− y)≥ (x2 − y2)2(1− x− y)
x2y2,
which is equivalent to (note that mn(u2+v2)−uv(m2+n2) = (um−vn)(un−vm))
(x− y)2(1− x− y)(x+ y − x2 − y2)(x+ y − 2xy)
x2y2(1− x)(1− y)≥ (x2 − y2)2(1− x− y)
x2y2.
This can be simplified into
(x+ y − x2 − y2)(x+ y − 2xy) ≥ (x+ y)2(1− x)(1− y).
Applying the Cauchy Schwarz Inequality, we have
(x+ y − x2 − y2)(x+ y − 2xy) = [x(1− x) + y(1− y)][x(1− y) + y(1− x)]
≥[x√
(1− x)(1− y) + y√
(1− y)(1− x)]2
= (x+ y)2(1− x)(1− y).
352
This proves our claim. Now, turning back to the main problem. Becausethe inequality is symmetric, we may assume without loss of generality thatd ≥ c ≥ b ≥ a. Then, applying the above claim by replacing (x, y, z, t) with(a, b, c, d) and (c, d, a, b), we get
(1− a)4 + (1− b)4
(1− a)(1− b)(1− c)(1− d)≤ (1− a)(1− b)(a4 + b4)
a2b2(1− c)(1− d)− M(a2 − b2)2
a2b2cd(1− c)(1− d),
(1− c)4 + (1− d)4
(1− a)(1− b)(1− c)(1− d)≤ (1− c)(1− d)(c4 + d4)
c2d2(1− a)(1− b)+
M(c2 − d2)2
abc2d2(1− a)(1− b),
where M = cd(1−a)(1−b)−ab(1−c)(1−d) ≥ 0. From these two inequalities,we can see that the original inequality is deduced from
cd(1− a)(1− b)(a4 + b4)
ab(1− c)(1− d)+ab(1− c)(1− d)(c4 + d4)
cd(1− a)(1− b)+
+M(c2 − d2)2
cd(1− a)(1− b)− M(a2 − b2)2
ab(1− c)(1− d)≤ a4 + b4 + c4 + d4.
Since
a4 + b4 − cd(1− a)(1− b)(a4 + b4)
ab(1− c)(1− d)= − M(a4 + b4)
ab(1− c)(1− d),
and
c4 + d4 − ab(1− c)(1− d)(c4 + d4)
cd(1− a)(1− b)=
M(c4 + d4)
cd(1− a)(1− b),
one can see that the last inequality is equivalent to
M(a4 + b4)
ab(1− c)(1− d)− M(a2 − b2)2
ab(1− c)(1− d)≤ M(c4 + d4)
cd(1− a)(1− b)− M(c2 − d2)2
cd(1− a)(1− b),
that is2abM
(1− c)(1− d)≤ 2cdM
(1− a)(1− b).
Of couse, this inequality is true because of M ≥ 0 and the fact x(1 − x) ≥y(1− y) for any
1
2> x ≥ y > 0. The proof is now completed.
?F?
02.19. Let x, y, z be positive real numbers such that x2 + y2 + z2 = 1. Provethat
x2yz + y2zx+ z2xy ≤ 1
3.
(United Kingdom 2002)
First solution: From the Cauchy Schwarz Inequality and the hypothesis, wehave
x2yz + y2zx+ z2xy = xyz(x+ y + z) ≤ xyz√
3(x2 + y2 + z2) =√
3xyz.
353
On the other hand,1
3=x2 + y2 + z2
3≥ 3√x2y2z2, using the AM-GM Inequal-
ity, hence xyz ≤ 1
3√
3. The desired inequality follows immediately by combin-
ing the two results. Note that the equality holds if and only if x = y = z =1√3.
Second solution: From the identity (a+b+c)2 = (a2+b2+c2)+2(ab+bc+ca),and the well-known inequality a2 + b2 + c2 ≥ ab+ bc+ ca, it follows that
(a+ b+ c)2 ≥ 3(ab+ bc+ ca).
Setting a = xy, b = yz, c = zx in this inequality yields
3(x2yz + y2zx+ z2xy)2 ≤ (xy + yz + zx)2 ≤ (x2 + y2 + z2) = 1,
from which we get
x2yz + y2zx+ z2xy ≤ 1
3,
as claimed.?F?
02.20. Let a, b, c be real numbers satisfying a2 + b2 + c2 = 9. Prove theinequality
2(a+ b+ c)− abc ≤ 10.
(Vietnam 2002)
First solution: If there is a negative number among a, b, c (for exampleb < 0). Then, from the AM-GM Inequality, we have
2(a+ b+ c)− abc ≤ 2(a+ b+ c)− b · a2 + c2
2
≤ a2 + 4
2+c2 + 4
2+ 2b− b · a
2 + c2
2
= 4 +9− b2
2+ 2b− b · 9− b2
2
= 10 +(b− 3)(b+ 1)2
2≤ 10.
Let us consider now the case a, b, c ≥ 0. Without loss of generality, we mayassume that a ≥ b ≥ c, then 3 ≥ a ≥
√3. There are two cases to consider
If ab ≥ 1, we have
2(a+ b+ c)− abc ≤ 2(a+ b+ c)− c = 2a+ 2b+ c
≤√
(22 + 22 + 12)(a2 + b2 + c2) = 9 < 10.
If ab < 1, then it follows that a2b2 + a2c2 ≤ 2a2b2 < 2, or b2 + c2 < 2a2. This
yields
2(a+ b+ c)− abc ≤ 2(a+ b+ c) ≤ 2a+ b2 + c2 + 2
< 2a+2
a2+ 2 < 2 · 3 +
2(√3)2 + 2 =
26
3< 10.
354
Therefore, in any cases, we always have
2(a+ b+ c)− abc ≤ 10.
The equality holds if and only if (a, b, c) is a permutation of (2, 2,−1).
Second solution: Without loss of generality, we may assume that a2 ≤ b2 ≤c2. And since a2 + b2 + c2 = 9, this assumption gives us c2 ≥ 3 and 2ab ≤ 6.From this, applying the Cauchy Schwarz Inequality, we have
[2(a+ b+ c)− abc]2 = [2(a+ b) + (2− ab)c]2 ≤ [4 + (2− ab)2][(a+ b)2 + c2]
= (8− 4ab+ a2b2)(9 + 2ab) = 2a3b3 + a2b2 − 20ab+ 72
= (ab+ 2)2(2ab− 7) + 100 ≤ 100.
According to this inequality, it follows that
2(a+ b+ c)− abc ≤ |2(a+ b+ c)− abc| ≤ 10,
as desired.
Third solution: Because max{a, b, c} ≤ 3 and |abc| ≤ 10, it is enough toconsider only the cases when a, b, c ≥ 0 or exactly one of the three numbers isnegative. First, we will suppose that a, b, c are nonnegative. If abc ≥ 1, thenwe are done because
2(a+ b+ c)− abc ≤ 2√
3(a2 + b2 + c2)− 1 < 10.
Otherwise, we may assume that a < 1. In this case, we have
2(a+ b+ c)− abc ≤ 2[a+
√2(b2 + c2)
]= 2a+ 2
√18− 2a2 ≤ 10.
Now, assume that not all three numbers are nonnegative and let c < 0. Thus,the problem reduces to proving that for any nonnegative x, y, z whose sum ofsquares is 9, we have 4(x+y−z)+2xyz ≤ 20. But we can write this inequalityas (x − 2)2 + (y − 2)2 + (z − 1)2 ≥ 2xyz − 6z − 2. Because 2xyz − 6z − 2 ≤z(x2 + y2) − 6z − 2 = −z3 + 3z − 2 = −(z − 1)2(z + 2) ≤ 0, the inequalityfollows.
Fourth solution: Of course, we have |a|, |b|, |c| ≤ 3 and |a+b+c|, |abc| ≤ 3√
3.Also, we may assume of course that a, b, c are non-zero and that a ≤ b ≤ c.If c < 0, then we have 2(a + b + c) − abc < −abc ≤ 3
√3 < 10. Also, if
a ≤ b < 0 < c, then we have 2(a + b + c) < 2c ≤ 6 < 10 + abc becauseabc > 0. If a < 0 < b ≤ c, using the Cauchy Schwarz Inequality, we find that2b+ 2c− a ≤ 9. Thus 2(a+ b+ c) = (2b+ 2c− a) + 3a ≤ 9 + 3a and it remainsto prove that 3a− 1 ≤ abc. But a < 0 and 2bc ≤ 9− a2, so that it remains to
show that9a− a3
2≥ 3a− 1 or (a+ 1)2(a− 2) ≤ 0, which follows. So, we just
have to threat the case 0 < a ≤ b ≤ c. In this case we have 2b + 2c + a ≤ 9and hence 2(a+ b+ c) ≤ 9 + a. So, we need to prove that a ≤ 1 + abc. This isclear if a < 1 and if a ≥ 1, we have b, c ≥ 1 and the inequality is again. Thus,the problem is solved.
355
?F?
03.1. If a, b, c > −1, then
1 + a2
1 + b+ c2+
1 + b2
1 + c+ a2+
1 + c2
1 + a+ b2≥ 2.
(Laurentiu Panaitopol, Balkan 2003)
Solution: We have 1 + b + c2 ≥ 1 + b > 0 and 1 + b + c2 ≤ 1 + b2
2+ 1 + c2,
hence1 + a2
1 + b+ c2≥ 2(1 + a2)
1 + b2 + 2(1 + c2).
Setting x = 1 + a2, y = 1 + b2, z = 1 + c2, it suffices to show that
x
y + 2z+
y
z + 2x+
z
x+ 2y≥ 1.
Using the Cauchy Schwarz Inequality, we have
x
y + 2z+
y
z + 2x+
z
x+ 2y≥ (x+ y + z)2
x(y + 2z) + y(z + 2x) + z(x+ 2y)
=(x+ y + z)2
3(xy + yz + zx)≥ 1.
Note that the equality holds if and only if a = b = c = 1.?F?
03.2. Prove that if a, b and c are positive real numbers with sum 3, then
a
b2 + 1+
b
c2 + 1+
c
a2 + 1≥ 3
2.
(Bulgaria 2003)
Solution: By the AM-GM Inequality, we have
a
b2 + 1= a− ab2
b2 + 1≥ a− ab2
2b= a− ab
2.
Adding this to the two analogous inequalities, we get
a
b2 + 1+
b
c2 + 1+
c
a2 + 1≥ a+b+c−ab+ bc+ ca
2≥ a+b+c− (a+ b+ c)2
6=
3
2,
as desired. Note that the equality holds if and only if a = b = c = 1.?F?
03.3. Let a, b, c, d be positive real numbers such that ab + cd = 1 and letx1, x2, x3, x4, y1, y2, y3, y4 be real numbers such that x21 + y21 = x22 + y22 =x23 + y23 = x24 + y24 = 1. Prove that the following inequality holds
(ay1 + by2 + cy3 + dy4)2 + (ax4 + bx3 + cx2 + dx1)
2 ≤ 2
(a2 + b2
ab+c2 + d2
cd
).
356
(China 2003)
First solution: Fix a number k > 0. Using the Cauchy Schwarz Inequality,we have
(ay1 + by2 + cy3 + dy4)2 ≤
(a2k + b2 + c2 + d2k
)(y21k
+ y22 + y23 +y24k
),
(ax4 + bx3 + cx2 + dx1)2 ≤
(a2k + b2 + c2 + d2k
)(x24k
+ x23 + x22 +x21k
),
and then, we deduce that
(ay1 + by2 + cy3 + dy4)2 + (ax4 + bx3 + cx2 + dx1)
2 ≤
≤ (a2k + b2 + c2 + d2k)
(x21 + y21
k+ x22 + y22 + x23 + y23 +
x24 + y24k
)= 2
(1
k+ 1
)[k(a2 + d2) + (b2 + c2)
]= 2(a2 + b2 + c2 + d2) + 2k(a2 + d2) +
2(b2 + c2)
k.
Now, we choose k > 0 such that
k(a2 + d2) +b2 + c2
k=cd(a2 + b2)
ab+ab(c2 + d2)
cd,
or
f(k) = (a2 + d2)k2 −[cd(a2 + b2)
ab+ab(c2 + d2)
cd
]k + b2 + c2 = 0.
We have
∆f =
[cd(a2 + b2)
ab+ab(c2 + d2)
cd
]2− 4(a2 + d2)(b2 + c2)
=
[a2d2(b2 + c2) + b2c2(a2 + d2)
]2a2b2c2d2
− 4(a2 + d2)(b2 + c2)
=
[a2d2(b2 + c2)− b2c2(a2 + d2)
]2a2b2c2d2
≥ 0.
Therefore, the number k > 0 satisfies the above equation is
k =
cd(a2 + b2)
ab+ab(c2 + d2)
cd+
∣∣a2d2(b2 + c2)− b2c2(a2 + d2)∣∣
abcd2(a2 + d2)
.
From this value of k and the above arguments, we have that
(ay1 + by2 + cy3 + dy4)2 + (ax4 + bx3 + cx2 + dx1)
2 ≤
≤ 2(a2 + b2 + c2 + d2) +2cd(a2 + b2)
ab+
2ab(c2 + d2)
cd
= 2(ab+ cd)
(a2 + b2
ab+c2 + d2
cd
)= 2
(a2 + b2
ab+c2 + d2
cd
),
357
as desired.
Second solution: Set u = ay1 + by2, v = cy3 + dy4, u1 = ax4 + bx3 andv1 = cx2 + dx1. Then
u2 ≤ (ay1 + by2)2 + (ax1 − bx2)2 = a2 + b2 + 2ab(y1y2 − x1x2),
that is
x1x2 − y1y2 ≤a2 + b2 − u2
2ab.
Similarly, we have
v21 ≤ (cx2 + dx1)2 + (dy1 − cy2)2 = c2 + d2 − 2cd(y1y2 − x1x2),
and then
y1y2 − x1x2 ≤c2 + d2 − v21
2cd.
From this, we deduce that
a2 + b2 − u2
2ab+c2 + d2 − v21
2cd≥ (x1x2 − y1y2) + (y1y2 − x1x2) = 0,
that isu2
ab+v21cd≤ a2 + b2
ab+c2 + d2
cd.
Similarly, we also have
v2
cd+u21ab≤ a2 + b2
ab+c2 + d2
cd.
Now, applying the Cauchy Schwarz Inequality, we get
(u+ v)2 + (u1 + v1)2 ≤ (ab+ cd)
(u2
ab+v2
cd
)+ (ab+ cd)
(u21ab
+v21cd
)=u2
ab+v21cd
+v2
cd+u21ab≤ 2
(a2 + b2
ab+c2 + d2
cd
),
as desired.
Third solution: By the Cauchy Schwarz Inequality, we get
(ay1 + by2 + cy3 + dy4)2 ≤ (ab+ cd)
[(ay1 + by2)
2
ab+
(cy3 + dy4)2
cd
]=
(ay1 + by2)2
ab+
(cy3 + dy4)2
cd
=a
by21 +
b
ay22 +
c
dy23 +
d
cy24 + 2y1y2 + 2y3y4,
and
(ax4 + bx3 + cx2 + dx1)2 ≤ (ab+ cd)
[(ax4 + bx3)
2
ab+
(cx2 + dx1)2
cd
]=
(ax4 + bx3)2
ab+
(cx2 + dx1)2
cd
=a
bx24 +
b
ax23 +
c
dx22 +
d
cx21 + 2x1x2 + 2x3x4.
358
Denote
P = (ay1+by2+cy3+dy4)2+(ax4+bx3+cx2+dx1)
2−2
(a2 + b2
ab+c2 + d2
cd
).
Then from the above inequalities, we find that
P ≤ a
by21 +
b
ay22 +
c
dy23 +
d
cy24 + 2y1y2 + 2y3y4 +
a
bx24 +
b
ax23 +
c
dx22+
+d
cx21 + 2x1x2 + 2x3x4 − 2
(a
b+b
a+c
d+d
c
)= −
(a
bx21 +
b
ax22
)−(c
dx23 +
d
cx24
)−(a
by24 +
b
ay23
)−(c
dy22 +
d
cy21
)+
+ 2x1x2 + 2x3x4 + 2y1y2 + 2y3y4
≤ −2x1x2 − 2x3x4 − 2y4y3 − 2y2y1 + 2x1x2 + 2x3x4 + 2y1y2 + 2y3y4 = 0,
as desired.?F?
03.4. Let a1, a2, . . . , a2n be real numbers such that
2n−1∑i=1
(ai − ai+1)2 = 1.
Determine the maximum value of
(an+1 + an+2 + · · ·+ a2n)− (a1 + a2 + . . .+ an).
(China 2003)
Solution: For n = 1, we have (a1 − a2)2 = 1, implying that a2 − a1 = ±1.Therefore, in this case, max {a2 − a1} = 1. Suppose that n ≥ 2, let x1 =a1, xi+1 = ai+1 − ai for i = 1, 2, . . . , 2n − 1. By this substitution, we haveai = x1 + · · ·+ xi for i = 1, 2, . . . , 2n and x22 + x23 + · · ·+ x22n = 1. Now, usingthe Cauchy Schwarz Inequality, we have
(an+1 + an+2 + · · ·+ a2n)− (a1 + a2 + · · ·+ an) =
= n(x1 + x2 + · · ·+ xn + xn+1) + (n− 1)xn+2 + · · ·+ x2n −n∑i=1
(n+ 1− i)xi
= x2 + 2x3 + · · ·+ (n− 1)xn + nxn+1 + (n− 1)xn+2 + · · ·+ x2n
≤√
[12 + 22 + · · ·+ (n− 1)2 + n2 + (n− 1)2 + · · ·+ 12](x22 + x23 + · · ·+ x22n
)=√
12 + 22 + · · ·+ (n− 1)2 + n2 + (n− 1)2 + · · ·+ 12 =
√n(2n2 + 1)
3.
The equality holds when
ai =
√3i(i− 1)
2√n(2n2 + 1)
, an+i =
√3[2n2 − (n− i)(n− i+ 1)
]2√n(2n2 + 1)
359
for all i = 1, 2, . . . , n. So, the maximum value of the desired expression is√n(2n2 + 1)
3.
?F?
03.5. Suppose x be a real number in the interval
[3
2, 5
]. Prove that
2√x+ 1 +
√2x− 3 +
√15− 3x < 2
√19.
(China 2003)
Solution: From the Cauchy Schwarz Inequality and the given hypothesis, wehave
2√x+ 1 +
√2x− 3 +
√15− 3x =
=√x+ 1 +
√x+ 1 +
√2x− 3 +
√15− 3x
≤√
[(x+ 1) + (x+ 1) + (2x− 3) + (15− 3x)] (12 + 12 + 12 + 12)
= 2√x+ 14 ≤ 2
√19.
We have equality if and only if√x+ 1 =
√2x− 3 =
√15− 3x and x = 5, but
this is impossible. Therefore, we conclude that
2√x+ 1 +
√2x− 3 +
√15− 3x < 2
√19,
as desired.?F?
03.6. Let x1, x2, . . . , x5 be nonnegative real numbers such that
5∑i=1
1
1 + xi= 1.
Prove that5∑i=1
xix2i + 4
≤ 1.
(China 2003)
First solution: The original inequality is equivalent to
5∑i=1
(1− 5xi
x2i + 4
)≥ 0,
or5∑i=1
(xi − 1)(xi − 4)
x2i + 4≥ 0.
This last inequality can be written as
5∑i=1
x2i − 1
x2i + 4· xi − 4
xi + 1≥ 0.
360
Now, due to symmetry, we may assume without loss of generality that x1 ≥x2 ≥ x3 ≥ x4 ≥ x5. By this assumption, it is easy to check that
x21 − 1
x22 + 4≥ x22 − 1
x22 + 4≥ x23 − 1
x23 + 4≥ x24 − 1
x24 + 4≥ x25 − 1
x25 + 4,
andx1 − 4
x1 + 1≥ x2 − 4
x2 + 1≥ x3 − 4
x3 + 1≥ x4 − 4
x4 + 1≥ x5 − 4
x5 + 1.
Therefore, by the Chebyshev’s Inequality, we have
5∑i=1
x2i − 1
x2i + 4· xi − 4
xi + 1≥ 1
5
(5∑i=1
x2i − 1
x2i + 4
)(5∑i=1
xi − 4
xi + 1
)
=
(5∑i=1
x2i − 1
x2i + 4
)(1−
5∑i=1
1
xi + 1
)= 0.
The equality holds if and only if x1 = x2 = x3 = x4 = x5 = 4.
Second solution: For each i (1 ≤ i ≤ 5), let yi =1
xi + 1, then xi =
1− yiyi
>
0 and y1 + y2 + y3 + y4 + y5 = 1. Replacing xi by1− yiyi
into the desired
inequality, we see that it is equivalent to
5∑i=1
yi − y2i5y2i − 2yi + 1
≤ 1, or5∑i=1
5yi − 5y2i5y2i − 2yi + 1
≤ 5,
which is true because
5∑i=1
5yi − 5y2i5y2i − 2yi + 1
=5∑i=1
(−1 +
3yi + 1
5y2i − 2yi + 1
)=
5∑i=1
3yi + 1
5y2i − 2yi + 1− 5
=
5∑i=1
3yi + 1
5
(yi −
1
5
)2
+4
5
− 5 ≤ 5
4
5∑i=1
(3yi + 1)− 5 = 5.
?F?
03.7. Find the greatest real number k such that, for any positive a, b, c witha2 > bc,
(a2 − bc)2 > k(b2 − ca)(c2 − ab).
(Japan 2003)
Solution: We will prove that the greatest k is 4. First suppose that (a2−bc)2 >k(b2−ca)(c2−ab) whenever a, b, c > 0 and a2 > bc. Let t ∈ (0, 1), Since 12 > t·t,we have
(1− t2)2 > k(t2 − t)(t2 − t),
from which we deduce that (1 + t
t
)2
> k.
361
It follows that
k ≤ limt→1
(1 + t
t
)2
= 4.
Now, we will show that (a2 − bc)2 > 4(b2 − ca)(c2 − ab) whenever a, b, c > 0and a2 > bc. Assume on the contrary that
(a2 − bc)2 ≤ 4(b2 − ca)(c2 − ab) (1)
for some positive a, b, c such that a2 > bc, and define
f(x) = (b2 − ca)x2 + (a2 − bc)x+ (c2 − ab).
From (1), either f(x) ≥ 0 for all real x or f(x) ≤ 0 for all real x. Actually, theformer holds since f(x) = a2 + b2 + c2 − ab− bc− ca > 0 (note that a = b = cis excluded by a2 > bc, and so a2 + b2 + c2 > ab + bc + ca). It follows thatb2 − ca is positive. Now, we write
f(x) = (bx− c)2 − ag(x)− x(a2 − bc),
where g(x) = cx2 − 2ax+ b. Since a2 − bc > 0 and
g(cb
)=c(c2 − ab) + b(b2 − ac)
b2> 0
(since c2 − ab has the same sign as b2 − ac by (1)), we have f(cb
)< 0, a
contradiction. This completes the proof.?F?
03.8. Prove that in any acute triangle ABC,
cot3A+ cot3B + cot3C + 6 cotA cotB cotC ≥ cotA+ cotB + cotC.
(MOSP 2003)
Solution: Let x = cotA, y = cotB, z = cotC. Because xy + yz + zx = 1, itsuffices to prove the homogeneous inequality
x3 + y3 + z3 + 6xyz ≥ (x+ y + z)(xy + yz + zx).
But it is equivalent to
x(x− y)(x− z) + y(y − z)(y − x) + z(z − x)(z − y) ≥ 0,
which is the Schur’s Inequality (in the special case third degree). Note that
the equality holds if and only if A = B = C =π
3.
?F?
03.9. Let ai be positive real numbers, for all i = 1, 2, . . . , n, satisfying
a1 + a2 + · · ·+ an =1
a1+
1
a2+ · · ·+ 1
an.
362
Prove that
1
n− 1 + a1+
1
n− 1 + a2+ · · ·+ 1
n− 1 + an≤ 1.
(Vasile Cirtoaje, MOSP 2003)
Solution: Notice that the inequality is equivalent to
n∑i=1
(1− n− 1
n− 1 + ai
)≥ 1, or
n∑i=1
ain− 1 + ai
≥ 1.
Set bi =1
ai, then the condition from the hypothesis remains invariant, but in
terms of bi’s rewrites as
b1 + b2 + · · ·+ bn =1
b1+
1
b2+ · · ·+ 1
bn.
On the other hand, the inequality to prove becomes now
n∑i=1
1
(n− 1)bi + 1≥ 1.
We proceed now by making use of the contradiction method. In this case,assume that there exist some numbers b1, b2, . . . , bn satisfying the above con-dition, but such that
n∑i=1
1
(n− 1)bi + 1< 1.
Then there also exists some k < 1 such that
n∑i=1
1
(n− 1)kbi + 1= 1.
Furthermore, setting xi =1
(n− 1)kbi + 1< 1, i.e. bi =
1− xi(n− 1)kxi
, we have
that x1 + x2 + · · ·+ xn = 1, and the given condition becomes
n∑i=1
1− xi(n− 1)kxi
=
n∑i=1
(n− 1)kxi1− xi
,
and since k < 1, we also have that
n∑i=1
1− xi(n− 1)xi
<n∑i=1
(n− 1)xi1− xi
.
On the other hand, from the Cauchy Schwarz Inequality, we get
n∑i=1
∑j 6=i
xj
xi=
n∑i=1
xi∑j 6=i
1
xj
≥ n∑i=1
xi · (n− 1)2∑j 6=i
xj
= (n− 1)2n∑i=1
xi∑j 6=i
xj,
363
and therefore
n∑i=1
∑j 6=i
xj
xi≥ (n−1)2
n∑i=1
xi∑j 6=i
xj, or equivalently,
n∑i=1
1− xi(n− 1)xi
≥n∑i=1
(n− 1)xi1− xi
.
This obviously comes in contradiction with
n∑i=1
1− xi(n− 1)xi
<
n∑i=1
(n− 1)xi1− xi
,
completing then our proof. Note that the equality holds if and only if a1 =a2 = · · · = an = 1.
?F?
03.10. Let a, b, c be nonnegative real numbers satisfying a2 + b2 + c2 = 1.Prove that
1 ≤ a
1 + bc+
b
1 + ca+
c
1 + ab≤√
2.
(Faruk Zejnulahi, MOSP 2003)
First solution: For all real numbers x contained in the interval
[0,
1
2
], we
have 1− 2
3x− 1
1 + x=x(1− 2x)
3(1 + x)≥ 0, and therefore
1
1 + x≤ 1− 2
3x.
Now, notice that max{ab, bc, ca} ≤ 1
2. In this case, we get
a
1 + bc+
b
1 + ca+
c
1 + ab≤ a
(1− 2
3bc
)+ b
(1− 2
3ca
)+ c
(1− 2
3ab
)= a+ b+ c− 2abc = a(1− 2bc) + b+ c
≤√
[a2 + (b+ c)2][(1− 2bc)2 + 1]
=√
2(2b2c2 − 2bc+ 1)(1 + 2bc)
=√
2[1− 2b2c2(1− 2bc)] ≤√
2.
For the right hand side of the inequality, note that1
1 + x−1 +x =
x2
1 + x≥ 0,
and therefore1
1 + x≥ 1− x,
for all positive real numbers x. Hence
a
1 + bc+
b
1 + ca+
c
1 + ab≥ a(1− bc) + b(1− ca) + c(1− ab)
= a+ b+ c− 3abc
=√a2 + b2 + c2 + 2(ab+ bc+ ca)− 3abc
=√
1 + 2(ab+ bc+ ca)− 3abc,
364
and since
√1 + 2u− 1 =
2u
1 +√
2u+ 1≥ 2u
1 +√
3=(√
3− 1)u
for any 0 ≤ u ≤ 1, we obtain√1 + 2(ab+ bc+ ca)− 3abc ≥ 1 +
(√3− 1
)(ab+ bc+ ca)− 3abc
≥ 1 +(√
3− 1)
(ab+ bc+ ca)3/2 − 3abc
≥ 1 +(√
3− 1)(
33√a2b2c2
)3/2− 3abc
= 1 + 3(
2−√
3)abc ≥ 1.
This yieldsa
1 + bc+
b
1 + ca+
1
1 + ab≥ 1,
and the proof is completed. Note that the equality in the left inequality occurs
iff (a, b, c) is a permutation of
(1√2,
1√2, 0
), while the equality in the right
inequality occurs iff (a, b, c) is a permutation of (1, 0, 0).
Second solution: First, we will show that
(a+ b+ c)2 ≤ 2(1 + bc)2,
or equivalently,2a(b+ c) ≤ 1 + 2bc+ 2b2c2.
This can be rewritten as
2a(b+ c) ≤ a2 + b2 + c2 + 2bc+ 2b2c2,
which is obviously true since a2 + b2 + c2 + 2bc− 2a(b+ c) = (b+ c− a)2 ≥ 0.Thus, we now have that
a
1 + bc+
b
1 + ca+
c
1 + ab≤
√2a
a+ b+ c+
√2b
a+ b+ c+
√2c
a+ b+ c=√
2,
and since the AM-GM Inequality gives us
a+ abc ≤ a+a(b2 + c2)
2= a+
a(1− a2)2
= 1− (a− 1)2(a+ 2)
2≤ 1
it follows thata
1 + bc− a2 =
a[1− (a+ abc)]
1 + bc≥ 0,
and therefore
a
1 + bc+
b
1 + ca+
c
1 + ab≥ a2 + b2 + c2 = 1.
?F?
365
03.11. Let a, b, c be positive real numbers so that abc = 1. Prove that
1 +3
a+ b+ c≥ 6
ab+ bc+ ca.
(Romania 2003)
Solution: We set x =1
a, y =
1
b, z =
1
cand observe that xyz = 1. The
inequality is equivalent to
1 +3
xy + yz + zx≥ 6
x+ y + z.
From (x+ y + z)2 ≥ 3(xy + yz + zx), we get
1 +3
xy + yz + zx≥ 1 +
9
(x+ y + z)2,
so it suffices to prove that
1 +9
(x+ y + z)2≥ 6
x+ y + z.
The last inequality is equivalent to
(1− 3
x+ y + z
)2
≥ 0 and this ends the
proof. Note that the equality holds if and only if a = b = c = 1.?F?
03.12. Let a, b, c be positive real numbers. Prove that
(2a+ b+ c)2
2a2 + (b+ c)2+
(2b+ c+ a)2
2b2 + (c+ a)2+
(2c+ a+ b)2
2c2 + (a+ b)2≤ 8.
(USA 2003)
First solution: Because the inequality is homogeneous, we can assume thata+ b+ c = 3. Then
(2a+ b+ c)2
2a2 + (b+ c)2=
a2 + 6a+ 9
3a2 − 6a+ 9=
1
3
[1 +
2(4a+ 3)
2 + (a− 1)2
]≤ 1
3
[1 +
2(4a+ 3)
2
]=
4a+ 4
3.
Thus ∑ (2a+ b+ c)2
2a2 + (b+ c)2≤ 1
3
∑(4a+ 4) = 8,
as desired. Note that the equality holds if and only if a = b = c.
Second solution: Observe that
3− (2a+ b+ c)2
2a2 + (b+ c)2=
2(b+ c− a)2
2a2 + (b+ c)2,
366
the desired inequality can be rewritten as
(b+ c− a)2
2a2 + (b+ c)2+
(c+ a− b)2
2b2 + (c+ a)2+
(a+ b− c)2
2c2 + (a+ b)2≥ 1
2.
By the Cauchy Schwarz Inequality, we find that[∑ (b+ c− a)2
2a2 + (b+ c)2
] [∑b2[2a2 + (b+ c)2
]]≥[∑
b(b+ c− a)]2
=(∑
a2)2.
Therefore, the above inequality is deduced from
2(∑
a2)2≥∑
b2[2a2 + (b+ c)2
].
By expanding, we see that it is equivalent to
a4 + b4 + c4 + a2b2 + b2c2 + c2a2 ≥ 2(a3b+ b3c+ c3a),
which is true since a4 + a2b2 ≥ 2a3b, b4 + b2c2 ≥ 2b3c, c4 + c2a2 ≥ 2c3a.
Third solution: Denote x =b+ c
a, y =
c+ a
b, z =
a+ b
c. We have to prove
that ∑ (x+ 2)2
x2 + 2≤ 8, or equivalently,
∑ (x− 1)2
x2 + 2≥ 1
2.
But, from the Cauchy Schwarz Inequality, we have∑ (x− 1)2
x2 + 2≥ (x+ y + z − 3)2
x2 + y2 + z2 + 6.
It remains to prove that
2(x2 + y2 + z2 + 2xy + 2yz + 2zx− 6x− 6y − 6z + 9) ≥ x2 + y2 + z2 + 6,
or(x+ y + z)2 + 2(xy + yz + zx)− 12(x+ y + z) + 12 ≥ 0.
Now xy + yz + zx ≥ 3 3√x2y2z2 ≥ 12 (because xyz ≥ 8), so we still have to
prove that(x+ y + z)2 + 24− 12(x+ y + z) + 12 ≥ 0,
which is equivalent to (x+ y + z − 6)2 ≥ 0, clearly true.?F?
03.13. Prove that for any a, b, c ∈(
0,π
2
), the following inequality holds
∑ sin a · sin(a− b) · sin(a− c)sin(b+ c)
≥ 0.
(USA 2003)
Solution: Let x = sin a, y = sin b, z = sin c. Then we have x, y, z > 0. It iseasy to see that the following relations are true
sin a · sin(a− b) · sin(a− c) · sin(a+ b) · sin(a+ c) = x(x2 − y2)(x2 − z2).
367
Using similar relations for the other terms, we have to prove that
x(x2 − y2)(x2 − z2) + y(y2 − z2)(y2 − x2) + z(z2 − x2)(z2 − y2) ≥ 0.
With the substitution x =√u, y =
√v, z =
√w, the inequality becomes
√u(u− v)(u− w) +
√v(v − w)(v − u) +
√w(w − u)(w − v) ≥ 0.
But this follows from the Schur’s Inequality and hence, our proof is completed.Note that the equality holds if and only if a = b = c.
?F?
04.1. Prove that
(a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab+ bc+ ca)
for any positive real numbers a, b, c.
(APMO 2004)
First solution: We denote x = a2 + 2, y = b2 + 2, z = c2 + 2, and apply theAM-GM Inequality to obtain
ab ≤ 1
2
[a2(b2 + 2)
a2 + 2+b2(a2 + 2)
b2 + 2
]=
1
2
[(1− 2
a2 + 2
)(b2 + 2) +
(1− 2
b2 + 2
)(a2 + 2)
]=
1
2
[(1− 2
x
)y +
(1− 2
y
)x
]=
1
2
(x+ y − 2x
y− 2y
x
).
Proceeding the same way as this, we can get two other similar estimations forbc and ca. Adding up these three inequalities, one can see that
ab+ bc+ ca ≤ x+ y + z − x
y− y
x− y
z− z
y− z
x− x
z.
And thus, we can reduce the original inequality into proving that
xyz ≥ 9
(x+ y + z − x
y− y
x− y
z− z
y− z
x− x
z
),
or equivalently,
9x
y+
9x
z+
9y
z+
9y
x+
9z
x+
9z
y+ xyz ≥ 9x+ 9y + 9z.
The rest is simple as we have (by the AM-GM Inequality)
9x
y+
9x
z+xyz
3≥ 3 3
√9x
y· 9x
z· xyz
3= 9x.
Note that the equality holds if and only if a = b = c = 1.
368
Second solution: Choose A,B,C ∈(0, π2
)with a =
√2 tanA, b =
√2 tanB,
and c =√
2 tanC. Using the well-known trigonometric identity 1 + tan2 θ =1
cos2 θ, we may rewrite our inequality as
4
9≥ cosA cosB cosC (cosA sinB sinC + sinA cosB sinC + sinA sinB cosC) .
We can easily check the following trigonometric identity
cos(A+B + C) = cosA cosB cosC −∑
cosA sinB sinC.
Then, the above trigonometric inequality takes the form
4
9≥ cosA cosB cosC [cosA cosB cosC − cos(A+B + C)] .
Let θ =A+B + C
3. Applying the AM-GM Inequality and Jensen’s Inequal-
ity, we have
cosA cosB cosC ≤(
cosA+ cosB + cosC
3
)3
≤ cos3 θ.
We now need to show that
4
9≥ cos3 θ(cos3 θ − cos 3θ).
Using the trigonometric identity
cos 3θ = 4 cos3 θ − 3 cos θ or cos3 θ − cos 3θ = 3 cos θ − 3 cos3 θ,
it becomes4
27≥ cos4 θ
(1− cos2 θ
),
which follows from the AM-GM Inequality[cos2 θ
2· cos2 θ
2·(1− cos2 θ
)] 13
≤ 1
3
[cos2 θ
2+
cos2 θ
2+(1− cos2 θ
)]=
1
3.
Third solution: Observe that 9(ab + bc + ca) ≤ 3(a + b + c)2, and so, itsuffices to prove that
(a2 + 2)(b2 + 2)(c2 + 2) ≥ 3(a+ b+ c)2.
On the other hand, one has
[(a2−1)(b2−1)]·[(b2−1)(c2−1)]·[(c2−1)(a2−1)] = (a2−1)2(b2−1)2(c2−1)2 ≥ 0.
Therefore, at least one of three expressions (a2 − 1)(b2 − 1), (b2 − 1)(c2 − 1),(c2 − 1)(a2 − 1) must be nonnegative. And since the inequality is symmetric,
369
we can suppose that (a2 − 1)(b2 − 1) ≥ 0. Now, applying the Cauchy SchwarzInequality, we have
(a+ b+ c)2 = (a · 1 + b · 1 + 1 · c)2 ≤ (a2 + b2 + 1)(12 + 12 + c2).
According to this inequality, one can see that the above inequality is deducedfrom
(a2 + 2)(b2 + 2) ≥ 3(a2 + b2 + 1).
However, this is obviously true since it is equivalent to (a2 − 1)(b2 − 1) ≥ 0.?F?
04.2. Let x, y, z, t be positive real numbers such that xyzt = 1. Prove that
1
(1 + x)2+
1
(1 + y)2+
1
(1 + z)2+
1
(1 + t)2≥ 1.
(China 2004)
First solution: From x, y, z, t > 0 and xyzt = 1, we can find that exist
positive numbers a, b, c, d satisfying x =b
a, y =
c
b, z =
d
c, t =
a
d. And by this
substitution, the desired inequality becomes
a2
(a+ b)2+
b2
(b+ c)2+
c2
(c+ d)2+
d2
(d+ a)2≥ 1.
Now, applying the Cauchy Schwarz Inequality, we have[∑ a2
(a+ b)2
] [∑(a+ b)2(a+ d)2
]≥[∑
a(a+ b)]2.
According to this estimation, it suffices to prove that[∑a(a+ b)
]2≥∑
(a+ b)2(a+ d)2.
Since∑
a(a + b) =1
2[(a + b)2 + (b + c)2 + (c + d)2 + (d + a)2], we find that
the last inequality is equivalent to
(m+ n+ p+ w)2 ≥ 4(mn+ np+ pw + wm),
where m = (a + b)2, n = (b + c)2, p = (c + d)2, w = (d + a)2. However, thisinequality is clearly true (according to the AM-GM Inequality), so our proofis completed. Note that the equality holds if and only if x = y = z = t = 1.
Second solution: According to the Cauchy Schwarz Inequality, we have
1
(1 + x)2+
1
(1 + y)2≥ 1
(1 + xy)
(1 +
x
y
) +1
(1 + xy)(
1 +y
x
) =1
1 + xy.
And in the same way, we also have
1
(1 + z)2+
1
(1 + t)2≥ 1
1 + zt=
xy
1 + xy.
370
Adding up these two inequalities, the result follows immediately.?F?
04.3. If a, b, c are positive real numbers, then
1 <a√
a2 + b2+
b√b2 + c2
+c√
c2 + a2≤ 3√
2
2.
(China 2004)
First solution: By applying the Cauchy Schwarz Inequality, we find that
(∑ a√a2 + b2
)2
=
[∑√a2
(a2 + b2)(a2 + c2)· (a2 + c2)
]2
≤[∑ a2
(a2 + b2)(a2 + c2)
] [∑(a2 + c2)
]=
4(a2 + b2 + c2)(a2b2 + b2c2 + c2a2)
(a2 + b2)(b2 + c2)(c2 + a2).
On the other hand, the AM-GM Inequality implies
(a2 + b2 + c2)(a2b2 + b2c2 + c2a2) =
= (a2 + b2)(b2 + c2)(c2 + a2) + a2b2c2
≤ (a2 + b2)(b2 + c2)(c2 + a2) +(a2 + b2)(b2 + c2)(c2 + a2)
8
=9
8(a2 + b2)(b2 + c2)(c2 + a2).
Using this in combination with the above inequality, we get(∑ a√a2 + b2
)2
≤ 9
2,
from which it follows that
a√a2 + b2
+b√
b2 + c2+
c√c2 + a2
≤ 3√
2
2.
This is the right hand side of the desired inequality.
We still have to prove the left hand side. However, this is actual easy and wecan prove it as follows
a√a2 + b2
+b√
b2 + c2+
c√c2 + a2
>a
a+ b+
b
b+ c+
c
c+ a
>a
a+ b+ c+
b
b+ c+ a+
c
c+ a+ b= 1.
Note that the equality in the right hand side occurs iff a = b = c.
371
Second solution: We provide another solution for the right hand side in-
equality. With the notations x =b
a, y =
c
b, z =
a
c, the inequality reduces to
proving that xyz = 1 implies√2
1 + x2+
√2
1 + y2+
√2
1 + z2≤ 3.
We presume that x ≤ y ≤ z, which implies xy ≤ 1 and z ≥ 1. We have(√2
1 + x2+
√2
1 + y2
)2
≤ 2
(2
1 + x2+
2
1 + y2
)= 4
[1 +
1− x2y2
(1 + x2)(1 + y2)
]≤ 4
[1 +
1− x2y2
(1 + xy)2
]=
8
1 + xy=
8z
z + 1,
so √2
1 + x2+
√2
1 + y2≤ 2
√2z
z + 1,
and we need to prove that
2
√2z
z + 1+
√2
1 + z2≤ 3.
Because
√2
1 + z2≤ 2
z + 1, we only need to prove that
2
√2z
z + 1+
2
z + 1≤ 3.
This inequality is equivalent to
1 + 3z − 2√
2z(1 + z) ≥ 0, or(√
2z −√
1 + z)2≥ 0,
which is obvious and we are done.
Third solution: We give another way to prove the right hand side inequality.Clearly, the inequality asks to prove that if xyz = 1, then√
2
1 + x+
√2
1 + y+
√2
1 + z≤ 3.
We have two cases. The first and easy one is when xy + yz + zx ≥ x+ y + z.In this case, we can apply the Cauchy Schwarz Inequality to get
∑√2
1 + x≤√
3∑ 2
1 + x.
But∑ 2
1 + x≤ 3, since it is equivalent to
∑2(xy + x+ y + 1) ≤ 3(2 + x+ y + z + xy + yz + zx),
372
orx+ y + z ≤ xy + yz + zx,
and so in this case, the inequality is proved. The second case is when xy +yz + zx < x+ y + z. Thus,
(x− 1)(y − 1)(z − 1) = x+ y + z − (xy + yz + zx) > 0,
and so exactly two of the numbers x, y, z are smaller than 1, let them be xand y. So, we must prove that if x and y are smaller than 1, then√
2
x+ 1+
√2
y + 1+
√2xy
xy + 1≤ 3.
Using the Cauchy Schwarz Inequality, we get√2
x+ 1+
√2
y + 1+
√2xy
xy + 1≤ 2
√1
x+ 1+
1
y + 1+
√2xy
xy + 1,
and so it is enough to prove that this last inequality is at most 3. But thiscomes down to
2 ·
1
x+ 1+
1
y + 1− 1
1 +
√1
x+ 1+
1
y + 1
≤1− 2xy
xy + 1
1 +
√2xy
xy + 1
.
Because1
x+ 1+
1
y + 1≥ 1, the left hand side is at most
1
x+ 1+
1
y + 1− 1 =
1− xy(x+ 1)(y + 1)
,
and so we are left with the inequality
1− xy(x+ 1)(y + 1)
≤ 1− xy
(1 + xy)
(1 +
√2xy
xy + 1
) ,or equivalently,
xy + 1 + (xy + 1)
√2xy
xy + 1≤ xy + 1 + x+ y,
that is,x+ y ≥
√2xy(xy + 1).
This last one follows from√
2xy(xy + 1) ≤ 2√xy ≤ x + y, and thus, our
inequality is proved.?F?
04.4. Let n be a positive integer with n greater than one, and let a1, a2, . . . , anbe positive integers such that a1 < a2 < · · · < an and
1
a1+
1
a2+ · · ·+ 1
an≤ 1.
373
Prove that, for any real number x, the following inequality holds(1
a21 + x2+
1
a22 + x2+ · · ·+ 1
a2n + x2
)2
≤ 1
2· 1
a1(a1 − 1) + x2.
(China 2004)
Solution: Applying the Cauchy Schwarz Inequality, we have(n∑i=1
1
a2i + x2
)2
≤
(n∑i=1
1
ai
)[n∑i=1
ai(a2i + x2)2
]
≤n∑i=1
ai(a2i + x2)2
≤n∑i=1
ai(a2i + x2)2 − a2i
=
n∑i=1
ai(a2i + x2 + ai)(a2i + x2 − ai)
=1
2
n∑i=1
(1
a2i − ai + x2− 1
a2i + ai + x2
).
Denoting an+1 = an + 1, we have that a1 < a2 < · · · < an < an+1 and becauseai is a positive integer, we can see that ai ≤ ai+1 − 1 for any i = 1, 2, . . . , n.By this argument, it follows that
1
a2i − ai + x2− 1
a2i + ai + x2≤ 1
a2i − ai + x2− 1
(ai+1 − 1)2 + (ai+1 − 1) + x2
=1
a2i − ai + x2− 1
a2i+1 − ai+1 + x2
for any i = 1, 2, . . . , n. Therefore, from the above estimation, we get(n∑i=1
1
a2i + x2
)2
≤ 1
2
n∑i=1
(1
a2i − ai + x2− 1
a2i+1 − ai+1 + x2
)=
1
2
(1
a21 − a1 + x2− 1
a2n+1 − an+1 + x2
)<
1
2· 1
a21 − a1 + x2=
1
2· 1
a1(a1 − 1) + x2,
which is just the desired result.?F?
04.5. Find all real numbers k such that the following inequality
a3 + b3 + c3 + d3 + 1 ≥ k(a+ b+ c+ d)
holds for all real numbers a, b, c, d ≥ −1.
(China 2004)
374
Solution: If a = b = c = d = −1, then −3 ≥ k · (−4), and hence k ≥ 3
4. If
a = b = c = d =1
2, then 4 · 1
8+ 1 ≥ k · 2. Thus k ≤ 3
4, and so k =
3
4. Now, we
want to prove that the inequality
a3 + b3 + c3 + d3 + 1 ≥ 3
4(a+ b+ c+ d)
holds for any a, b, c, d ∈ [−1,∞). At first, we prove that 4x3 + 1 ≥ 3x forx ∈ [−1,∞). In fact, from (x+ 1)(2x− 1)2 ≥ 0, we have 4x3 + 1 ≥ 3x for allx ∈ [−1,∞). Therefore
4a3 + 1 ≥ 3a, 4b3 + 1 ≥ 3b, 4c3 + 1 ≥ 3c, 4c3 + 1 ≥ 3d.
By adding theses four inequalities and dividing each side of the resulting in-equality by 4, we get
a3 + b3 + c3 + d3 + 1 ≥ 3
4(a+ b+ c+ d),
as claimed. Thus, the real number we want to find is k =3
4.
?F?
04.6. Determine the maximum constant λ such that
x+ y + z ≥ λ,
where x, y, z are positive real numbers with x√yz + y
√zx+ z
√xy ≥ 1.
(China 2004)
Solution: The maximum value of λ is√
3. It is not difficult to see that for
x = y = z =
√3
3, x√yz + y
√zx + z
√xy = 1 and x + y + z =
√3. Hence,
the maximum value of λ is less than or equal to√
3. It suffices to show thatx+ y+ z ≥
√3. Indeed, applying the AM-GM Inequality in combination with
the well-known inequalities xy + yz + zx ≤ (x+ y + z)2
3, we have
1 ≤ x√yz + y√zx+ z
√xy ≤ x · y + z
2+ y · z + x
2+ z · x+ y
2
= xy + yz + zx ≤ (x+ y + z)2
3.
From this, we deduce that
x+ y + z ≥√
3,
as desired.?F?
04.7. Let a, b, c be positive real numbers. Determine the minimal value of thefollowing expression
a+ 3c
a+ 2b+ c+
4b
a+ b+ 2c− 8c
a+ b+ 3c.
375
(China 2004)
Solution: Setting x = a+ 2b+ c, y = a+ b+ 2c and z = a+ b+ 3c. It is easyto see that c = z − y, b = z + x − 2y, and a = 5y − x − 3z. Now, using theAM-GM Inequality, we have
a+ 3c
a+ 2b+ c+
4b
a+ b+ 2c− 8c
a+ b+ 3c=
=(5y − x− 3z) + 3(z − y)
x+
4(z + x− 2y)
y− 8(z − y)
z
= −17 + 2
(2x
y+y
x
)+ 4
(2y
z+z
y
)≥ 4√
2 + 8√
2− 17 = 12√
2− 17.
The equality holds if and only if2x
y=
y
xand
2y
z=
z
y, or 2x =
√2y = z.
Hence the equality holds if and only if a + b + 2c =√
2(a + 2b + c) anda+ b+3c = 2(a+2b+ c). Solving this system of equations for b and c in termsof a gives b =
(1 +√
2)a and c =
(4 + 3
√2)a. From now, we conclude that
the expressiona+ 3c
a+ 2b+ c+
4b
a+ b+ 2c− 8c
a+ b+ 3c
has minimum value 12√
2−17 if and only if (a, b, c) =(a,(1 +√
2)a,(4 + 3
√2)a).
?F?
04.8. Determine the largest constant M such that the following inequalityholds for all real numbers x, y, z,
x4 + y4 + z4 + xyz (x+ y + z) ≥M (xy + yz + zx)2 .
(Hellenic 2004)
Solution: We will show that2
3is the largest value of M. Indeed, take x =
y = z = 1, we get M ≤ 2
3. It suffices to prove that
x4 + y4 + z4 + xyz (x+ y + z) ≥ 2
3(xy + yz + zx)2 ,
oor equivalently,
3(x4 + y4 + z4
)≥ 2
(x2y2 + y2z2 + z2x2
)+ xyz (x+ y + z) .
Applying the well-known inequalities a2 + b2 + c2 ≥ ab + bc + ca when thetriple (a, b, c) equals (x2, y2, z2) and (xy, yz, zx), we get
3(x4 + y4 + z4) ≥ 3(x2y2 + y2z2 + z2x2),
andx2y2 + y2z2 + z2z2 ≥ xyz(x+ y + z).
376
Adding up these two inequalities, we get the result.?F?
04.9. Let n ≥ 3 be an integer and let t1, t2, . . . , tn be positive real numberssuch that
n2 + 1 > (t1 + t2 + · · ·+ tn)
(1
t1+
1
t2+ · · ·+ 1
tn
).
Show that ti, tj , tk are the side lengths of a triangle for all i, j, k with 1 ≤ i <j < k ≤ n.
(IMO 2004)
Solution: By symmetry, it suffices to show that t1 < t2 + t3. If n > 3, apply-ing the Cauchy Schwarz Inequality, we have
n2 + 1 >
(n∑i=1
ti
)(n∑i=1
1
ti
)=
(t1 + t2 + t3 +
n∑i=4
ti
)(1
t1+
1
t2+
1
t3+
n∑i=4
1
ti
)
≥
√(t1 + t2 + t3)
(1
t1+
1
t2+
1
t3
)+
√√√√( n∑i=4
ti
)(n∑i=4
1
ti
)2
≥
[√(t1 + t2 + t3)
(1
t1+
1
t2+
1
t3
)+ n− 3
]2.
Therefore, it follows that√(t1 + t2 + t3)
(1
t1+
1
t2+
1
t3
)<√n2 + 1− n+ 3.
If n = 3, this inequality holds as well (according to the hypothesis). Since√n2 + 1 − n =
1
n+√n2 + 1
≤ 1
3 +√
10=√
10 − 3, the above inequality
implies
(t1 + t2 + t3)
(1
t1+
1
t2+
1
t3
)< 10.
In order to prove t1 < t2 + t3, it suffices to prove that if t1 ≥ t2 + t3, then thisinequality cannot hold. Indeed, if t1 ≥ t2 + t3, we have
(t1 + t2 + t3)
(1
t1+
1
t2+
1
t3
)= 1 + t1
(1
t2+
1
t3
)+t2 + t3t1
+ (t2 + t3)
(1
t2+
1
t3
)≥ 1 +
4t1t2 + t3
+t2 + t3t1
+ 4
= 5 +3t1
t2 + t3+
(t1
t2 + t3+t2 + t3t1
)≥ 5 + 3 + 2 = 10.
This completes our proof.
377
Remark: It is not hard to determine the greatest number f(n) such that, forpositive t1, . . . , tn, the inequality(
n∑i=1
ti
)(n∑i=1
1
ti
)< f(n)
implies that each three of the ti’s are the side lengths of a triangle. The answer
is f(n) =(n+√
10− 3)2
, and the proof is quite similar to the above solution.?F?
04.10. For a, b, c positive reals, find the minimum value of
b2 + c2
a2 + bc+c2 + a2
b2 + ca+a2 + b2
c2 + ab.
(India 2004)
Solution: We will show that
b2 + c2
a2 + bc+c2 + a2
b2 + ca+a2 + b2
c2 + ab≥ 3,
with equality iff a = b = c. Indeed, after using the AM-GM Inequality, itsuffices to prove that
(a2 + b2)(b2 + c2)(c2 + a2) ≥ (a2 + bc)(b2 + ca)(c2 + ab),
which is true since by the Cauchy Schwarz Inequality, we have
(a2 + b2)(b2 + c2)(c2 + a2) =
=√
(a2 + b2)(a2 + c2) ·√
(b2 + c2)(b2 + a2) ·√
(c2 + a2)(c2 + b2)
≥ (a2 + bc)(b2 + ca)(c2 + ab).
This completes our proof, and so, the searched minimum is 3, which is attainedfor a = b = c.
?F?
04.11. Let x1, x2, . . . , xn be real numbers in the interval
(0,
1
2
). Prove that
n∏i=1
xi(n∑i=1
xi
)n ≤n∏i=1
(1− xi)[n∑i=1
(1− xi)
]n .
(India 2004)
Solution: The desired inequality is equivalent to
n∑i=1
lnxi − n ln
(n∑i=1
xi
)≤
n∑i=1
ln(1− xi)− n ln
[n∑i=1
(1− xi)
],
378
orn∑i=1
[ln(1− xi)− lnxi] ≥ n ln
[n∑i=1
(1− xi)
]− n ln
(n∑i=1
xi
).
Consider the function f(x) = ln(1− x)− lnx with x ∈(
0,1
2
). We have
f ′(x) = − 1
1− x− 1
x, f ′′(x) =
1
x2− 1
(1− x)2=
1− 2x
x2(1− x)2> 0.
Therefore f(x) is (strictly) convex on
(0,
1
2
). And thus, we are allowed to
apply the Jensen’s Inequality and get
f(x1) + f(x2) + · · ·+ f(xn) ≥ nf(x1 + x2 + · · ·+ xn
n
),
which is equivalent to
n∑i=1
[ln(1− xi)− lnxi] ≥ n ln
[n∑i=1
(1− xi)
]− n ln
(n∑i=1
xi
).
Our proof is completed. Note that the equality holds if and only if x1 = x2 =· · · = xn.
?F?
04.12. Prove that for all positive real numbers a, b, the following inequalityholds √
2a(a+ b)3 + b√
2(a2 + b2) ≤ 3(a2 + b2).
(Ireland 2004)
First solution: By applying the AM-GM Inequality, we get√2a(a+ b)3 ≤ 2a(a+ b) + (a+ b)2
2, and b
√2(a2 + b2) ≤ 2b2 + a2 + b2
2.
Therefore√2a(a+ b)3 + b
√2(a2 + b2) ≤ 2a2 + 2b2 + 2ab ≤ 3(a2 + b2).
The equality holds if and only if a = b.
Second solution: Denoting P as the left hand side of the original inequality.According to the Cauchy Schwarz Inequality, we have that
P 2 =[√
2a(a+ b) · (a+ b) +√
2b2 ·√a2 + b2
]2≤[2a(a+ b) + 2b2
] [(a+ b)2 + a2 + b2
]= 4(a2 + ab+ b2)2,
and thus
P =√
2a(a+ b)3 + b√
2(a2 + b2) ≤ 2a2 + 2b2 + 2ab ≤ 3(a2 + b2).
379
?F?
04.13. If a, b, c are positive reals such that a+ b+ c = 1, prove that
1 + a
1− a+
1 + b
1− b+
1 + c
1− c≤ 2
(b
a+c
b+a
c
).
(Japan 2004)
First solution: By the Cauchy Schwarz Inequality, it is easy to see that
b
a+c
b+a
c=b2
ab+c2
bc+a2
ca≥ (a+ b+ c)2
ab+ bc+ ca.
So, the original inequality is deduced from
2(a+ b+ c)2
ab+ bc+ ca≥ 1 + a
1− a+
1 + b
1− b+
1 + c
1− c.
Noting that1 + a
1− a=
(a+ b+ c) + a
(a+ b+ c)− a=
2a
b+ c+ 1, the last inequality is equiv-
alent to2(a+ b+ c)2
ab+ bc+ ca≥ 2a
b+ c+
2b
c+ a+
2c
a+ b+ 3.
Multplying both sides for ab+ bc+ ca > 0 and using the fact that
2a(ab+ bc+ ca)
b+ c= 2a2 +
2abc
b+ c≤ 2a2 +
a(b+ c)
2,
we can see that the above inequality follows from
2(a+b+c)2 ≥ 2a2+a(b+ c)
2+2b2+
b(c+ a)
2+2c2+
c(a+ b)
2+3(ab+bc+ca).
which is trivial since it leads to an identity. It is easy to see that the equality
occurs if and only if a = b = c =1
3.
Second solution: Similar to the above solution, we see that1 + a
1− a=
2a
b+ c+
1, hence the desired inequality can be written as(2b
a− 2b
c+ a
)+
(2c
b− 2c
a+ b
)+
(2a
c− 2a
b+ c
)≥ 3,
orbc
a(c+ a)+
ca
b(a+ b)+
ab
c(b+ c)≥ 3
2.
Now, applying the Cauchy Schwarz Inequality, we have
bc
a(c+ a)+
ca
b(a+ b)+
ab
c(b+ c)=
b2c2
abc(c+ a)+
c2a2
abc(a+ b)+
a2b2
abc(b+ c)
≥ (ab+ bc+ ca)2
2abc(a+ b+ c).
380
Therefore, in order to prove the original inequality, it suffices to prove that
(ab+ bc+ ca)2
2abc(a+ b+ c)≥ 3
2,
which can be written as (ab + bc + ca)2 ≥ 3abc(a + b + c). We have alreadyproved this nice inequality in the previous problem.
?F?
04.14. Let a, b be real numbers in the interval [0, 1]. Prove that
a√2b2 + 5
+b√
2a2 + 5≤ 2√
7.
(Lithuania 2004)
Solution: For a = b = 0, the inequality is trivial. Let us consider now thecase a2 + b2 > 0. Using the Cauchy Schwarz Inequality in combination withthe given hypothesis, we have(
a√2b2 + 5
+b√
2a2 + 5
)2
≤ 2
(a2
2b2 + 5+
b2
2a2 + 5
)≤ 2
(a2
2b2 + 5a2+
b2
2a2 + 5b2
).
So it is enough to prove that
a2
2b2 + 5a2+
b2
2a2 + 5b2≤ 2
7.
Becausea2
2b2 + 5a2=
1
5− b2
5(2b2 + 5a2), the last inequality can be written as
a2
2a2 + 5b2+
b2
2b2 + 5a2≥ 2
7.
Now, we apply first the Cauchy Schwarz Inequality and then the AM-GMInequality to get
a2
2a2 + 5b2+
b2
2b2 + 5a2≥ (a2 + b2)2
a2(2a2 + 5b2) + b2(2b2 + 5a2)=
(a2 + b2)2
2(a2 + b2)2 + 6a2b2
≥ (a2 + b2)2
2(a2 + b2)2 + 6 · (a2 + b2)2
4
=2
7,
as desired. It is easy to see that the equality holds if and only if a = b = 1.?F?
04.15. Prove that for any positive real numbers a, b, c,∣∣∣∣a3 − b3a+ b+b3 − c3
b+ c+c3 − a3
c+ a
∣∣∣∣ ≤ (a− b)2 + (b− c)2 + (c− a)2
4.
381
(Moldova 2004)
Solution: With notice that∣∣∣∣a3 − b3a+ b+b3 − c3
b+ c+c3 − a3
c+ a
∣∣∣∣ =(ab+ bc+ ca) |(a− b)(b− c)(c− a)|
(a+ b)(b+ c)(c+ a),
one can write the original inequality as
(a+ b)(b+ c)(c+ a)
ab+ bc+ ca[(a− b)2 + (b− c)2 + (c− a)2] ≥ 4 |(a− b)(b− c)(c− a)| .
We denote with P the left hand side of this inequality, then by applying theAM-GM Inequality, we find that
P ≥ 3(a+ b)(b+ c)(c+ a)
ab+ bc+ ca3√
(a− b)2(b− c)2(c− a)2
≥3 3√
(a+ b)2(b+ c)2(c+ a)2
ab+ bc+ ca|(a− b)(b− c)(c− a)| .
Therefore, in order to prove the original inequality, it suffices to show that
(a+ b)2(b+ c)2(c+ a)2 ≥ 64
27(ab+ bc+ ca)3.
This is obviously true because
(a+ b)2(b+ c)2(c+ a)2 ≥ 64
81(a+ b+ c)2(ab+ bc+ ca)2 ≥ 64
27(ab+ bc+ ca)3.
Note that the inequality becomes equality if and only if a = b = c.
Second solution: It is easy to see that the inequality is not only cyclic, butsymmetric. That is why we may assume that a ≥ b ≥ c > 0. The idea is touse the inequality
x+y
2≥ x2 + xy + y2
x+ y≥ y +
x
2,
which is true if x ≥ y > 0. The proof of this inequality is easy and we won’tinsist. Now, because a ≥ b ≥ c > 0, we have the three inequalities
a+b
2≥ a2 + ab+ b2
a+ b≥ b+
a
2, b+
c
2≥ b2 + bc+ c2
b+ c≥ c+
b
2,
and of course
a+c
2≥ a2 + ac+ c2
a+ c≥ c+
a
2.
That is why we can write∑ a3 − b3
a+ b= (a− b)a
2 + ab+ b2
a+ b+ (b− c)b
2 + bc+ c2
b+ c− (a− c)a
2 + ac+ c2
a+ c
≥ (a− b)(b+
a
2
)+ (b− c)
(c+
b
2
)− (a− c)
(a+
c
2
)= −
∑(a− b)2
4.
382
In the same manner, we can prove that
∑ a3 − b3
a+ b≤
∑(a− b)2
4,
and the conclusion follows.?F?
04.16. Prove that if n > 3 and x1, x2, . . . , xn > 0 have product 1, then
1
1 + x1 + x1x2+
1
1 + x2 + x2x3+ · · ·+ 1
1 + xn + xnx1> 1.
(Russia 2004)
Solution: We use a similar form of the classical substitution x1 =a2a1, x2 =
a3a2, . . . , x3 =
a1an. In this case, the inequality becomes
a1a1 + a2 + a3
+a2
a2 + a3 + a4+ · · ·+ an
an + a1 + a2> 1,
and it is clear, because n > 3 and ai + ai+1 + ai+2 < a1 + a2 + · · ·+ an for alli.
?F?
04.17. (a) Given real numbers a, b, c with a+ b+ c = 0, prove that
a3 + b3 + c3 > 0 if and only if a5 + b5 + c5 > 0.
(b) Given real numbers a, b, c, d with a+ b+ c+ d = 0, prove that
a3 + b3 + c3 + d3 > 0 if and only if a5 + b5 + c5 + d5 > 0.
(United Kingdom 2004)
Solution: It is easy to show that
a3 + b3 + c3 − (a+ b+ c)3 = −3(a+ b)(b+ c)(c+ a),
and
a5 + b5 + c5 − (a+ b+ c)5 = −5
2(a+ b)(b+ c)(c+ a)
∑(a+ b)2.
We will use these identities to solve our problem.
(a) Using the above identities with noting that a+ b+ c = 0, we have
a3 + b3 + c3 = −3(a+ b)(b+ c)(c+ a) = −3(−c)(−a)(−b) = 3abc,
and
a5 + b5 + c5 = −5
2(a+ b)(b+ c)(c+ a)
∑(a+ b)2
= −5
2(−c)(−a)(−b)
∑(−c)2 =
5
2abc∑
a2.
383
It follows that
(a3 + b3 + c3)(a5 + b5 + c5) =15
2a2b2c2
∑a2.
The last quantity is obvious nonnegative, from which we deduce that a5 +b5 +c5 ≥ 0 if and only if a3 + b3 + c3 ≥ 0. Since we need to prove a5 + b5 + c5 > 0if and only if a3 + b3 + c3 > 0, we have to disprove the following two casesThe first case is when it exists some numbers a, b, c such that a5 + b5 + c5 > 0and a3 + b3 + c3 = 0. Since a3 + b3 + c3 = 0 and a3 + b3 + c3 = 3abc, we get
abc = 0, and hence a5 + b5 + c5 =5
2abc∑
a2 = 0, a contradiction.
The second case is when it exists some numbers a, b, c such that a5+b5+c5 = 0
and a3 + b3 + c3 > 0. Since a5 + b5 + c5 = 0 and a5 + b5 + c5 =5
2abc∑
a2, we
get abc = 0 or∑
a2 = 0. If abc = 0, then we have a3 + b3 + c3 = 3abc = 0,
contradiction. If a2 + b2 + c2 = 0, then we have a = b = c = 0 and hencea3 + b3 + c3 = 0, contradiction.From these arguments, we conclude that
a3 + b3 + c3 > 0 if and only if a5 + b5 + c5 > 0,
as desired.
(b) We use the same arguments to prove this claim. Indeed, we have
a3 + b3 + c3 + d3 = a3 + b3 + c3 − (a+ b+ c)3 = −3(a+ b)(b+ c)(c+ a),
and
a5 + b5 + c5 + d5 = a5 + b5 + c5 − (a+ b+ c)5
= −5
2(a+ b)(b+ c)(c+ a)[(a+ b)2 + (b+ c)2 + (c+ a)2].
It follows that
(a3 + b3 + c3 + d3)(a5 + b5 + c5 + d5) =15
2(a+ b)2(b+ c)2(c+ a)2
∑(a+ b)2.
The last quantity is obvious nonnegative, from which we deduce that a5 +b5 + c5 + d5 ≥ 0 if and only if a3 + b3 + c3 + d3 ≥ 0. Since we need to provea5 + b5 + c5 + d5 > 0 if and only if a3 + b3 + c3 + d3 > 0, we have to disprovethe following two casesThe first case is when it exists some numbers a, b, c, d such that a5 + b5 +c5 + d5 > 0 and a3 + b3 + c3 + d5 = 0. Since a3 + b3 + c3 + d3 = 0 anda3 + b3 + c3 +d3 = −3(a+ b)(b+ c)(c+a), we get (a+ b)(b+ c)(c+a) = 0, and
hence a5+b5+c5+d3 = −5
2(a+b)(b+c)(c+a)
∑(a+b)2 = 0, a contradiction.
The second case is when it exists some numbers a, b, c, d such that a5 + b5 +c5 + d5 = 0 and a3 + b3 + c3 + d3 > 0. Since a5 + b5 + c5 + d5 = 0 and
a5+b5+c5+d5 = −5
2(a+b)(b+c)(c+a)
∑(a+b)2, we get (a+b)(b+c)(c+a) = 0
or∑
(a+ b)2 = 0. If (a+ b)(b+ c)(c+a) = 0, then we have a3 + b3 + c3 +d3 =
384
−3(a+b)(b+c)(c+a) = 0, contradiction. If (a+b)2+(b+c)2+(c+a)2 = 0, thenwe have a = b = c = 0 and hence a3+b3+c3+d3 = −3(a+b)(b+c)(c+a) = 0,contradiction.From these arguments, we conclude that
a3 + b3 + c3 + d3 > 0 if and only if a5 + b5 + c5 + d5 > 0,
as desired.?F?
04.18. Let a, b, c be positive real numbers. Prove that
(a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a+ b+ c)3.
(USA 2004)
Solution: For any positive number x, the quantities x2 − 1 and x3 − 1 havethe same sign. Thus, we have 0 ≤ (x3 − 1)(x2 − 1) = x5 − x3 − x2 + 1, or
x5 − x2 + 2 ≥ x3 + 2.
It follows that
(a5 − a2 + 3)(b5 − b2 + 3)(c5 − c2 + 3) ≥ (a3 + 2)(b3 + 2)(c3 + 2),
and hence it suffices to prove that
(a3 + 2)(b3 + 2)(c3 + 2) ≥ (a+ b+ c)3.
Of course, this is true since by the Holder’s Inequality, we have
(a3 + 1 + 1)(1 + b3 + 1)(1 + 1 + c3) ≥(
3√a3 · 1 · 1 +
3√
1 · b3 · 1 +3√
1 · 1 · c3)3
= (a+ b+ c)3.
Note that the equality holds if and only if a = b = c = 1.?F?
04.19. Let x, y, z > 0 such that (x+ y+ z)3 = 32xyz. Find the minimum and
maximum ofx4 + y4 + z4
(x+ y + z)4.
(Vietnam 2004)
First solution: We may of course assume that x + y + z = 4 and xyz = 2.
Thus, we have to find the extremal values ofx4 + y4 + z4
44. Now, we have
x4 + y4 + z4 = (x2 + y2 + z2)2 − 2(x2y2 + y2z2 + z2x2)
=(
16− 2∑
xy)2− 2
(∑xy)2
+ 4xyz(x+ y + z)
= 2a2 − 64a+ 288,
385
where a = xy + yz + zx. Because y + z = 4 − x and yz =2
x, we must have
(4− x)2 ≥ 8
x, which implies that 3−
√5 ≤ x ≤ 2. Due to symmetry, we have
x, y, z ∈[3−√
5, 2]. This means that (x− 2)(y − 2)(z − 2) ≤ 0, and also(
x− 3 +√
5)(
y − 3 +√
5)(
z − 3 +√
5)≥ 0.
Clearing parenthesis, we deduce that
a ∈
[5,
5√
5− 1
2
].
But becausex4 + y4 + z4
44=
(a− 16)2 − 112
128, we find that the extremal values
of the expression are9
128,
383− 165√
5
256, attained for the triples (2, 1, 1) and(
3−√
5,1 +√
5
2,1 +√
5
2
), respectively.
Second solution: As in the above solution, we must find the extremal values
of x2 + y2 + z2 when x+ y+ z = 1, xyz =1
32, because after that the extremal
values of the expression x4 + y4 + z4 can be immediately found. Let us make
substitution x =a
4, y =
b
4, z =
c
2, where abc = 1, a + b + 2c = 4. Then
x2 + y2 + z2 =a2 + b2 + 4c2
16and so we must find the extremal values of
a2 + b2 +4c2. Now, a2 + b2 +4c2 = (4−2c)2− 2
c+4c2 and the problem reduces
to find the maximum and minimum of 4c2 − 8c − 1
cwhere there are positive
real numbers a, b, c such that abc = 1, a + b + 2c = 4. Of course, this comes
down to (4− 2c)2 ≥ 4
c, or to c ∈
[3−√
5
2, 1
]. But this reduces to the study
of the function f(x) = 4x2−8x− 1
xdefined for
[3−√
5
2, 1
], which is an easy
task.?F?
05.1. For any positive real numbers a, b, c such that abc = 8, then
a2√(a3 + 1) (b3 + 1)
+b2√
(b3 + 1) (c3 + 1)+
c2√(c3 + 1) (a3 + 1)
≥ 4
3.
(APMO 2005)
Solution: For any x > 0, the AM-GM Inequality implies
√x3 + 1 =
√(x+ 1) (x2 − x+ 1) ≤
(x+ 1) +(x2 − x+ 1
)2
=x2 + 2
2.
386
Applying this inequality, we get
a2√(a3 + 1) (b3 + 1)
+b2√
(b3 + 1) (c3 + 1)+
c2√(c3 + 1) (a3 + 1)
≥
≥ 4
[a2
(a2 + 2) (b2 + 2)+
b2
(b2 + 2) (c2 + 2)+
c2
(c2 + 2) (a2 + 2)
].
Therefore, in order to prove the desired inequality, it suffices to show that
a2
(a2 + 2) (b2 + 2)+
b2
(b2 + 2) (c2 + 2)+
c2
(c2 + 2) (a2 + 2)≥ 1
3.
By some easy computations (with noting that abc = 8), we can write thisinequality as
a2b2 + b2c2 + c2a2 + 2(a2 + b2 + c2) ≥ 72.
This last inequality is obviously true since from the AM-GM Inequality, wehave
a2b2 + b2c2 + c2a2 ≥ 33√a4b4c4 = 48,
and
a2 + b2 + c2 ≥ 33√a2b2c2 = 12.
The equality holds if and only if a = b = c = 2.?F?
05.2. Let a, b, c, d be positive real numbers. Show that
1
a3+
1
b3+
1
c3+
1
d3≥ a+ b+ c+ d
abcd.
(Austria 2005)
Solution: Setting x =1
a, y =
1
b, z =
1
c, t =
1
d, then we can rewrite the
inequality as
x3 + y3 + z3 + t3 ≥ xyz + yzt+ ztx+ txy.
By the AM-GM Inequality, we have that
x3+y3+z3 ≥ 3xyz, y3+z3+t3 ≥ 3yzt, z3+t3+x3 ≥ 3ztx, t3+x3+y3 ≥ 3txy.
Adding up these four inequalities and dividing each side of the resulting in-equality by 3, we get the result. It is easy to see that the equality holds if andonly if a = b = c = d.
?F?
05.3. Let a, b, c be positive real numbers. Prove that
a2
b+b2
c+c2
a≥ a+ b+ c+
4(a− b)2
a+ b+ c.
(Balkan 2005)
387
Solution: The inequality in question is equivalent to(a2
b+ b− 2a
)+
(b2
c+ c− 2b
)+
(c2
a+ a− 2c
)≥ 4(a− b)2
a+ b+ c,
or(a− b)2
b+
(c− b)2
c+
(a− c)2
a≥ 4(a− b)2
a+ b+ c.
Now, applying the Cauchy Schwarz Inequality, we have
(a− b)2
b+
(c− b)2
c+
(a− c)2
a≥ [(a− b) + (c− b) + (a− c)]2
b+ c+ a=
4(a− b)2
a+ b+ c,
which is just the desired inequality. The equality holds if and only if a = b = c.?F?
05.4. If a, b, c are positive real numbers such that abc = 1, then the inequalityholds
a
a2 + 2+
b
b2 + 2+
c
c2 + 2≤ 1.
(Baltic Way 2005)
Solution: Let x =1
a, y =
1
b, z =
1
c, then xyz = 1. Now, applying the
AM-GM Inequality, we have
a
a2 + 2+
b
b2 + 2+
c
c2 + 2≤ a
2a+ 1+
b
2b+ 1+
c
2c+ 1
=1
2 + x+
1
2 + y+
1
2 + z.
Thus, it suffices to show that
1
2 + x+
1
2 + y+
1
2 + z≤ 1,
which is equivalent to
(2 + x) (2 + y) + (2 + y) (2 + z) + (2 + z) (2 + x) ≤ (2 + x) (2 + y) (2 + z) ,
or3 ≤ xy + yz + zx.
However, this is true since xy + yz + zx ≥ 3 3√x2y2z2 = 3 by the AM-GM
Inequality. Therefore, our proof is completed. Note that the equality holds ifand only if a = b = c = 1.
?F?
05.5. Let a, b, c be positive real numbers. Prove that(a2 + b+
3
4
)(b2 + a+
3
4
)≥(
2a+1
2
)(2b+
1
2
).
(Belarus 2005)
388
Solution: Applying the AM-GM Inequality, we get(a2 + b+
3
4
)(b2 + a+
3
4
)≥
(2
√a2 · 1
4+ b+
1
2
)(2
√b2 · 1
4+ a+
1
2
)= (a+ b)2 + a+ b+
1
4≥ 4ab+ a+ b+
1
4
=
(2a+
1
2
)(2b+
1
2
),
as desired. The equality holds if and only if a = b =1
2.
?F?
05.6. Given positive real numbers a, b, c such that a+ b+ c = 1. Prove that
a√b+ b
√c+ c
√a ≤ 1√
3.
(Bosnia and Hercegovina 2005)
Solution: Using the well-known inequality (a+ b+ c)2 ≥ 3(ab+ bc+ ca) and
the hypothesise a+ b+ c = 1, we deduce that ab+ bc+ ca ≤ 1
3. Thus, by the
Cauchy Schwarz Inequality, we get(a√b+ b
√c+ c
√a)2
=(√
a√ab+
√b√bc+
√c√ca)2
≤ (a+ b+ c)(ab+ bc+ ca) ≤ 1 · 1
3=
1
3.
This implies that
a√b+ b
√c+ c
√a ≤ 1√
3,
which is just the desired inequality. It is easy to see that the equality holds iff
a = b = c =1
3.
?F?
05.7. Let x, y be positive real numbers such that x3 + y3 = x− y. Prove that
x2 + 4y2 < 1.
(China 2005)
Solution: From the given hypothesis, we have
(x3 + y3)(1− x2 − 4y2) = x3 + y3 − (x3 + y3)(x2 + 4y2)
= x3 + y3 − (x− y)(x2 + 4y2)
= y(x2 − 4xy + 5y2) = y[(x− 2y)2 + y2
]> 0.
Since x3 + y3 > 0, this inequality implies that
x2 + 4y2 < 1,
389
as desired.?F?
05.8. Let a, b, c be positive real numbers such that ab + bc + ca =1
3. Prove
that1
a2 − bc+ 1+
1
b2 − ca+ 1+
1
c2 − ab+ 1≤ 3.
(China 2005)
Solution: From 3 (ab+ bc+ ca) = 1, we have
a2 − bc+ 1 = a2 + 3 (ab+ ac) + 2bc = a (a+ b+ c) + 2 (ab+ bc+ ca) ,
and so, we can rewrite our inequality as∑ 2 (ab+ bc+ ca)
a (a+ b+ c) + 2 (ab+ bc+ ca)≤ 2.
This is equivalent to∑[1− 2 (ab+ bc+ ca)
a (a+ b+ c) + 2 (ab+ bc+ ca)
]≥ 1,
or(a+ b+ c)
∑ a
a (a+ b+ c) + 2 (ab+ bc+ ca)≥ 1.
Now, denoting with P the left hand side of the above inequality and applyingthe Cauchy Schwarz Inequality, we have
P = (a+ b+ c)∑ a2
a2 (a+ b+ c) + 2a (ab+ bc+ ca)
≥ (a+ b+ c) · (a+ b+ c)2∑[a2 (a+ b+ c) + 2a (ab+ bc+ ca)]
.
On the other hand, it is easy to verify that∑[a2(a+ b+ c) + 2a(ab+ bc+ ca)] = (a+ b+ c)3.
Therefore, from the above Cauchy Schwarz step, the result follows immedi-
ately. Note that the equality holds if and only if a = b = c =1
3.
?F?
05.9. Given positive real numbers a, b, c such that a + b + c = 1. Prove thatthe following inequality holds
10(a3 + b3 + c3)− 9(a5 + b5 + c5) ≥ 1.
(China 2005)
Solution: Replacing 1 by a+ b+ c, we can rewrite the original inequality as
(10a3 − 9a5 − a) + (10b3 − 9b5 − b) + (10c3 − 9c5 − c) ≥ 0,
390
or equivalently,
a(1− a2)(9a2 − 1) + b(1− b2)(9b2 − 1) + c(1− c2)(9c2 − 1) ≥ 0.
Now, with noting that for any nonnegative number x,
(1 + x)(9x2 − 1)− 8
3(3x− 1) =
1
3(3x+ 5)(3x− 1)2 ≥ 0,
we can find that the above inequality is deduced from
a(1− a)(3a− 1) + b(1− b)(3b− 1) + c(1− c)(3c− 1) ≥ 0.
Expanding (with noting that a+ b+ c = 1), one may write it into
4(a2 + b2 + c2)− 3(a3 + b3 + c3) ≥ 1,
or
4(a2 + b2 + c2)(a+ b+ c)− 3(a3 + b3 + c3) ≥ (a+ b+ c)3.
Expanding and simplifying once more time, it can be written in the form
ab(a+ b) + bc(b+ c) + ca(c+ a) ≥ 6abc.
Of course, this last inequality is obviously true by the AM-GM Inequality and
so, our proof is completed. The equality holds if and only if a = b = c =1
3.
?F?
05.10. Let ABC be an acute triangle. Determine the least value of thefollowing expression
P =cos2A
cosA+ 1+
cos2B
cosB + 1+
cos2C
cosC + 1.
(China 2005)
First solution: Put x = cotA, y = cotB, z = cotC, then x, y, z > 0 andxy + yz + zx = 1. Now, we see that
cos2A
cosA+ 1=
x2
x2 + 1
1 +x√
x2 + 1
=x2
√x2 + 1
(√x2 + 1 + x
)=x2(√
x2 + 1− x)
√x2 + 1
= x2 − x3√x2 + 1
= x2 − x3√x2 + xy + yz + zx
= x2 − x3√(x+ y)(x+ z)
≥ x2 − x3
2(x+ y)− x3
2(x+ z),
391
Similarly, we also have
cos2B
cosB + 1≥ y2 − y3
2(y + z)− y3
2(y + x),
cos2C
cosC + 1≥ z2 − z3
2(z + x)− z3
2(z + y).
Hence
P =cos2A
cosA+ 1+
cos2B
cosB + 1+
cos2C
cosC + 1
≥ x2 + y2 + z2 − x3 + y3
2(x+ y)− y3 + z3
2(y + z)− z3 + x3
2(z + x)
= x2 + y2 + z2 − 1
2(x2 − xy + y2)− 1
2(y2 − yz + z2)− 1
2(z2 − zx+ x2)
=1
2(xy + yz + zx) =
1
2.
Moreover, we can see that the equality holds for an equilateral triangle. There-
fore, the minimum value of P is1
2.
Second solution: Similarly, we need to prove the inequality
cos2A
cosA+ 1+
cos2B
cosB + 1+
cos2C
cosC + 1≥ 1
2.
Put x = cosA, y = cosB, z = cosC then x, y, z > 0 and x2+y2+z2+2xyz = 1.Our inequality becomes
x2
x+ 1+
y2
y + 1+
z2
z + 1≥ 1
2,
which is equivalent to(2x2
x+ 1− x2
)+
(2y2
y + 1− y2
)+
(2z2
z + 1− z2
)≥ 1− x2 − y2 − z2,
orx2(1− x)
1 + x+y2(1− y)
1 + y+z2(1− z)
1 + z≥ 2xyz.
By the AM-GM Inequality, we get
x2(1− x)
1 + x+y2(1− y)
1 + y+z2(1− z)
1 + z≥ 3 3
√x2y2z2(1− x)(1− y)(1− z)
(1 + x)(1 + y)(1 + z).
Hence, it suffices to prove that
(1− x)(1− y)(1− z) ≥ 8
27xyz(1 + x)(1 + y)(1 + z),
or (1
x− 1
)(1
y− 1
)(1
z− 1
)≥ 8
27(1 + x)(1 + y)(1 + z).
392
Now, we can see that this inequality follows from combining the two inequal-ities
8
27(1 + x)(1 + y)(1 + z) ≤ 1,
and (1
x− 1
)(1
y− 1
)(1
z− 1
)≥ 1.
In addition, it is easy to see that the first inequality follows immediately fromthe AM-GM Inequality and the well-known x+y+z = cosA+cosB+cosC ≤3
2. So, it suffices to prove the second. Without loss of generality, we may
assume that x ≥ y ≥ z. From this assumption and the inequality x+y+z ≤ 3
2,
we get y + z ≤ 1, and hence it follows that(1
y− 1
)(1
z− 1
)−(
2
y + z− 1
)2
=(1− y − z)(y − z)2
yz(y + z)2≥ 0.
On the other hand, we have
(y + z)2 − 2(1− x) = y2 + z2 + 2yz + 2x− 2(x2 + y2 + z2 + 2xyz)
= 2x(1− x− 2yz)− (y − z)2
=2x(1 + x)(1− x− 2yz)
1 + x− (y − z)2
=2x(1− x2 − 2yz − 2xyz)
1 + x− (y − z)2
=2x(y − z)2
1 + x− (y − z)2 =
(x− 1)(y − z)2
x+ 1≤ 0.
This implies that
2
y + z− 1 ≥ 2√
2(1− x)− 1 =
√2
1− x− 1 ≥ 0.
Combining this with the above inequality, we deduce that(1
x− 1
)(1
y− 1
)(1
z− 1
)≥(
1
x− 1
)(√2
1− x− 1
)2
=
(√2−√
1− x)2
x.
Furthermore, note that the inequality
(√2−√
1− x)2
x≥ 1 is equivalent to(
1−√
2− 2x)2 ≥ 0, which is true. So, from the above inequality, can conclude
that (1
x− 1
)(1
y− 1
)(1
z− 1
)≥ 1,
as desired.
Third solution: We will now give another proof for the inequality
cos2A
cosA+ 1+
cos2B
cosB + 1+
cos2C
cosC + 1≥ 1
2.
393
Since cosA =b2 + c2 − a2
2bc, we get
cos2A
cosA+ 1=
(b2 + c2 − a2)2
4b2c2+ 1
b2 + c2 − a2
2bc+ 1
=(b2 + c2 − a2)2
2[bc(b+ c)2 − a2bc].
Therefore, our inequality is equivalent to
(b2 + c2 − a2)2
bc(b+ c)2 − a2bc+
(c2 + a2 − b2)2
ca(c+ a)2 − b2ca+
(a2 + b2 − c2)2
ab(a+ b)2 − c2ab≥ 1.
By the Cauchy Schwarz Inequality, we have
(b2 + c2 − a2)2
bc(b+ c)2 − a2bc+
(c2 + a2 − b2)2
ca(c+ a)2 − b2ca+
(a2 + b2 − c2)2
ab(a+ b)2 − c2ab≥
≥ (a2 + b2 + c2)2
ab(a+ b)2 + bc(b+ c)2 + ca(c+ a)2 − abc(a+ b+ c).
So, it is enough to prove that
(a2 + b2 + c2)2 ≥ ab(a+ b)2 + bc(b+ c)2 + ca(c+ a)2 − abc(a+ b+ c),
which is equivalent to
a2(a− b)(a− c) + b2(b− c)(b− a) + c2(c− a)(c− b) ≥ 0,
It is the Schur’s Inequality (in the special case fourth degree).?F?
05.11. Let a, b, c be positive real numbers such that1
a+
1
b+
1
c= 1. Prove
that(a− 1) (b− 1) (c− 1) ≥ 8.
(Croatia 2005)
Solution: From the given hypothesis, we have ab+ bc+ ca = abc, and
1 =1
a+
1
b+
1
c≥ 3
3
√1
abc,
using the AM-GM Inequality. This implies that abc ≥ 27, and so we get
(a− 1) (b− 1) (c− 1) = abc− (ab+ bc+ ca) + a+ b+ c− 1
= abc− abc+ a+ b+ c− 1 ≥ 33√abc− 1 ≥ 8,
as desired. It is easy to see that the equality holds if and only if a = b = c = 3.?F?
05.12. Let a, b, c > 0 such that abc = 1. Prove that
a
(a+ 1) (b+ 1)+
a
(b+ 1) (c+ 1)+
a
(c+ 1) (a+ 1)≥ 3
4.
394
(Czech-Slovak 2005)
Solution: The desired inequality is equivalent to
4a (c+ 1) + 4b (a+ 1) + 4c (b+ 1) ≥ 3 (a+ 1) (b+ 1) (c+ 1) ,
or
ab+ bc+ ca+ a+ b+ c ≥ 6,
which is true since
ab+ bc+ ca+ a+ b+ c ≥ 66√a3b3c3 = 6,
from the AM-GM Inequality. The equality holds if and only if a = b = c = 1.?F?
05.13. If a, b, c are three positive real numbers such that ab + bc + ca = 1,prove that
3
√1
a+ 6b+
3
√1
b+ 6c+
3
√1
c+ 6a ≤ 1
abc.
(Germany 2005)
First solution: According to the AM-GM Inequality, we have 1 = ab+ bc+
ca ≥ 33√a2b2c2, from which it follows that abc ≤ 1
3√
3. On the other hand, for
any real numbers x, y, z, we have
(x+ y + z)2 − 3(xy + yz + zx) =1
2[(x− y)2 + (y − z)2 + (z − x)2] ≥ 0.
Setting x = ab, y = bc, z = ca in this inequality, we get 1 = (ab+ bc+ ca)2 ≥3(ab · bc+ bc · ca+ ca ·ab) = 3abc(a+ b+ c), or a+ b+ c ≤ 1
3abc. Now, applying
the AM-GM Inequality again, we have
3
√1
a+ 6b =
1
3·√
3 ·√
3 · 3
√1
a+ 6b ≤ 1
3·
3√
3 + 3√
3 +
(1
a+ 6b
)3
=1
9
(6√
3 +1
a+ 6b
).
Therefore
∑3
√1
a+ 6b ≤ 1
9
[18√
3 +
(1
a+
1
b+
1
c
)+ 6(a+ b+ c)
]=
1
9
[18√
3 +ab+ bc+ ca
abc+ 6(a+ b+ c)
]=
1
9
[18√
3 +1
abc+ 6(a+ b+ c)
].
395
Now, from what we have shown above, it is easy to see that 18√
3 ≤ 6
abcand
6(a+ b+ c) ≤ 2
abc. Thus, from the last inequality, we get
3
√1
a+ 6b+
3
√1
b+ 6c+
3
√1
c+ 6a ≤ 1
9
(6
abc+
1
abc+
2
abc
)=
1
abc,
which is just the desired inequality. The equality holds if and only if three
numbers a, b, c are equal and they are equal to1√3.
Second solution: By the Power Mean Inequality
u+ v + w
3≤ 3
√u3 + v3 + w3
3,
the left hand side of the original inequality does not exceed
33√
3
3
√1
a+ 6b+
1
b+ 6c+
1
c+ 6a =
33√
3
3
√ab+ bc+ ca
abc+ 6(a+ b+ c). (1)
The condition ab+ bc+ ca = 1 enables us to write
a+ b =1− abc
=ab− (ab)2
abc, b+ c =
bc− (bc)2
abc, c+ a =
ca− (ca)2
abc.
Hence
ab+ bc+ ca
abc+ 6(a+ b+ c) =
1
abc+ 3[(a+ b) + (b+ c) + (c+ a)]
=4− 3[(ab)2 + (bc)2 + (ca)2]
abc.
Now, we have 3[(ab)2 + (bc)2 + (ca)2] ≥ (ab+ bc+ ca)2 = 1 by the well-knowninequality 3(u2 + v2 + w2) ≥ (u + v + w)2. Hence an upper bound for the
right hand side of (1) is3
3√abc
. So it suffices to check3
3√abc≤ 1
abc, which is
equivalent to (abc)2 ≤ 1
27. This follows from the AM-GM Inequality, in view
of ab+ bc+ ca = 1 again
(abc)2 = (ab)(bc)(ca) ≤(ab+ bc+ ca
3
)3
=
(1
3
)3
=1
27.
?F?
05.14. Given positive real numbers x, y, z such that xyz ≥ 1. Prove that
x5 − x2
x5 + y2 + z2+
y5 − y2
y5 + z2 + x2+
z5 − z2
z5 + x2 + y2≥ 0.
(IMO 2005)
396
First solution: Note that
x5 − x2
x5 + y2 + z2≥ x5 − x2
x3 (x2 + y2 + z2)
is equivalent to (x3 − 1
)2x2(y2 + z2
)x3 (x2 + y2 + z2) (x5 + y2 + z2)
≥ 0,
which is true for all positive x, y, z. Hence
x5 − x2
x5 + y2 + z2≥
x2 − 1
xx2 + y2 + z2
.
Summing the above inequality with its analogous cyclic inequalities, we seethat the desired result follows from
x2 + y2 + z2 − 1
x− 1
y− 1
z≥ 0.
Since xyz ≥ 1,
x2 + y2 + z2 − 1
x− 1
y− 1
z≥ x2 + y2 + z2 − yz − zx− xy ≥ 0,
so we are done. It is easy to see that the equality holds if and only if x = y =z = 1.
Second solution: Note that
x5 − x2
x5 + y2 + z2= 1− x2 + y2 + z2
x5 + y2 + z2,
and its cyclic analogous forms. The given inequality is equivalent to
x2 + y2 + z2
x5 + y2 + z2+x2 + y2 + z2
y5 + z2 + x2+x2 + y2 + z2
z5 + x2 + y2≤ 3.
In view of the Cauchy Schwarz Inequality and the condition xyz ≥ 1, we have
(x5 + y2 + z2)(yz + y2 + z2) ≥(x2√xyz + y2 + z2
)2 ≥ (x2 + y2 + z2)2,
orx2 + y2 + z2
x5 + y2 + z2≤ yz + y2 + z2
x2 + y2 + z2.
Taking the cyclic sum of the above inequality and analogous forms gives
x2 + y2 + z2
x5 + y2 + z2+x2 + y2 + z2
y5 + z2 + x2+x2 + y2 + z2
z5 + x2 + y2≤ 2 +
xy + yz + zx
x2 + y2 + z2.
It suffices to show that xy + yz + zx ≤ x2 + y2 + z2, which is well-known.
Third solution: We may write our inequality as∑ x5
x5 + y2 + z2≥∑ x2
x5 + y2 + z2.
397
Notice that for all m,n ≥ 0 the function f(t) =t
mt+ nis always increasing.
Therefore since x ≥ 1
yz, we have
x
x5 + y2 + z2=
x
x4 · x+ y2 + z2≥
1
yz
x4 · 1
yz+ y2 + z2
=1
x4 + yz(y2 + z2).
It follows thatx5
x5 + y2 + z2≥ x4
x4 + yz(y2 + z2),
and hence∑ x5
x5 + y2 + z2≥∑ x4
x4 + yz(y2 + z2)≥∑ x4
x4 + y4 + z4= 1. (1)
Also, since 1 ≤ xyz, we have
x2
x5 + y2 + z2=
x2 · 1(y2 + z2) · 1 + x5
≤ x2 · xyz(y2 + z2) · xyz + x5
=x2yz
x4 + yz(y2 + z2)
≤ x2yz
x4 + 2y2z2=
x2yz
x4 + y2z2 + y2z2≤ x2yz
2x2yz + y2z2=
x2
2x2 + yz.
From this, we deduce that
∑ x2
x5 + y2 + z2≤∑ x2
2x2 + yz. (2)
Combining (1) and (2), we see that it is enough to check that
x2
2x2 + yz+
y2
2y2 + zx+
z2
2z2 + xy≤ 1,
or equivalently,
yz
yz + 2x2+
zx
zx+ 2y2+
xy
xy + 2z2≥ 1.
By the Cauchy Schwarz Inequality, we get
∑ yz
yz + 2x2≥ (xy + yz + zx)2
yz(yz + 2x2) + zx(zx+ 2y2) + xy(xy + 2z2)
=(xy + yz + zx)2
(xy + yz + zx)2= 1,
as desired.
Fourth solution: We have to prove that
x2 + y2 + z2
x5 + y2 + z2+x2 + y2 + z2
y5 + z2 + x2+x2 + y2 + z2
z5 + x2 + y2≤ 3.
398
Because y2 + z2 ≥ 2yz and xyz ≥ 1, we have
1
x5 + y2 + z2≤ 1
x4
yz+ y2 + z2
≤ 1
2x4
y2 + z2+ y2 + z2
,
and the cyclic analogous forms. Thus it suffices to show that
x2 + y2 + z2
2x4
y2 + z2+ y2 + z2
+x2 + y2 + z2
2y4
z2 + x2+ z2 + x2
+x2 + y2 + z2
2z4
x2 + y2+ x2 + y2
≤ 3.
However, since this is a homogeneous inequality, the condition xyz ≥ 1 is notrelevant anymore. Furthermore, we may assume that x2 + y2 + z2 = 3. Thenthe inequality reduces to
1
2x4
3− x2+ 3− x2
+1
2y4
3− y2+ 3− y2
+1
2z4
3− z2+ 3− z2
≤ 1,
or3− x2
3x4 − 6x2 + 9+
3− y2
3y4 − 6y2 + 9+
3− z2
3z4 − 6z2 + 9≤ 1,
where x, y, z are positive real numbers with x2 + y2 + z2 = 3. Because 3x4 −6x2 + 9 = 3(x2 − 1)2 + 6 ≥ 6 and 3− x2 = y2 + z2 ≥ 0, we obtain
3− x2
3x4 − 6x2 + 9≤ 3− x2
6.
Adding the above inequality and the cyclic analogous forms gives
3− x2
3x4 − 6x2 + 9+
3− y2
3y4 − 6y2 + 9+
3− z2
3z4 − 6z2 + 9≤ 9− (x2 + y2 + z2)
6= 1,
as desired.
Fifth solution: From the given condition, we have1
x≤ yz. Using it and the
AM-GM Inequality, we find that
4(x5 − x2)x5 + y2 + z2
+ 1 =5x5 − 4x2 + y2 + z2
x5 + y2 + z2=
3x5 + (2x5 + y2 + z2 − 4x2)
x5 + y2 + z2
≥ 3x5 + 4 4√x10y2z2 − 4x2
x5 + y2 + z2≥ 3x5
x5 + y2 + z2
=3x4
x4 +1
x· (y2 + z2)
≥ 3x4
x4 + yz(y2 + z2)≥ 3x4
x4 + y4 + z4.
From this, it follows that
x5 − x2
x5 + y2 + z2≥ 3
4· x4
x4 + y4 + z4− 1
4.
399
Adding the above inequality and the cyclic analogous forms gives
x5 − x2
x5 + y2 + z2+
y5 − y2
y5 + z2 + x2+
z5 − z2
z5 + x2 + y2≥ 3
4· x
4 + y4 + z4
x4 + y4 + z4− 3
4= 0,
as desired.?F?
05.15. Let a1 ≤ a2 ≤ · · · ≤ an be positive real numbers such that
a21 + a22 + · · ·+ a2nn
= 1,a1 + a2 + · · ·+ an
n= m,
where 1 ≥ m > 0. Prove that for all i satisfying ai ≤ m, we have
n− i ≥ n(m− ai)2.
(Iran 2005)
First solution: By the Cauchy Schwarz Inequality, we have
m =a1 + a2 + · · ·+ ai + ai+1 + ai+2 + · · ·+ an
n
=a1 + a2 + · · ·+ ai
n+ai+1 + ai+2 + · · ·+ an
n
≤ a1 + a2 + · · ·+ ain
+
√(n− i)(a2i+1 + a2i+2 + · · ·+ a2n)
n
≤ ai +
√(n− i)(a2i+1 + a2i+2 + · · ·+ a2n)
n
= ai +
√(n− i)(n− a21 − a22 − · · · − a2i )
n
≤ ai +
√n(n− i)n
,
and therefore
m− ai ≤√n− in
.
This means that n− i ≥ n(m− ai)2, which yields our conclusion.
Second solution: Let bk = m− ak for all k = 1, 2, . . . , n, we then have that
m ≥ b1 ≥ b2 ≥ · · · ≥ bn,n∑k=1
bk = 0,
n∑k=1
b2k = n(1−m2).
Since ai ≤ m, we can note that m ≥ b1 ≥ b2 ≥ · · · ≥ bi ≥ 0, and thus itremains to prove that
b2i ≤n− in
.
By the Cauchy Schwarz Inequality, we get
b21 + b22 + · · ·+ b2i ≥1
i(b1 + b2 + · · ·+ bi)
2,
400
and
b2i+1 + b2i+2 + · · ·+ b2n ≥1
n− i(bi+1 + bi+2 + · · ·+ bn)2
=1
n− i(−b1 − b2 − · · · − bi)2
=1
n− i(b1 + b2 + · · ·+ bi)
2.
Summing up these two inequalities, it follows that
n(1−m2) ≥(
1
i+
1
n− i
)(b1 + b2 + · · ·+ bi)
2 ≥(
1
i+
1
n− i
)· i2b2i =
nib2in− i
,
and therefore
b2i ≤(n− i)(1−m2)
i.
On the other hand, we also have b2i ≤ m2, and so we can write that
b2i ≤ min
{m2,
(n− i)(1−m2)
i
}.
Note that this proves our inequality, since (it is easy to check that)
min
{m2,
(n− i)(1−m2)
i
}≤ n− i
n.
?F?
05.16. If three nonnegative real numbers a, b, c satisfy the condition
1
a2 + 1+
1
b2 + 1+
1
c2 + 1= 2,
prove that
ab+ bc+ ca ≤ 3
2.
(Iran 2005)
First solution: Set the following substitutions x =1
a2 + 1, y =
1
b2 + 1, and
z =1
c2 + 1, then 1 > x, y, z > 0 and x+ y + z = 2. We furthermore get
a2 =1− xx
=y + z − x
2x≥ 0,
b2 =1− yy
=z + x− y
2y≥ 0,
c2 =1− zz
=x+ y − z
2z≥ 0,
and therefore x, y, z are the sidelengths of a triangle. Next, putm =y + z − x
2, n =
z + x− y2
, p =x+ y − z
2. We then have that m+ n+ p = 2 and
a =
√m
n+ p, b =
√n
p+m, c =
√p
m+ n.
401
Now, by using the AM-GM Inequality,
2ab = 2
√m
n+ p· n
p+m= 2
√m
m+ p· n
n+ p≤ m
m+ p+
n
n+ p,
and similarly, we get
2bc ≤ n
m+ n+
p
m+ p, 2ca ≤ m
m+ n+
p
n+ p.
Thus
2(ab+ bc+ ca) ≤ m
m+ p+
n
n+ p+
n
m+ n+
p
m+ p+
m
m+ n+
p
n+ p= 3.
The equality holds if and only if a = b = c = 1√2.
Second solution: From the given hypothesis, we get
1 =
(1− 1
a2 + 1
)+
(1− 1
b2 + 1
)+
(1− 1
c2 + 1
)=
a2
a2 + 1+
b2
b2 + 1+
c2
c2 + 1.
On the other hand, the Cauchy Schwarz Inequality yields
a2
a2 + 1+
b2
b2 + 1+
c2
c2 + 1≥ (a+ b+ c)2
a2 + b2 + c2 + 3.
Thus, we have a2 + b2 + c2 + 3 ≥ (a+ b+ c)2, which leads us to
ab+ bc+ ca ≤ 3
2.
?F?
05.17. If x, y, z are real numbers satisfying xyz = −1, prove that
x4 + y4 + z4 + 3(x+ y + z) ≥ y2 + z2
x+z2 + x2
y+x2 + y2
z.
(Iran 2005)
Solution: According to the condition from the hypothesis, we have 3(x+ y+z) = −3xyz(x+ y + z), and thus
y2 + z2
x+z2 + x2
y+x2 + y2
z= −yz(y2 + z2)− zx(z2 + x2)− xy(x2 + y2)
= −x3(y + z)− y3(z + x)− z3(x+ y).
The inequality in question is now equivalent to
x4 + y4 + z4 − 3xyz(x+ y + z) ≥ −x3(y + z)− y3(z + x)− z3(x+ y),
which can be rewritten as
[x4 + y4 + z4 + x3(y + z) + y3(z + x) + z3(x+ y)]− 3xyz(x+ y + z) ≥ 0,
402
or(x+ y + z)(x3 + y3 + z3 − 3xyz) ≥ 0,
which is obviously true according to
(x+ y + z)(x3 + y3 + z3 − 3xyz) = (x+ y + z)2(x2 + y2 + z2 − xy − yz − zx),
and to the well-known
x2 + y2 + z2 ≥ xy + yz + zx.
?F?
05.18. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
a3√
1 + b− c+ b 3√
1 + c− a+ c3√
1 + a− b ≤ 1.
(Japan 2005)
Solution: Because 1 + b − c > 0, we can apply the AM-GM Inequality andget
a3√
1 + b− c = a 3√
1 · 1 · (1 + b− c) ≤ a(
1 + 1 + 1 + b− c3
)= a+
ab− ac3
.
Adding this to the two analogous inequalities and using the fact that a+b+c =1, we get the result. It is easy to see that the equality holds if and only if
a = b = c =1
3.
?F?
05.19. For any real numbers x1, x2, . . . , xn with x21 + x22 + · · ·+ x2n = 1, provethat
x11 + x21
+x2
1 + x21 + x22+ · · ·+ xn
1 + x21 + x22 + · · ·+ x2n<
√n
2.
(Korea 2005)
Solution: For convenience, let us denote x0 = 0. Then, by applying theCauchy Schwarz Inequality, we can see that the left hand side of the originalinequality does not exceed√√√√n
n∑i=1
x2i(1 + x20 + · · ·+ x2i )
2≤
√√√√n
n∑i=1
x2i(1 + x20 + · · ·+ x2i )
2
≤
√√√√n
n∑i=1
x2i(1 + x20 + · · ·+ x2i−1)(1 + x20 + · · ·+ x2i )
=
√√√√n
n∑i=1
(1
1 + x20 + · · ·+ x2i−1− 1
1 + x20 + · · ·+ x2i
)
=
√n
(1
1 + x20− 1
1 + x20 + · · ·+ x2n
)=
√n
2.
403
It is easy to see that the equality does not occur. So, we must have
n∑i=1
xi1 + x21 + · · ·+ x2i
<
√n
2.
The proof is completed.?F?
05.20. Given positive real numbers a, b, c such that a4 + b4 + c4 = 3. Provethat
1
4− ab+
1
4− bc+
1
4− ca≤ 1.
(Moldova 2005)
Solution: Since a4 + b4 + c4 = 3, it is easy to see that a2, b2, c2 < 2. Now,applying the Cauchy Schwarz Inequality and the AM-GM Inequality, we findthat
1
4− a2+
1
4− b2≥ 4
4− a2 + 4− b2≥ 4
8− 2ab=
2
4− ab.
Adding this to the two analogous inequalities and dividing each side of theresulting inequality by 2, we get
1
4− ab+
1
4− bc+
1
4− ca≤ 1
4− a2+
1
4− b2+
1
4− c2.
Therefore, it suffices to prove that
1
4− a2+
1
4− b2+
1
4− c2≤ 1.
To prove it, we observe that for each x2 < 2,
1
4− x2≤ x4 + 5
18,
since it is equivalent to (x2 − 1)2(2− x2) ≥ 0, true. So we have
1
4− a2+
1
4− b2+
1
4− c2≤ a4 + b4 + c4 + 15
18= 1,
as desired. The proof is completed. Note that the equality holds if and onlyif a = b = c = 1.
?F?
05.21. Let a, b, c ∈ [0, 1]. Prove that
a
bc+ 1+
b
ca+ 1+
c
ab+ 1≤ 2.
(Poland 2005)
404
Solution: Since a, b, c ∈ [0, 1], we have (1− a)(1− b) + (1− ab)(1− c) ≥ 0. Itfollows that 2 + abc ≥ a+ b+ c. Using this inequality, we get
a
bc+ 1+
b
ca+ 1+
c
ab+ 1≤ a
abc+ 1+
b
abc+ 1+
c
abc+ 1
=a+ b+ c
1 + abc≤ 2 + abc
1 + abc≤ 2 + 2abc
1 + abc= 2,
as desired. Note that the equality holds if and only if (a, b, c) equals (1, 1, 0),or (1, 0, 1), or (0, 1, 1).
?F?
05.22. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
√ab (1− c) +
√bc (1− a) +
√ca (1− b) ≤
√2
3.
(Republic of Srpska 2005)
Solution: Applying the Cauchy Schwarz Inequality in combination with thewell-known inequality 3(ab+ bc+ ca) ≤ (a+ b+ c)2, we get[∑√
ab (1− c)]2≤ (ab+ bc+ ca) (1− c+ 1− a+ 1− b)
≤ 2 · (a+ b+ c)2
3=
2
3,
as desired. The equality holds iff (a, b, c) =
(1
3,1
3,1
3
).
?F?
05.23. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that
1
1 + a+ b+
1
1 + b+ c+
1
1 + c+ a≤ 1.
(Romania 2005)
Solution: By expanding, we can rewrite the inequality as∑(1 + b+ c) (1 + c+ a) ≤ (1 + a+ b) (1 + b+ c) (1 + c+ a) ,
or equivalently,
2 + 2 (a+ b+ c) ≤ 2abc+ a2 (b+ c) + b2 (c+ a) + c2 (a+ b) .
Since abc ≥ 1, it suffices to show that
a2 (b+ c) + b2 (c+ a) + c2 (a+ b) ≥ 2 (a+ b+ c) .
405
Now, applying the AM-GM Inequality and the Chebyshev’s Inequality, wehave
a2 (b+ c) + b2 (c+ a) + c2 (a+ b) ≥ 2a2√bc+ 2b2
√ca+ 2c2
√ab
≥ 2(a√a+ b
√b+ c
√c)
≥ 2
3(a+ b+ c)
(√a+√b+√c)
≥ 2
3· 3
3
√√abc (a+ b+ c) ≥ 2(a+ b+ c).
Therefore, the last inequality is obviously true. And so, our proof is completed.Note that the equality holds if and only if a = b = c = 1.
?F?
05.24. Let n is a positive integer. Prove that if x is a positive real numbers,then
1 + xn+1 ≥ (2x)n
(1 + x)n−1.
(Russia 2005)
Solution: By the AM-GM Inequality, we have
1 + xn+1 ≥ 2√xn+1 = 2x
n+12 , 1 + x ≥ 2
√x = 2x
12 .
Therefore (1 + xn+1
)(1 + x)n−1 ≥ 2x
n+12 ·
(2x
12
)n−1= (2x)n .
Dividing both sides of this inequality by (1 + x)n−1, we get the result. It iseasy to see that the equality holds if and only if x = 1.
?F?
05.25. Given positive real numbers x, y, z such that x2 + y2 + z2 = 1. Provethe following inequality
x
x3 + yz+
y
y3 + zx+
z
z3 + xy> 3.
(Russia 2005)
Solution: Setting a =yz
x, b =
zx
y, c =
xy
z, then we have ab+ bc+ ca = 1 and
x
x3 + yz=
1
a+ bc.
Therefore, the original inequality is equivalent to
1
a+ bc+
1
b+ ca+
1
c+ ab> 3.
406
Without loss of generality, we may assume that a = max{a, b, c}. If b+ c > 1,then we have a + b ≥ c + b > 1 and a + c ≥ b + c > 1, from which it followsthat
1
a+ bc>
1
a(b+ c) + bc= 1,
1
b+ ca>
1
b(c+ a) + ca= 1,
1
c+ ab>
1
c(a+ b) + ab= 1.
Therefore, it is clear that
1
a+ bc+
1
b+ ca+
1
c+ ab> 3.
Let us consider now the case b + c ≤ 1. In this case, we apply the CauchySchwarz Inequality and obtain
1
b+ ca+
1
c+ ab≥ 4
b+ ca+ c+ ab=
4
a(b+ c) + b+ c=
4
1− bc+ b+ c>
4
1 + b+ c.
On the other hand, we have
a+ bc =1− bcb+ c
+ bc =1
b+ c+bc(b+ c− 1)
b+ c≤ 1
b+ c,
which yields that1
a+ bc≥ b + c. Using this in combination with the above
inequality, it suffices to prove that
(b+ c) +4
b+ c+ 1≥ 3.
However, this inequality is obvious since from the AM-GM Inequality, we get
(b+ c) +4
b+ c+ 1= (b+ c+ 1) +
4
b+ c+ 1− 1 ≥ 4− 1 = 3.
?F?
05.26. Let a, b, c be positive real numbers. Prove that
a√b+ c
+b√c+ a
+c√a+ b
≥√
3
2(a+ b+ c).
(Serbia and Montenegro 2005)
Solution: Without loss of generality, we may assume that a ≥ b ≥ c. Fromthis assumption, we have
1√b+ c
≥ 1√c+ a
≥ 1√a+ b
.
407
Therefore, applying the Chebyshev’s Inequality and the AM-GM Inequality,we get
a√b+ c
+b√c+ a
+c√a+ b
≥ 1
3(a+ b+ c)
(1√b+ c
+1√c+ a
+1√a+ b
)≥ a+ b+ c
3· 3
3
√√(a+ b) (b+ c) (c+ a)
≥ a+ b+ c√(a+ b) + (b+ c) + (c+ a)
3
=
√3
2(a+ b+ c),
as desired. The equality holds if and only if a = b = c.?F?
05.27. If a, b, c are positive real numbers such that ab + bc + ca = 1. Provethat
33
√1
abc+ 6 (a+ b+ c) ≤
3√
3
abc.
(Slovenia 2005)
Solution: The original inequality can be written as
a2b2c2[1 + 6abc (a+ b+ c)] ≤ 1
9.
Now, using the AM-GM Inequality, we find that 1 = ab+ bc+ ca ≥ 33√a2b2c2,
and hence a2b2c2 ≤ 1
27. On the other hand, applying the well-known inequality
(x+ y + z)2 ≥ 3(xy + yz + zx) for the triple (x, y, z) = (ab, bc, ca), we get
1 = (ab+ bc+ ca)2 ≥ 3abc(a+ b+ c).
From these inequalities, we conclude that
a2b2c2[1 + 6abc (a+ b+ c)] ≤ 1
27(1 + 2) =
1
9,
as desired. Note that the equality holds if and only if a = b = c =1√3.
?F?
05.28. Let a1, a2, . . . , a95 be positive real numbers. Prove that
95∑k=1
ak ≤ 94 +
95∏k=1
max {1, ak} .
(Taiwan 2005)
Solution: Let bk = max{ak, 1}, then we have
95∏k=1
max {1, ak} =
95∏k=1
bk, and
95∑k=1
ak ≤95∑k=1
bk.
408
So it suffices to prove that
95∑k=1
bk ≤ 94 +
95∏k=1
bk.
This inequality can be written as
(1− b1)(1− b2) + (1− b1b2)(1− b3) + · · ·+ (1− b1b2 · · · b94)(1− b95) ≥ 0,
which is obviously true because bk ≥ 1 for all k = 1, 2 . . . , 95. Our proof iscompleted.
?F?
05.29. Let a, b, c be positive real numbers. Prove that(a
b+b
c+c
a
)2
≥ (a+ b+ c)
(1
a+
1
b+
1
c
).
(United Kingdom 2005)
First solution: Due to the cyclicity, we may assume that c is the smallestnumbers among a, b, c. Then, applying the AM-GM Inequality, we have
(a+ b+ c)
(1
a+
1
b+
1
c
)=a+ b+ c
b
(b
a+ 1 +
b
c
)≤ 1
4
(a+ b+ c
b+b
a+b
c+ 1
)2
.
Using this inequality, it suffices to prove that
2
(a
b+b
c+c
a
)≥ a+ b+ c
b+b
a+b
c+ 1,
which is equivalent to(ab− c
b
)+
(b
c− b
a
)+
(2c
a− 2
)≥ 0,
or
(a− c)(
1
b+
b
ac− 2
a
)≥ 0,
and this is true since a− c ≥ 0 and1
b+
b
ac≥ 2√
ac≥ 2
a. It is easy to see that
the equality holds if and only if a = b = c.
Second solution: Denote x =a
b, y =
b
c, z =
c
a, then we have xyz = 1. Now,
with noting that
(a+ b+ c)
(1
a+
1
b+
1
c
)=a
b+b
c+c
a+b
a+c
b+a
c+ 3
= x+ y + z +1
x+
1
y+
1
z+ 3,
409
we can rewrite the original inequality in the following form
(x+ y + z)2 ≥ 3 + x+ y + z +1
x+
1
y+
1
z.
Because1
x+
1
y+
1
z= xy + yz + zx, it is equivalent to
(x+ y + z)2 ≥ 3 + x+ y + z + xy + yz + zx,
orx2 + y2 + z2 + xy + yz + zx ≥ 3 + x+ y + z.
By the AM-GM Inequality, we have
xy + yz + zx ≥ 3 3√x2y2z2 = 3,
and
x2 + y2 + z2 = (x2 + 1) + (y2 + 1) + (z2 + 1)− 3 ≥ 2(x+ y + z)− 3
≥ x+ y + z + 3 3√xyz − 3 ≥ x+ y + z.
Therefore, the last inequality is obviously true and so, our proof is completed.?F?
05.30. Let a, b, c be positive real numbers. Prove that(a
a+ b
)3
+
(b
b+ c
)3
+
(c
c+ a
)3
≥ 3
8.
(Vietnam 2005)
First solution: By the Cauchy Schwarz Inequality, we get[∑ a3
(a+ b)3
] [∑a(a+ b)
]≥(∑ a2
a+ b
)2
,
and since∑
a(a+ b) =∑
a2 +∑
ab ≤ 2∑
a2, it suffices to prove that(∑ a2
a+ b
)2
≥ 3
4
∑a2,
or equivalently, ∑ 2a2
a+ b≥√
3∑
a2.
Since∑ 2a2
a+ b=∑ a2 + b2
a+ b+∑ a2 − b2
a+ b=∑ a2 + b2
a+ b, the above inequality
can be rewritten as ∑ a2 + b2
a+ b≥√
3∑
a2,
or equivalently, ∑(a2 + b2
a+ b− a+ b
2
)≥√
3∑
a2 −∑
a,
410
i.e. ∑ (a− b)2
2(a+ b)≥
∑(a− b)2√
3∑
a2 +∑
a.
From the Cauchy Schwarz Inequality, we have√
3∑
a2 ≥∑
a, hence it is
enough to show that
∑ (a− b)2
2(a+ b)≥
∑(a− b)2
2∑
a,
which is obviously valid since it can be written as∑ c(a− b)2
(a+ b)(a+ b+ c)≥ 0.
Note that the equality holds if and only if a = b = c.
Second solution: By the Cauchy Schwarz Inequality, we get[∑ a3
(a+ b)3
] [∑c3(a+ b)3
]≥(∑
c3/2a3/2)2,
and therefore we are left to prove that
8(∑
c3/2a3/2)2≥ 3
∑c3(a+ b)3.
We continue by setting the substitutions x =√ab, y =
√bc, z =
√ca. In this
case, the inequality to prove becomes
8(x3 + y3 + z3)2 ≥ 3(x2 + y2)3 + 3(y2 + z2)3 + 3(z2 + x2)3,
or equivalently,∑(x6 + y6)− 9
∑x2y2(x2 + y2) + 16
∑x3y3 ≥ 0,
which is valid because
x6 + y6 − 9x2y2(x2 + y2) + 16x3y3 = (x2 + 4xy + y2)(x− y)4 ≥ 0.
Third solution: Using the Power Mean Inequality, we get[1
3
∑ a3
(a+ b)3
]1/3≥[
1
3
∑ a2
(a+ b)2
]1/2,
hence we must show that
a2
(a+ b)2+
b2
(b+ c)2+
c2
(c+ a)2≥ 3
4.
411
Consider the substitutions x =b
a, y =
c
b, z =
a
c. Then xyz = 1, and thus the
previous inequality becomes
1
(1 + x)2+
1
(1 + y)2+
1
(1 + z)2≥ 3
4.
Since
1
(1 + x)2+
1
(1 + y)2− 1
1 + xy=
xy(x− y)2 + (xy − 1)2
(1 + xy)(1 + x)2(1 + y)2≥ 0,
we have that1
(1 + x)2+
1
(1 + y)2≥ 1
1 + xy=
z
z + 1,
and therefore
1
(1 + x)2+
1
(1 + y)2+
1
(1 + z)2≥ z
z + 1+
1
(1 + z)2=
(z − 1)2
4(z + 1)2+
3
4≥ 3
4.
Fourth solution: Like in the third solution, we reach the point where wemust show that
a2
(a+ b)2+
b2
(b+ c)2+
c2
(c+ a)2≥ 3
4.
This time we proceed as follows: from the Cauchy Schwarz Inequality, we have[∑ a2
(a+ b)2
] [∑(a+ b)2(a+ c)2
]≥[∑
a(a+ b)]2
=1
4
[∑(a+ b)2
]2,
and therefore, we are left to prove that[∑(a+ b)2
]2≥ 3
∑(a+ b)2(a+ c)2,
which is immediate, according to the AM-GM Inequality.?F?
06.1. Let x1, x2, . . . , xn be positive real numbers such that x1+x2+ · · ·+xn =1. Prove that (
n∑i=1
√xi
)(n∑i=1
1√1 + xi
)≤ n2√
n+ 1.
(China 2006)
Firs solution: From the AM-GM Inequality, we get(n∑i=1
√xi
)(n∑i=1
1√1 + xi
)≤√n+ 1
4n
(n√n+ 1
n∑i=1
√xi +
n∑i=1
1√1 + xi
)2
,
and thus it is enough to show that
n∑i=1
(n√n+ 1
√xi +
1√1 + xi
)≤ 2n
√n
n+ 1.
412
We shall now make use of the following auxiliary resultLemma. For any positive real number x, the following inequality holds
n√n+ 1
√x+
1√1 + x
≤√n(n+ 1)
2(n+ 1)2(n2x+ 3n+ 4
).
Proof. This rewrites as(n√n+ 1
√x+
1√1 + x
)2
≤ n
4(n+ 1)3(n2x+ 3n+ 4)2,
and since from the Cauchy Schwarz Inequality, we have(n√n+ 1
√x+
1√1 + x
)2
≤ (nx+ 1)
(n
n+ 1+
1
x+ 1
)=
(nx+ 1)(nx+ 2n+ 1)
(n+ 1)(x+ 1),
the problem reduces to show that
4(n+ 1)2(nx+ 1)(nx+ 2n+ 1) ≤ n(x+ 1)(n2x+ 3n+ 4)2.
On the other hand, the AM-GM Inequality gives us that
(n2x+ 3n+ 4)2 ≥ 4(n2x+ n+ 2)(2n+ 2) = 8(n+ 1)(n2x+ n+ 2),
and therefore, we are left to prove that
(n+ 1)(nx+ 1)(nx+ 2n+ 1) ≤ 2n(x+ 1)(n2x+ n+ 2),
which is valid, since it reduces to (1− n)(t− 2)2 ≤ 0.Returning to our inequality, we now conclude that
n∑i=1
(n√n+ 1
√xi +
1√1 + xi
)≤√n(n+ 1)
2(n+ 1)2
n∑i=1
(n2xi + 3n+ 4
)= 2n
√n
n+ 1.
The equality holds if and only if x1 = x2 = · · · = xn =1
n.
Second solution: Put yi = xi + 1, for all i = 1, 2, . . . , n. In this case, wehave yi ≥ 1 and y1 + y2 + · · · + yn = n + 1, and therefore the inequality inquestion becomes (
n∑i=1
√yi − 1
)(n∑i=1
1√yi
)≤ n2√
n+ 1,
or equivalently,
n∑i=1
n∑j=1
√yj − 1
√yi
≤ n2√n+ 1
.
413
From the Cauchy Schwarz Inequality, we get
n∑i=1
√yi − 1
√y1
=
=
√n
(1√n·√y1 − 1 +
√y2 − 1 · 1√
n+ · · ·+
√yn − 1 · 1√
n
)√y1
≤
√n
√[1
n+ (y2 − 1) + · · ·+ (yn − 1)
] [(y1 − 1) +
1
n+ · · ·+ 1
n
]√y1
=
√−ny1 + 2(n+ 1)− 2n+ 1
ny1,
and similarly, one can prove that
n∑j=1
√yj − 1
√yi
≤√−nyi + 2(n+ 1)− 2n+ 1
nyi
for any i = 1, 2, . . . , n. This yields
n∑i=1
n∑j=1
√yj − 1
√yi
≤n∑i=1
√−nyi + 2(n+ 1)− 2n+ 1
nyi
≤
√√√√nn∑i=1
[−nyi + 2(n+ 1)− 2n+ 1
nyi
]
=
√√√√n
[n(n+ 1)− 2n+ 1
n
n∑i=1
1
yi
]
≤
√√√√√√√√√n
n(n+ 1)− 2n+ 1
n· n2
n∑i=1
yi
=
√n
[n(n+ 1)− 2n+ 1
n· n2
n+ 1
]=
n2√n+ 1
.
?F?
06.2. Let a, b, c be positive real numbers satisfying a+ b+ c = 1. Prove that
ab√ab+ bc
+bc√
bc+ ca+
ca√ca+ ab
≤√
2
2.
414
(China 2006)
First solution: From the Cauchy Schwarz Inequality, we get
∑ ab√ab+ bc
=∑ a
√b√
c+ a≤∑ a
√2b√
c+√a,
hence it suffices to prove that
2a√b√
c+√a
+2b√c
√a+√b
+2c√a√
b+√c≤ 1.
Furthermore, set√a = x,
√b = y,
√c = z. We thus have to prove that
2x2y
z + x+
2y2z
x+ y+
2z2x
y + z≤ x2 + y2 + z2,
or equivalently,
x2 + y2 + z2 +∑(
2xy − 2x2y
z + x
)≥ 2(xy + yz + zx).
In other words,
x2 + y2 + z2 + 2xyz
(1
x+ y+
1
y + z+
1
z + x
)≥ 2(xy + yz + zx).
Now since the Cauchy Schwarz Inequality gives us that
2xyz
(1
x+ y+
1
y + z+
1
z + x
)≥ 9xyz
x+ y + z,
the problem reduces to proving that
x2 + y2 + z2 +9xyz
x+ y + z≥ 2(xy + yz + zx),
which is simply a particular case of the Schur’s Inequality written for the third
degree. Note that the equality occurs iff a = b = c =1
3.
Second solution: By the Cauchy Schwarz Inequality, we have(∑ ab√ab+ bc
)2
=
[∑√ab
ab+ bc+ ca·√a(ab+ bc+ ca)
c+ a
]2≤(∑ ab
ab+ bc+ ca
)[∑ a(ab+ bc+ ca)
c+ a
]=∑ a(ab+ bc+ ca)
c+ a,
and thus, it is enough to prove that∑ a(ab+ bc+ ca)
c+ a≤ 1
2(a+ b+ c)2,
415
or equivalently,2ab2
a+ b+
2bc2
b+ c+
2ca2
c+ a≤ a2 + b2 + c2.
This can be rewritten into(2b2 − 2ab2
a+ b
)+
(2c2 − 2bc2
b+ c
)+
(2a2 − 2ca2
c+ a
)≥ a2 + b2 + c2,
or in other words,
2
(b3
a+ b+
c3
b+ c+
c3
c+ a
)≥ a2 + b2 + c2.
This last inequality is valid, since from the Cauchy Schwarz Inequality, we get
2
(b3
a+ b+
c3
b+ c+
c3
c+ a
)≥ 2(a2 + b2 + c2)2
a2 + b2 + c2 + ab+ bc+ ca
≥ 2(a2 + b2 + c2)2
2(a2 + b2 + c2)= a2 + b2 + c2.
Third solution: By same Cauchy Schwarz Inequality, we have
(∑ ab√ab+ bc
)2
=
[∑√a+ b ·
√a2b
(a+ b)(a+ c)
]2
≤[∑
(a+ b)] [∑ a2b
(a+ b)(a+ c)
]= 2
∑ a2b
(a+ b)(a+ c),
hence it suffices to prove that
4∑ a2b
(a+ b)(a+ c)≤ a+ b+ c,
which is equivalent to
4a2b(b+ c) + 4b2c(c+ a) + 4c2a(a+ b) ≤ (a+ b)(b+ c)(c+ a)(a+ b+ c),
this last one being true, because it can be written as
ab(a− b)2 + bc(b− c)2 + ca(c− a)2 ≥ 0.
?F?
06.3. Suppose that a1, a2, . . . , an are real numbers with sum 0. Prove that thefollowing inequality holds
max1≤i≤n
a2i ≤n
3
n−1∑i=1
(ai − ai+1)2.
416
(China 2006)
Solution: It is sufficient to prove that for every 1 ≤ i ≤ n, we have
(a1 − a2)2 + (a2 − a3)2 + · · ·+ (an−1 − an)2 ≥ 3
na2i .
We will the Cauchy Schwarz Inequality to prove it. Indeed, from the CauchySchwarz Inequality, we have that for all m0,m1,m2, . . . ,mn−1,mn ∈ R (m0 =mn = 0),[
n−1∑k=1
(ak − ak+1)2
](n∑k=0
m2k
)=
[n−1∑k=1
(ak − ak+1)2
](n−1∑k=1
m2k
)
≥
[n−1∑k=1
mk(ak − ak+1)
]2
=
[n∑k=1
(mk −mk−1)ak
]2.
Now, choosing mk = k for k = 0, . . . , i − 1 and mk = k − n for k = i, . . . , n,we have
n∑k=0
m2k =
i−1∑k=0
k2 +n∑k=i
(k − n)2,
and
n∑k=1
(mk −mk−1)ai =i−1∑k=1
(mk −mk−1)ak + (mi −mi−1)ai +n∑
k=i+1
(mk −mk−1)ak
=
i−1∑k=1
ak + (1− n)ai +
n∑k=i+1
ak = −nai.
So, from the above inequality, we get[n−1∑k=1
(ak − ak+1)2
][i−1∑k=0
k2 +
n∑k=i
(k − n)2
]≥ n2a2i .
On the other hand, it is clear that
i−1∑k=0
k2+n∑k=i
(k−n)2 =i(i− 1)(2i− 1) + (n− i)(n− i+ 1)(2n− 2i+ 1)
6≤ n3
3.
Combining this with the above inequality, we get the result.?F?
06.4. Let a, b, c be the sidelengths of a triangle. Prove that
√b+ c− a√
b+√c−√a
+
√c+ a− b
√c+√a−√b
+
√a+ b− c
√a+√b−√c≤ 3.
417
(IMO Shortlist 2006)
First solution: Since a, b, c are the sidelengths of a triangle, the numbersx =
√a, y =
√b, z =
√c are also the sidelengths of a triangle (note that the
numbers −x+y+z, x−y+z and x+y−z are positive). In this case, accordingto the Cauchy Schwarz Inequality,(∑√
−x2 + y2 + z2
−x+ y + z
)2
≤ 3∑ −x2 + y2 + z2
(−x+ y + z)2,
and thus, it is sufficient to prove that∑ −x2 + y2 + z2
(−x+ y + z)2≤ 3.
Moving all into one side and distributing the 3 to each of the fractions equally,the last inequality can be rewritten as∑[
1− y2 + z2 − x2
(y + z − x)2
]≥ 0,
or equivalently,
2∑ 1
(y + z − x)2(x− y)(x− z) ≥ 0.
Without loss of generality, we can assume that x ≥ y ≥ z, then 0 < y+z−x <z + x− y, from which it follows that
1
(y + z − x)2≥ 1
(z + x− y)2.
Therefore, with notice that (x− y)(x− z) ≥ 0, we get
1
(y + z − x)2(x− y)(x− z) +
1
(z + x− y)2(y − z)(y − x) ≥
≥ 1
(z + x− y)2(x− y)(x− z) +
1
(z + x− y)2(y − z)(y − x)
=(x− y)2
(z + x− y)2≥ 0.
On the other hand, it is clear that1
(x+ y − z)2(z − x)(z − y) ≥ 0, so
∑ 1
(y + z − x)2(x− y)(x− z) ≥ 0,
and it completes our proof. Note that the equality holds if and only if a =b = c.
Second solution: Since the inequality is symmetric in the three variables,we may assume that a ≥ b ≥ c. We claim that
√a+ b− c
√a+√b−√c≤ 1,
418
and √b+ c− a√
b+√c−√a
+
√c+ a− b
√c+√a−√b≤ 2.
It is clear that the denominators are positive. So, the first inequality is equiv-alent to √
a+√b ≥√a+ b− c+
√c.
Squaring both sides, we can rewrite it as(√a+√b)2≥(√
a+ b− c+√c)2,
or equivalently, √ab ≥
√c(a+ b− c).
This inequality follows immediately from the inequality (a − c)(b − c) ≥ 0.Now, we prove the second inequality. Setting p =
√a+√b and q =
√a−√b,
we obtain a− b = pq and p ≥ 2√c. It now becomes
√c− pq√c− q
+
√c+ pq√c+ q
≤ 2.
We now apply the Cauchy Schwarz Inequality to deduce(√c− pq√c− q
+
√c+ pq√c+ q
)2
≤(c− pq√c− q
+c+ pq√c+ q
)(1√c− q
+1√c+ q
)=
2(c√c− pq2
)c− q2
· 2√c
c− q2=
4(c2 −
√cpq2
)(c− q2)2
≤4(c2 − 2cq2
)(c− q2)2
≤4(c2 − 2cq2 + q4
)(c− q2)2
= 4.
?F?
06.5. Determine the least real number M such that the inequality∣∣ab(a2 − b2) + bc(b2 − c2) + ca(c2 − a2)∣∣ ≤M(a2 + b2 + c2)2
holds for all real numbers a, b, c.(IMO 2006)
Solution: The given inequality can be rewritten as
|(a− b)(b− c)(c− a)(a+ b+ c)| ≤M(a2 + b2 + c2)2.
Now, let a = 1− 3√2, b = 1, c = 1 +
3√2
, we get M ≥ 9√
2
32. We will show that
the constant M =9√
2
32works. Indeed, put x = a− b, y = b− c, z = c− a and
s = a+ b+ c, then x+ y + z = 0, and it becomes
|xyzs| ≤ M
9(x2 + y2 + z2 + s2)2.
419
Since x + y + z = 0, there exist two numbers, supposed x and y, having thesame sign. Now, denote x + y = 2m, then z = −2m. Applying the AM-GMInequality with noting that xy ≥ 0, we have
|xyzs| = xy|zs| ≤(x+ y
2
)2
|zs| =∣∣2m3s
∣∣ .On the other hand, from the Cauchy Schwarz Inequality and the AM-GMInequality, we deduce that
(x2 + y2 + z2 + s2)2 ≥[
(x+ y)2
2+ z2 + s2
]2= (2m2 + 2m2 + 2m2 + s2)2
≥(
44√
8m6s2)2
= 32√
2∣∣m3s
∣∣ .Combining these two inequalities, we get
|xyzs| ≤ 1
16√
2(x2 + y2 + z2 + s2)2 =
M
9(x2 + y2 + z2 + s2)2.
We have equality if and only if x = y and 2m2 = s2, i.e. when 2b = a + cand (c − a)2 = 18b2. There are many triples (a, b, c) satisfying this system of
equations (for example, we can take a = 1− 3√2, b = 1, c = 1 +
3√2
as above).
From now, we conclude that the minimum value of M is9√
2
32.
?F?
06.6. Let x1, x2, x3, y1, y2, y3, z1, z2, z3 be positive real numbers. Find themaximum value of real number A if
M = (x31 + x32 + x33 + 1)(y31 + y32 + y33 + 1)(z31 + z32 + z33 + 1),
and
N = A(x1 + y1 + z1)(x2 + y2 + z2)(x3 + y3 + z3),
then M ≥ N always holds, and find the condition that the equality holds.
(Japan 2006)
Solution: Let x1 = x2 = x3 = y1 = y2 = y3 = z1 = z2 = z3 =13√
6, we get
A ≤ 3
4. From this, if we can show that the constant A =
3
4works, then we can
also conclude that it is value we need to find. However, this constant indeedworks. It suffices to prove
(x31 + x32 + x33 + 1)(y31 + y32 + y33 + 1)(z31 + z32 + z33 + 1)
(x1 + y1 + z1)(x2 + y2 + z2)(x3 + y3 + z3)≥ 3
4.
Put a =x1 + x2 + x3
3, b =
y1 + y2 + y33
and c =z1 + z2 + z3
3. Using the
420
Holder’s Inequality, we have
(x31 + x32 + x33 + 1)(y31 + y32 + y33 + 1)(z31 + z32 + z33 + 1) ≥≥ (3a3 + 1)(3b3 + 1)(3c3 + 1)
=
(3a3 +
1
2+
1
2
)(1
2+ 3b3 +
1
2
)(1
2+
1
2+ 3c3
)≥
(3
√3a3 · 1
2· 1
2+
3
√1
2· 3b3 · 1
2+
3
√1
2· 1
2· 3c3
)3
=3
4(a+ b+ c)3.
On the other hand, from the AM-GM Inequality, we get
(x1 + y1 + z1)(x2 + y2 + z2)(x3 + y3 + z3) ≤
≤ [(x1 + y1 + z1) + (x2 + y2 + z2) + (x3 + y3 + z3)]3
27
=[(x1 + x2 + x3) + (y1 + y2 + y3) + (z1 + z2 + z3)]
3
27= (a+ b+ c)3.
Combining these two inequalities, we deduce that
(x31 + x32 + x33 + 1)(y31 + y32 + y33 + 1)(z31 + z32 + z33 + 1)
(x1 + y1 + z1)(x2 + y2 + z2)(x3 + y3 + z3)≥ 3
4,
as desired. And so, we conclude that the maximum value of A is3
4.
?F?
06.7. Let a, b, c, d be real numbers with sum 0. Prove the inequality
(ab+ ac+ ad+ bc+ bd+ cd)2 + 12 ≥ 6(abc+ abd+ acd+ bcd).
(Kazakhstan 2006)
First solution: Replacing d = −a− b− c, our inequality becomes
(a2 + b2 + c2 + ab+ bc+ ca)2 + 12 + 6(a+ b+ c)(ab+ bc+ ca) ≥ 6abc,
or equivalently,
(a2 + b2 + c2 + ab+ bc+ ca)2 + 12 + 6(a+ b)(b+ c)(c+ a) ≥ 0.
This rewrites as
1
4
[(a+ b)2 + (b+ c)2 + (c+ a)2
]2+ 12 + 6(a+ b)(b+ c)(c+ a) ≥ 0.
Setting x =b+ c
2y =
c+ a
2, z =
a+ b
2, then the previous inequality goes into
(x2 + y2 + z2)2 + 48 + 24xyz ≥ 0.
421
Now, by the AM-GM Inequality, we have
(x2 + y2 + z2)2 ≥ 9 |xyz|4/3 .
and because 24xyz ≥ −24 |xyz|, it suffices to prove that
9t4 + 48− 24t3 ≥ 0,
where t = |xyz|1/3. This is true, since
9t4 + 48− 24t3 = 3(3t2 + 4t+ 4)(t− 2)2 ≥ 0.
Secdon solution: From the Rolle’s theorem, we know that there exist realnumbers x, y, z such that
x+ y + z =3
4(a+ b+ c+ d) = 0
ab+ ac+ ad+ bc+ bd+ cd = 2(xy + yz + zx)abc+ abd+ acd+ bcd = 4xyz.
Therefore, the inequality in question is equivalent to
(xy + yz + zx)2 + 3 ≥ 6xyz.
Without loss of generality, assume now that z = min {x, y, z}; then sincex+ y + z = 0, we have z = −x− y ≤ 0, and so our inequality becomes
(x2 + xy + y2)2 + 3 + 6xy(x+ y) ≥ 0.
Let us now put s = x + y ≥ 0, p = xy. In this case, the previous inequalityrewrites as
(s2 − p)2 + 3 + 6sp ≥ 0,
orf(p) = p2 + 2(3s− s2)p+ s4 + 3 ≥ 0.
We have now that
∆f = (3s− s2)2 − s4 − 3 = −3(2s+ 1)(s− 1)2 ≤ 0,
and we thus conclude that f(p) ≥ 0.?F?
06.8. Let a, b, c be sides of the triangle. Prove that
a2(b
c− 1
)+ b2
( ca− 1)
+ c2(ab− 1)≥ 0.
(Moldova 2006)
First solution: The original inequality is equivalent to
a2b
c+b2c
a+c2a
b≥ a2 + b2 + c2.
422
By the Cauchy Schwarz Inequality, we have
a2b
c+b2c
a+c2a
b=a4b2
a2bc+b4c2
b2ca+c4a2
c2ab≥ (a2b+ b2c+ c2a)2
abc(a+ b+ c).
Therefore, it suffices to prove that
(a2b+ b2c+ c2a)2 ≥ abc(a+ b+ c)(a2 + b2 + c2).
Due to the cyclicity, we can suppose without loss of generality that b is thenumber between a and c, hence (a− b)(b− c) ≥ 0. Now, applying the AM-GMInequality, we obtain
4abc(a+ b+ c)(a2 + b2 + c2) ≤ [ac(a+ b+ c) + b(a2 + b2 + c2)]2.
On the other hand,
2(a2b+ b2c+ c2a)− ac(a+ b+ c)− b(a2 + b2 + c2) = (a− b)(b− c)(a+ b− c),
which is obviously nonnegative. Therefore the inequality
(a2b+ b2c+ c2a)2 ≥ abc(a+ b+ c)(a2 + b2 + c2)
is valid and it completes our proof. The equality holds if and only if a = b = c.
Second solution: Using the substitutions a =1
x, b =
1
yand c =
1
z, the
inequality becomes
1
x2
(z
y− 1
)+
1
y2
(xz− 1)
+1
z2
(yx− 1)≥ 0,
oryz2(z − y) + zx2(x− z) + xy2(y − x) ≥ 0.
Without loss of generality, we can assume that a = min{a, b, c}, and hencex = max{a, b, c}. Denoting the left hand side of the last inequality as E(x, y, z),we will show that
E(x, y, z) ≥ E(y, y, z) ≥ 0.
We have
E(x, y, z)− E(y, y, z) = z(x3 − y3)− z2(x2 − y2) + y3(x− y)− y2(x2 − y2)= (x− y)(x− z)(xz + yz − y2).
Since (x− y)(x− z) ≥ 0 and
xz + yz − y2 ≥ y(2z − y) =2b− cb2c
=(b− a) + (a+ b− c)
b2c> 0,
it follows that E(x, y, z)− E(y, y, z) ≥ 0. On the other hand, we have
E(y, y, z) = yz(y − z)2 ≥ 0.
423
So our statement is proved.
Third solution: Writing the inequality as E(a, b, c) ≥ 0, where
E(a, b, c) = a3b2 + b3c2 + c3a2 − abc(a2 + b2 + c2).
Since 2E(a, b, c) =∑
a3(b−c)2−∑
a2(b3−c3) and∑
a2(b3−c3) =∑
a2(b−c)3, we have directly
2E(a, b, c) =∑
a2(b− c)2(a− b+ c) ≥ 0.
?F?
06.9. Let a, b, c be positive real numbers with ab+ bc+ ca = abc. Prove that
a4 + b4
ab(a3 + b3)+
b4 + c4
bc(b3 + c3)+
c4 + a4
ca(c3 + a3)≥ 1.
(Poland 2006)
Solution: We first notice that the constraint can be written as
1
a+
1
b+
1
c= 1.
It is now enough to establish the auxiliary inequality
x4 + y4
xy(x3 + y3)≥ 1
2
(1
x+
1
y
),
or2(x4 + y4
)≥(x3 + y3
)(x+ y) ,
where x, y > 0. However, we obtain
2(x4 + y4
)−(x3 + y3
)(x+ y) = x4 + y4−x3y−xy3 =
(x3 − y3
)(x− y) ≥ 0.
Therefore, we have established the inequalityx4 + y4
xy(x3 + y3)≥ 1
2
(1
x+
1
y
).
And using it, we can get the result. Note that the equality holds if and onlyif a = b = c = 3.
?F?
06.10. Find the maximum value of
(x3 + 1)(y3 + 1),
for all real numbers x, y, satisfying the condition that x+ y = 1.(Romania 2006)
Solution: Put xy = t, as x + y = 1 we get (x3 + 1)(y3 + 1) = t3 − 3t + 2.
From x + y = 1, we obtain t = xy ≤(x+ y
2
)2
=1
4. It is easy to prove that
424
t3 − 3t + 2 ≤ 4 for t ≤ 1
4, with equality if and only if t = −1. We infer that
(x3 + 1)(y3 + 1) ≤ 4 for x, y ∈ R with x+ y = 1 and (ψ3 + 1)(−1/ψ3 + 1) = 4,where ψ is one of the roots of z2 − z − 1 = 0.
?F?
06.11. Let a, b, c be three positive real numbers with sum 3. Prove that
1
a2+
1
b2+
1
c2≥ a2 + b2 + c2.
(Romania 2006)
First solution: From the AM-GM Inequality, we get
1
a2+
1
b2+
1
c2≥ 1
ab+
1
bc+
1
ca=
3
abc,
hence it suffices to prove that
3 ≥ abc(a2 + b2 + c2).
Again, the AM-GM Inequality gives us that
(ab+ bc+ ca)2 ≥ 3abc(a+ b+ c) = 9abc,
and therefore, we are left to show that
(a2 + b2 + c2)(ab+ bc+ ca)2 ≤ 27,
which is obviously valid, since
(a2 + b2 + c2)(ab+ bc+ ca)2 ≤[a2 + b2 + c2 + 2(ab+ bc+ ca)
3
]3=
(a+ b+ c)6
27= 27.
Note that the equality holds iff a = b = c = 1.
Second solution: Write the inequality in the form∑(1
a2− a2 + 4a− 4
)≥ 0,
which is equivalent to ∑ (a− 1)2(1 + 2a− a2)a2
≥ 0.
Without loss of generality, we may assume that a ≥ b ≥ c. We have to considertwo casesCase 1. a ≤ 1 +
√2. Since c ≤ b ≤ a ≤ 1 +
√2, we have 1 + 2a − a2 ≥
0, 1 + 2b − b2 ≥ 0 and 1 + 2c − c2 ≥ 0. Therefore, the inequality is obviouslytrue.
425
Case 2. a > 1 +√
2. Since b+ c = 3− a < 2−√
2 <2
3, we have
bc ≤ (b+ c)2
4<
1
9,
and hence
1
a2+
1
b2+
1
c2>
1
b2+
1
c2≥ 2
bc> 18 > (a+ b+ c)2 > a2 + b2 + c2.
Remark: Actually, we can show more:For any two positive integers n, p satisfying, n ≥ 4 and p ≥ 4, the propositionP(n, p) is false:
n∑i=1
1
xpi≥
n∑i=1
xpi for xi ∈ R, xi > 0, i = 1, . . . , n ,n∑i=1
xi = n.
This variation was considered as a problem in the Romanian IMO Team Se-lection Tests from 2007. We continue with its solutionSolution: Notice first that it is enough to find a set of values xi for n = 4 suchthat
E =n∑i=1
1
xpi−
n∑i=1
xpi < 0,
as then for any n > 4 we can extend this set of values by taking the extran− 4 ones to be equal to 1.Now that we have reduced it to the case n = 4, it makes sense to look for”simple” cases:• some xi very small - it yields E > 0, no good;• all xi equal - it yields common value 1, for which E = 0, no good;• let’s then try taking the smallest three xi equal to some value 0 < x < 1,the fourth one, denoted by y, will be 1 < y < 4, y = 4− 3x.Then,
E =3
xp+
1
yp− 3xp − yp =
1
xp
[3 +
(x
y
)p− 3x2p − (xy)p
]=
1
xp[3− (xy)p] +
1
yp[1− 3(xy)p].
It seems natural now to look for the maximum possible value for xy, it is not
difficult to see that xy =1
3· (3x) · (4 − 3x) ≤ 1
3· (2)2 =
4
3(by the AM-GM
Inequality), with equality for 3x = 4 − 3x i.e. x =2
3and y = 2. Then, as
4
3> 1 and p ≥ 4, we have
(4
3
)p≥(
4
3
)4
=256
81> 3, hence E < 0 for the set
of values (2
3,2
3,2
3, 2 and 1 repeated n− 4 times).
You might now wonder what can we say about the propositions P(4, 3) andP(3, 4). As a matter of fact, they are true. However, we will omit their proofhere.
426
?F?
06.12. Consider real numbers a, b, c contained in the interval
[1
2, 1
]. Prove
that
2 ≤ a+ b
1 + c+b+ c
1 + a+c+ a
1 + b≤ 3.
(Romania 2006)
Solution: We begin by proving the left hand side of the inequality. Since
a, b ≥ 1
2, we have a+ b ≥ 1, and thus
a+ b
1 + c≥ a+ b
a+ b+ c.
By adding the other two similar relations to the inequality from above, weobtain
2 =(a+ b) + (b+ c) + (c+ a)
a+ b+ c≤ a+ b
1 + c+b+ c
1 + a+c+ a
1 + b.
For the second inequality, note that the considered expression can be writtenas ∑(
a
1 + c+
c
1 + a
).
As a, c ≤ 1, we have
a
1 + c≤ a
a+ c, and
c
1 + a≤ c
c+ a,
and soa
1 + c+
c
1 + a≤ a
a+ c+
c
c+ a= 1.
The other two cyclic relations occur (again) similarly. Summing up the allthree, we get the desired result.
?F?
06.13. Let a, b be positive real numbers. Determine the largest constant Msuch that for all k ∈ [0, π] , we have
1
ka+ b+
1
kb+ a≥ M
a+ b.
(Thailand 2006)
Solution: Let a = b, k = π, we get M ≤ 4
π + 1. We claim that
4
π + 1is our
answer. Indeed, from the Cauchy Schwarz Inequality, we have
1
ka+ b+
1
kb+ a≥ 4
ka+ b+ kb+ a=
4
k + 1· 1
a+ b≥ 4
π + 1· 1
a+ b.
This proves that the inequality holds for M =4
π + 1. And since the equality
can occur, we conclude that the maximum value of M is4
π + 1.
427
?F?
06.14. If x, y, z are positive numbers satisfying the condition xy+yz+zx = 1,show that
27
4(x+ y)(y + z)(z + x) ≥
(√x+ y +
√y + z +
√z + x
)2 ≥ 6√
3.
(Turkey 2006)
Solution: By the Cauchy Schwarz Inequality,(√x+ y +
√y + z +
√z + x
)2 ≤ 3 (x+ y + y + z + z + x) = 6(x+ y + z),
and hence, in order to prove the left inequality, it suffices to show that
(x+ y)(y + z)(z + x) ≥ 8
9(x+ y + z).
For this, we proceed as follows
(x+ y)(y + z)(z + x) =
= (x+ y + z)(xy + yz + zx)− xyz
≥ (x+ y + z)(xy + yz + zx)− 1
9(x+ y + z)(xy + yz + zx)
=8
9(x+ y + z)(xy + yz + zx) =
8
9(x+ y + z).
For the right hand side of the inequality, we can make use of the AM-GMInequality in combination with the Minkowsky’s Inequality(√
x+ y +√y + z +
√z + x
)2 ≥≥ 3
[√(x+ y)(x+ z) +
√(y + z)(y + x) +
√(z + x)(z + y)
]= 3
(√x2 + 1 +
√y2 + 1 +
√z2 + 1
)≥ 3√
(x+ y + z)2 + (1 + 1 + 1)2
≥ 3√
3(xy + yz + zx) + 9 = 6√
3.
?F?
06.15. Let a, b, c be real numbers. Prove that the following inequality holds∑√(a2 − ab+ b2)(b2 − bc+ c2) ≥ a2 + b2 + c2.
(VMEO 2006)
Solution: By the Cauchy Schwarz Inequality, we have
(a2 − ab+ b2)(b2 − bc+ c2) =
[(b− a
2
)2+
3a2
4
] [(b− c
2
)2+
3c2
4
]≥[(b− a
2
)(b− c
2
)+
3ac
4
]2.
428
It follows that√(a2 − ab+ b2)(b2 − bc+ c2) ≥
∣∣∣∣(b− a
2
)(b− c
2
)+
3ac
4
∣∣∣∣≥(b− a
2
)(b− c
2
)+
3ac
4
= b2 +1
2(2ca− ab− bc).
Therefore∑√(a2 − ab+ b2)(b2 − bc+ c2) ≥
∑[b2 +
1
2(2ca− ab− bc)
]= a2 + b2 + c2.
Our proof is completed. It is easy to see that the equality holds if and only if(a, b, c) equals (t, t, t), or (t, 0, 0), or (0, t, 0), or (0, 0, t), where t is an arbitraryreal numbers.
?F?
07.1. Let x, y, z be positive real numbers such that√x+√y+√z = 1. Prove
that the following inequality holds
x2 + yz√2x2(y + z)
+y2 + zx√2y2(z + x)
+z2 + xy√2z2(x+ y)
≥ 1.
(APMO 2007)
Solution: According to the Cauchy Schwarz Inequality and the well-knowninequality (a+ b+ c)2 ≥ 3(ab+ bc+ ca), we have∑ yz√
2x2(y + z)=
1√2xyz
∑ y2z2√y + z
≥ 1√2· (xy + yz + zx)2
xyz(√y + z +
√z + x+
√x+ y
)≥ 1
2√
3· (xy + yz + zx)2
xyz√x+ y + z
≥ 1
2√
3· 3xyz(x+ y + z)
xyz√x+ y + z
=1
2
√3(x+ y + z) ≥ 1
2
(√x+√y +√z)
=1
2,
and∑ x2√2x2(y + z)
=1√2
∑ x√y + z
≥ 1√2· (x+ y + z)2
x√y + z + y
√z + x+ z
√x+ y
≥ 1√2· (x+ y + z)2√
(x+ y + z) [x(y + z) + y(z + x) + z(x+ y)]
=1
2
√(x+ y + z)3
xy + yz + zx≥ 1
2
√(x+ y + z) · 3(xy + yz + zx)
xy + yz + zx
=1
2
√3(x+ y + z) ≥ 1
2
(√x+√y +√z)
=1
2.
Adding up these two inequalities, we get the desired result. It is easy to see
that the equality holds if and only if x = y = z =1
9.
429
?F?
07.2. If a, b, c ∈ R such that abc = 1, then
a2+b2+c2+1
a2+
1
b2+
1
c2+2
(a+ b+ c+
1
a+
1
b+
1
c
)≥ 6+2
(b+ c
a+c+ a
b+a+ b
c
).
(Brazil 2007)
Solution: We have
a2 + b2 + c2 + 2
(1
a+
1
b+
1
c
)= a2 + b2 + c2 + 2(ab+ bc+ ca)
= (a+ b+ c)2,
1
a2+
1
b2+
1
c2+ 2 (a+ b+ c) = a2b2 + b2c2 + c2a2 + 2abc(a+ b+ c)
= (ab+ bc+ ca)2,
and
6 + 2
(b+ c
a+c+ a
b+a+ b
c
)=
2(a+ b+ c)(ab+ bc+ ca)
abc
= 2(a+ b+ c)(ab+ bc+ ca).
Therefore, our inequality is equivalent to
(a+ b+ c)2 + (ab+ bc+ ca)2 ≥ 2(a+ b+ c)(ab+ bc+ ca),
which is obviously true by the AM-GM Inequality.?F?
07.3. Given an integer n ≥ 2, find the largest constant C(n) for which theinequality
n∑i=1
xi ≥ C(n)∑
1≤j<i≤n
(2xixj +
√xixj
)holds for all real numbers xi ∈ (0, 1) satisfying (1 − xi)(1 − xj) ≥
1
4for
1 ≤ j < i ≤ n.
(Bulgaria 2007)
Solution: Let us see what happens if x1 = x2 = · · · = xn =1
2. We have that
n
2≥ C(n)
(2 · 1
4· n(n− 1)
2+
1
2· n(n− 1)
2
),
and thus,
C(n) ≤ 1
n− 1.
430
We will now show that this is the value we are looking for. That is, we areleft to prove that
(n− 1)n∑i=1
xi ≥∑
1≤j<i≤n
(2xixj +
√xixj
).
By the AM-GM Inequality, we have
∑1≤j<i≤n
√xixj ≤
1
2
∑1≤j<i≤n
(xi + xj) =n− 1
2
n∑i=1
xi,
and furthermore,
2∑
1≤j<i≤nxixj=
(n∑i=1
xi
)2
−n∑i=1
x2i
≤
(n∑i=1
xi
)2
− 1
n
(n∑i=1
xi
)2
=n− 1
n
(n∑i=1
xi
)2
.
Hence, the problem reduces to
(n− 1)
n∑i=1
xi ≥n− 1
n
(n∑i=1
xi
)2
+n− 1
2
n∑i=1
xi,
or equivalently,n∑i=1
xi ≤n
2.
For this, note that the AM-GM Inequality is giving us
2− xj − xi = (1− xj) + (1− xi) ≥ 2√
(1− xj)(1− xi) ≥ 1,
for all integers i, j satisfying 1 ≤ j < i ≤ n. This yields
1− xj − xi ≥ 0,
and therefore ∑1≤j<i≤n
(1− xi − xj) ≥ 0
from which we now can conclude that
n∑i=1
xi ≤n
2.
This proves our inequality, and therefore maxC(n) =1
n− 1.
?F?
431
07.4. Let a, b, c be the sidelengths of a triangle such that a + b + c = 3.Determine the minimum value of
a2 + b2 + c2 +4abc
3.
(China 2007)
Solution: Because
[(a− 1)(b− 1)][(b− 1)(c− 1)][(c− 1)(a− 1)] = (a− 1)2(b− 1)2(c− 1)2 ≥ 0,
we see that at least one of the numbers (a−1)(b−1), (b−1)(c−1), (c−1)(a−1)must be nonnegative. Due to symmetry, we may assume that (a−1)(b−1) ≥ 0,or ab ≥ a + b − 1 = 2 − c, and hence abc ≥ c(2 − c). Using this inequality incombination with the Cauchy Schwarz Inequality, we get
a2 + b2 + c2 +4abc
3≥ a2 + b2 + c2 +
4
3c(2− c) ≥ 1
2(a+ b)2 + c2 +
4
3c(2− c)
=1
2(3− c)2 + c2 +
4
3c(2− c) =
1
6
[(c− 1)2 + 26
]≥ 13
3.
We have equality when a = b = c = 1, therefore the searched minimum is13
3.
?F?
07.5. Let α, β be acute angles. Determine the maximal value of(1−√
tanα tanβ)2
cotα+ cotβ.
(China 2007)
Solution: Denote with P the desired expression. By the AM-GM Inequality,we have
P ≤(1−√
tanα tanβ)2
2√
cotα cotβ=
1
2
√tanα tanβ
(1−
√tanα tanβ
)2=
1
4· 2√
tanα tanβ ·(
1−√
tanα tanβ)·(
1−√
tanα tanβ)
≤ 1
4
[2√
tanα tanβ +(1−√
tanα tanβ)
+(1−√
tanα tanβ)
3
]3=
2
27.
We have equality when α = β = arctan1
3, therefore the searched maximum is
2
27.
?F?
07.6. Let a, b, c be positive real numbers such that abc = 1. Prove that for allk ≥ 2, we have
ak
a+ b+
bk
b+ c+
ck
c+ a≥ 3
2.
432
(China 2007)
Solution: By the Cauchy Schwarz Inequality, we have
ak
a+ b+
bk
b+ c+
ck
c+ a≥
(a
k+12 + b
k+12 + c
k+12
)2a(a+ b) + b(b+ c) + c(c+ a)
=
(a
k+12 + b
k+12 + c
k+12
)2a2 + b2 + c2 + ab+ bc+ ca
.
So, it suffices to prove that
2(a
k+12 + b
k+12 + c
k+12
)2≥ 3(a2 + b2 + c2 + ab+ bc+ ca),
or equivalently,
2(ak+1+bk+1+ck+1)+4
(1
ak+12
+1
bk+12
+1
ck+12
)≥ 3
(a2 + b2 + c2 +
1
a+
1
b+
1
c
).
Now, using the Chebyshev’s Inequality, we have
1
ak+12
+1
bk+12
+1
ck+12
≥ 1
3
(1
ak−12
+1
bk−12
+1
ck−12
)(1
a+
1
b+
1
c
)≥ 1
3· 3 3
√1
ak−12 b
k−12 c
k−12
·(
1
a+
1
b+
1
c
)=
1
a+
1
b+
1
c,
and
ak+1 + bk+1 + ck+1 ≥ 1
3(ak−2 + bk−2 + ck−2)(a3 + b3 + c3)
≥ 1
3· 3 3√ak−2bk−2ck−2 · (a3 + b3 + c3)
= a3 + b3 + c3.
According to these two inequalities, we see that it is enough to check that
2(a3 + b3 + c3) + 4
(1
a+
1
b+
1
c
)≥ 3
(a2 + b2 + c2 +
1
a+
1
b+
1
c
),
which is equivalent to ∑(2a3 +
1
a− 3a2
)≥ 0.
By the AM-GM Inequality, we have 2a3 − 3a2 + 1 ≥ 0. It follows that∑(2a3 +
1
a− 3a2
)≥∑(
1
a− 1
)=
1
a+
1
b+
1
c− 3 ≥ 3
3√abc− 3 = 0,
using the AM-GM Inequality again. The equality holds if and only if a = b =c = 1.
433
?F?
07.7. Let a, b, c > 0 such that a+ b+ c = 1. Prove that
a2
b+b2
c+c2
a≥ 3(a2 + b2 + c2).
(Croatia 2007)
First solution: Applying the AM-GM Inequality, we have1
b+ 9b ≥ 6. It
follows thata2
b≥ 3a2(2− 3b). According to this inequality, it suffices to prove
that
a2(2− 3b) + b2(2− 3c) + c2(2− 3a) ≥ a2 + b2 + c2,
which is equivalent to
a2 + b2 + c2 ≥ 3(a2b+ b2c+ c2a),
or
(a2 + b2 + c2)(a+ b+ c) ≥ 3(a2b+ b2c+ c2a).
After some simple computations, we can rewrite it as
a(a− b)2 + b(b− c)2 + c(c− a)2 ≥ 0,
which is obviously true. It is easy to see that the equality holds if and only if
a = b = c =1
3.
Second solution: From the Cauchy Schwarz Inequality, we have
a2
b+b2
c+c2
a≥ (a2 + b2 + c2)2
a2b+ b2c+ c2a.
It suffices to prove that
(a2 + b2 + c2)2
a2b+ b2c+ c2a≥ 3(a2 + b2 + c2),
or equivalently,
a2 + b2 + c2 ≥ 3(a2b+ b2c+ c2a).
This inequality has been proved in the above solution.?F?
07.8. Let a, b, c, d be positive real numbers such that a+ b+ c+ d = 1. Provethat
6(a3 + b3 + c3 + d3) ≥ a2 + b2 + c2 + d2 +1
8.
(France 2007)
434
Solution: According to the Chebyshev’s Inequality, we have
a3 + b3 + c3 + d3 ≥ 1
4(a+ b+ c+ d)(a2 + b2 + c2 + d2)
=1
4(a2 + b2 + c2 + d2).
It suffices to show that
3
2(a2 + b2 + c2 + d2) ≥ a2 + b2 + c2 + d2 +
1
8,
or equivalently,
a2 + b2 + c2 + d2 ≥ 1
4.
This is true since by the Cauchy Schwarz Inequality, we have
a2 + b2 + c2 + d2 ≥ 1
4(a+ b+ c+ d)2 =
1
4.
The equality holds if and only if a = b = c = d =1
4.
?F?
07.9. Let a, b, c be sides of a triangle, show that
(c+ a− b)4
a(a+ b− c)+
(a+ b− c)4
b(b+ c− a)+b(b+ c− a)
c(c+ a− b)≥ ab+ bc+ ca.
(Greece 2007)
First solution: Because a, b, c are the sidelengths of a triangle, we see thata+ b− c, b+ c− a, a+ b− c are positive real numbers. And thus, the CauchySchwarz Inequality yields
(c+ a− b)4
a(a+ b− c)+
(a+ b− c)4
b(b+ c− a)+
(b+ c− a)4
c(c+ a− b)≥
≥ [(c+ a− b)2 + (a+ b− c)2 + (b+ c− a)2]2
a(a+ b− c) + b(b+ c− a) + c(c+ a− b).
On the other hand, it is easy to verify that a(a+b−c)+b(b+c−a)+c(c+a−b) =a2 + b2 + c2, and
(c+ a− b)2+(a+ b− c)2 + (b+ c− a)2 =
= (a2 + b2 + c2) + 2(a2 + b2 + c2 − ab− bc− ca)
≥ a2 + b2 + c2.
Therefore, from the above Cauchy Schwarz step, we can get a stronger esti-mation than the original statement is
(c+ a− b)4
a(a+ b− c)+
(a+ b− c)4
b(b+ c− a)+
(b+ c− a)4
c(c+ a− b)≥ a2 + b2 + c2 ≥ ab+ bc+ ca.
435
The equality holds if and only if a = b = c.
Second solution: By the AM-GM Inequality, we have
(c+ a− b)4
a(a+ b− c)+ a(a+ b− c) ≥ 2(c+ a− b)2.
It follows that∑ (c+ a− b)4
a(a+ b− c)≥∑
[2(c+ a− b)2 − a(a+ b− c)].
On the other hand, it is easy to verify that∑
[2(c+ a− b)2 − a(a+ b− c)] ≥ab+ bc+ ca, (as it is equivalent to a2 + b2 + c2 ≥ ab+ bc+ ca after expanding),we can obtain ∑ (c+ a− b)4
a(a+ b− c)≥ ab+ bc+ ca.
?F?
07.10. Let a, b, c, d be real numbers such that
a2 ≤ 1, a2 + b2 ≤ 5, a2 + b2 + c2 ≤ 14, a2 + b2 + c2 + d2 ≤ 30.
Prove that
a+ b+ c+ d ≤ 10.
(Hungary-Isarel 2007)
Solution: Applying the Cauchy Schwarz Inequality, we have
(a+ b+ c+ d)2 =
(1 · a+
√2 · b√
2+√
3 · c√3
+ 2 · d2
)2
≤ (1 + 2 + 3 + 4)
(a2 +
b2
2+c2
3+d2
4
)=
5
6
(12a2 + 6b2 + 4c2 + 3d2
).
On the other hand, from the given hypothesis, we find that
12a2 + 6b2 + 4c2 + 3d2 =
= 6a2 + 2(a2 + b2) + (a2 + b2 + c2) + 3(a2 + b2 + c2 + d2)
≤ 6 · 1 + 2 · 5 + 14 + 3 · 30 = 120.
Using this in combination with the above inequality, we deduce that
(a+ b+ c+ d)2 ≤ 5
6· 120 = 100,
and hence, it follows that
a+ b+ c+ d ≤ |a+ b+ c+ d| ≤ 10,
436
as desired. It is easy to see that the equality holds if and only if (a, b, c, d) =(1, 2, 3, 4).
?F?
07.11. Let a1, a2, . . . , a100 be nonnegative eral numbers such that a21 + a22 +· · ·+ a2100 = 1. Prove that
a21a2 + a22a3 + · · ·+ a2100a1 <12
25.
(IMO Shortlist 2007)
Solution: According to the Cauchy Schwarz Inequality,
1
3[a1(a
2100 + 2a1a2)+a2(a
21 + 2a2a3) + · · ·+ a100(a
299 + 2a100a1)]
≤ 1
3
(100∑k=1
a2k
) 12
·
[100∑k=1
(a2k + 2ak+1ak+2)2
] 12
.
Thus, it is sufficient to show that
100∑k=1
(a2k + 2ak+1ak+2)2 ≤ 2.
Each term of this last sum can be seen as
a4k + 4a2k+1a2k+2 + 4a2k(ak+1 · ak+2) ≤ (a4k + 2a2ka
2k+1 + 2a2ka
2k+2) + 4a2k+1a
2k+2.
The required inequality now follows from
100∑k=1
(a4k + 2a2ka2k+1 + 2a2ka
2k+2) ≤
(100∑k=1
a2k
)2
= 1,
and
100∑k=1
a2ka2k+1 ≤ (a21 + a23 + · · ·+ a299) · (a22 + a24 + · · ·+ a2100)
≤ 1
4
(100∑k=1
a2k
)2
=1
4.
?F?
07.12. Let n be a positive integer, and let x and y be positive real numberssuch that xn + yn = 1. Prove that(
n∑k=1
1 + x2k
1 + x4k
)(n∑k=1
1 + y2k
1 + y4k
)<
1
(1− x)(1− y).
(IMO Shortlist 2007)
437
Solution: The inequality1 + t2
1 + t4<
1
tholds for all t ∈ (0, 1) because it is
equivalent to 0 < t4 − t3 − t + 1 = (1 − t)(1 − t3). Applying it to t = xk andsumming over k = 1, 2, . . . , n, we get
n∑k=1
1 + x2k
1 + x4k<
n∑k=1
1
xk=
xn − 1
xn(x− 1)=
yn
xn(1− x).
Similarly, we obtain that
n∑k=1
1 + x2k
1 + x4k<
xn
yn(1− y),
and therefore, by multiplying them up, yields(n∑k=1
1 + x2k
1 + x4k
)(n∑k=1
1 + y2k
1 + y4k
)<
1
(1− x)(1− y),
as desired.?F?
07.13. Real numbers a1, a2, . . . , an are given. For each i (1 ≤ i ≤ n) define
di = max {aj : 1 ≤ j ≤ i} −min {aj : i ≤ j ≤ n},
and letd = max {di : 1 ≤ i ≤ n}.
(a) Prove that for any real numbers x1 ≤ x2 ≤ · · · ≤ xn, we have
max {|xi − ai| : 1 ≤ i ≤ n} ≥ d
2.
(b) Show that there are real numbers x1 ≤ x2 ≤ · · · ≤ xn such that we haveequality in (a).
(IMO 2007)
Solution: (a) Let 1 ≤ p ≤ q ≤ r ≤ n be indices for which
d = dq, ap = max{aj : 1 ≤ j ≤ q}, ar = min{aj : q ≤ j ≤ n},
and thus d = ap − aq. (These indices are not necessarily unique.)For arbitrary real numbers x1 ≤ x2 ≤ · · · ≤ xn, consider just the two quantities|xp − ap| and |xr − ar|. Since
(ap − xp) + (xr − ar) = (ap − ar) + (xr − xp) ≥ ap − ar = d,
we have either ap − xp ≥d
2or xr − ar ≥
d
2. Hence,
max{|xi−ai| : 1 ≤ i ≤ n} ≥ max{|xp−ap|, |xr−ar|} ≥ max{xp−ap, xr−ar} ≥d
2.
438
(b) For each 1 ≤ i ≤ n, letMi = max {aj : 1 ≤ j ≤ i} andmi = min {aj : i ≤ j ≤ n}.
Set xi =mi +Mi
2. Clearly, mi ≤ ai ≤ Mi and both (mi) and (Mi) are non-
decreasing. Furthermore, from di = Mi −mi, we obtain that
−di2
=mi −Mi
2= xi −Mi ≤ xi − ai ≤ xi −mi =
Mi −mi
2=di2.
Therefore,
max {|xi − ai| : 1 ≤ i ≤ n} ≤ max
{di2
: 1 ≤ i ≤ n}
=d
2.
Since the opposite inequality has been proved in part (a), we must have equal-ity.
?F?
07.14. If x, y, z are positive real numbers, prove that the following inequalityholds
(x+ y + z)2(yz + zx+ xy)2 ≤ 3(y2 + yz + z2)(z2 + zx+ x2)(x2 + xy + y2).
(India 2007)
Solution: By the AM-GM Inequality, we have that
y2 + yz + z2 =3
4(y + z)2 +
1
4(y − z)2 ≥ 3
4(y + z)2,
hence it suffices to show that
81
64(y + z)2(z + x)2(x+ y)2 ≥ (x+ y + z)2(yz + zx+ xy)2.
This rewrites as
(x+ y)(y + z)(z + x) ≥ 8
9(x+ y + z)(xy + yz + zx),
which we have proved at the preceding problem.
Remark: Actually, we can show that the original inequality is valid for anyreal numbers x, y, z and they are not necessary to be nonnegative. The resultis followed from an interesting identity
3(x2 + xy+ y2)(y2 + yz + z2)(z2 + zx+ x2)− (x+ y+ z)2(xy+ yz + zx)2 =
=1
2(xy + yz + zx)2
∑(x− y)2 +
1
2(x+ y + z)2
∑(zx− yz)2.
?F?
07.15. Let a, b, c be distinct positive real numbers. Show that∣∣∣∣a+ b
a− b+b+ c
b− c+c+ a
c− a
∣∣∣∣ > 1.
439
(Iran 2007)
Solution: Observe that
a+ b
a− b· b+ c
b− c+b+ c
b− c· c+ a
c− a+c+ a
c− a· a+ b
a− b= −1.
From this, it follows that(a+ b
a− b+b+ c
b− c+c+ a
c− a
)2
=∑ (a+ b)2
(a− b)2+ 2
∑ a+ b
a− b· b+ c
b− c
=∑ (a+ b)2
(a− b)2− 2 =
∑[(a+ b)2
(a− b)2− 1
]+ 1
= 4∑ ab
(a− b)2+ 1 > 1.
Taking square root of each side of this inequality, we get the desired result.?F?
07.16. Find the largest constant T such that for all nonnegative real numbersa, b, c, d, e satisfying a+ b = c+ d+ e, we have√
a2 + b2 + c2 + d2 + e2 ≥ T(√
a+√b+√c+√d+√e)2.
(Iran 2007)
Solution: Let a = b =1
2, c = d = e =
1
3, then we get T ≤
√5
6
(√3−√
2)2.
Now, we will prove that the desired inequality holds for T =
√5
6
(√3−√
2)2
.
Indeed, by the Cauchy Schwarz Inequality, we have√a2 + b2 + c2 + d2 + e2 ≥
√(a+ b)2
2+
(c+ d+ e)2
3=
√5
6(a+ b),
and(√a+√b+√c+√d+√e)2
=
=
(14√
2· 4√
2a2 +14√
2· 4√
2b2 +14√
3· 4√
3c2 +14√
3· 4√
3d2 +14√
3· 4√
3e2)2
≤(
1√2
+1√2
+1√3
+1√3
+1√3
)(√2a+
√2b+
√3c+
√3d+
√3e)
=(√
2 +√
3)2
(a+ b).
It follows that
T(√
a+√b+√c+√d+√e)2≤ T
(√2 +√
3)2
(a+ b)
=
√5
6
(√3−√
2)2 (√
2 +√
3)2
(a+ b)
=
√5
6(a+ b) ≤
√a2 + b2 + c2 + d2 + e2,
440
as desired. So, for T =
√5
6
(√3−√
2)2
, the inequality holds and it attains
the equality for a = b =3
2c =
3
2d =
3
2e. This allows us to conclude that the
largest constant T satisfies the required question is T =
√5
6
(√3−√
2)2
.
?F?
07.17. Prove that for any positive real numbers a, b, c, we have
a+ b+ c
3≤√a2 + b2 + c2
3≤ 1
3
(bc
a+ca
b+ab
c
).
(Ireland 2007)
Solution: The left inequality is just the Cauchy Schwarz Inequality. So itis enough to prove the right inequality. Applying the well-known inequality
(x+ y + z)2 ≥ 3(xy + yz + zx) for the triple (x, y, z) =
(ab
c,bc
a,ca
b
), we get
(bc
a+ca
b+ab
c
)2
≥ 3
(bc
a· cab
+ca
b· abc
+ab
c· bca
)= 3(a2 + b2 + c2).
Taking square root of each side of this inequality, we get the desired inequality.It is easy to see that the equality (in both inequalities) holds if and only ifa = b = c.
?F?
07.18. Let n ≥ 2 be a given integer. Determine(a) the largest real cn such that
1
1 + a1+
1
1 + a2+ · · ·+ 1
1 + an≥ cn
holds for any positive numbers a1, a2, . . . , an with a1a2 · · · an = 1.(b) the largest real dn such that
1
1 + 2a1+
1
1 + 2a2+ · · ·+ 1
1 + 2an≥ dn
holds for any positive numbers a1, a2, . . . , an with a1a2 · · · an = 1.(Italy 2007)
Solution: (a) Let us see what happens if a1 = a2 = · · · = an−1 = x and
an =1
xn−1, where x is an arbitrary positive real number. In this case, the
inequality becomesn− 1
x+ 1+
xn−1
xn−1 + 1≥ cn,
and by setting x near to infinity, we get cn ≤ 1. Now, we are left to show thatindeed we have
1
1 + a1+
1
1 + a2+ · · ·+ 1
1 + an≥ 1.
441
For this, we can proceed by assuming (without loss of generality) that a1 ≤a2 ≤ · · · ≤ an, then a1a2 ≤ 1, and therefore
1
1 + a1+
1
1 + a2≥ 1
1 + a1+
1
1 + 1a1
=1
1 + a1+
a1a1 + 1
= 1.
This proves (a).
(b) If n = 2 then the inequality becomes
1
1 + 2a1+
1
1 + 2a2≥ d2, i.e.
1
1 + 2a1+
a1a1 + 2
≥ d2.
This can be rewritten as
2(a1 − 1)2
3(a1 + 2)(2a1 + 1)+
2
3≥ d2,
and it easily follows that d2 =2
3is the ”sharpest choice”.
If n ≥ 3, we set a1 = a2 = · · · = an−1 = x and an =1
xn−1, where x is an
arbitrary positive real number. In this case, the inequality becomes
n− 1
1 + 2x+
xn−1
xn−1 + 2≥ dn,
and thus, by letting (again) x near to infinity, we get get dn ≤ 1. We are nowleft to prove that
1
1 + 2a1+
1
1 + 2a2+ · · ·+ 1
1 + 2an≥ 1.
Without loss of generality, we can assume that a1 ≤ a2 ≤ · · · ≤ an, thena1a2a3 ≤ 1, and therefore there exists a positive number k such that k < 1
and satisfying a1a2a3 = k3. Now, by setting a1 =knp
m2, a2 =
kpm
n2, a3 =
kmn
p2,
we have
1
1 + 2a1+
1
1 + 2a2+
1
1 + 2a3=
m2
m2 + 2knp+
n2
n2 + 2kpm+
p2
p2 + 2kmn
≥ m2
m2 + 2np+
n2
n2 + 2pm+
p2
p2 + 2mn
≥ (m+ n+ p)2
m2 + 2np+ n2 + 2pm+ p2 + 2mn= 1.
This yields
dn =
{23 , if n = 21, if n > 2
.
?F?
07.19. If a, b are positive real numbers such that ab ≥ 1, then
1
(2a+ 3)2+
1
(2b+ 3)2≥ 2
5(2ab+ 3).
442
(Kiev 2007)
Solution: By the Cauchy Schwarz Inequality, we have
1
(2a+ 3)2+
1
(2b+ 3)2≥ (a+ b)2
b2(2a+ 3)2 + a2(2b+ 3)2
=(a+ b)2
9(a+ b)2 + 12ab(a+ b) + 8a2b2 − 18ab.
It suffices to prove that
(a+ b)2
9(a+ b)2 + 12ab(a+ b) + 8a2b2 − 18ab≥ 2
5(2ab+ 3),
or equivalently,
5
2(2ab+ 3) ≥ 9 +
12ab
a+ b+
8a2b2 − 18ab
(a+ b)2.
Now, we see that
5
2(2ab+ 3)−
(9 + 6
√ab+
8a2b2 − 18ab
4ab
)= 3
(√ab− 1
)2≥ 0,
so it is enough to check that
12ab
a+ b+
8a2b2 − 18ab
(a+ b)2≤ 6√ab+
8a2b2 − 18ab
4ab,
which is equivalent to
8a2b2 − 18ab
(a+ b)2− 8a2b2 − 18ab
4ab≤
6√ab(√
a−√b)2
a+ b,
or
(9− 4ab)(a− b)2
2(a+ b)2≤
6√ab(√
a−√b)2
a+ b.
Since
(√a−√b)2
a+ b≥ 0, we see that the above inequality follows from
(9− 4ab)(√
a+√b)2
2(a+ b)≤ 6√ab,
which is true because
(9− 4ab)(√
a+√b)2
2(a+ b)≤
5(√
a+√b)2
2(a+ b)≤ 5 < 6
√ab.
This completes our proof. It is easy to see that the equality holds iff a = b = 1.?F?
443
07.20. For all positive real numbers a, b, c, find all values of positive numberk such that the following inequality holds
a
c+ kb+
b
a+ kc+
c
b+ ka≥ 1
2007.
(Korea 2007)
Solution: Let a = b = c, we get k ≤ 6020. We claim that the answer is0 < k ≤ 6020. To prove this claim, it suffices to prove that for 0 < k ≤ 6020,the above inequality holds. Indeed, from the Cauchy Schwarz Inequality andthe well-known inequality (a+ b+ c)2 ≥ 3(ab+ bc+ ca), we have
a
c+ kb+
b
a+ kc+
c
b+ ka≥ (a+ b+ c)2
a(c+ kb) + b(a+ kc) + c(b+ ka)
=(a+ b+ c)2
(k + 1)(ab+ bc+ ca)≥ 3(ab+ bc+ ca)
(k + 1)(ab+ bc+ ca)
=3
k + 1≥ 1
2007,
as desired.?F?
07.21. Let a, b, c be positive real numbers. Prove that
1 +3
ab+ bc+ ca≥ 6
a+ b+ c.
(Macedonia 2007)
Solution: We use first the well-known inequality ab+ bc+ ca ≤ (a+ b+ c)2
3and then the AM-GM Inequality to get
1 +3
ab+ bc+ ca≥ 1 +
9
(a+ b+ c)2≥ 2
√9
(a+ b+ c)2=
6
a+ b+ c,
as desired. The equality holds if and only if a = b = c = 1.?F?
07.22. Let a, b, c, d be nonnegative real numbers such that a+ b+ c+ d = 4.Prove that
a2bc+ b2cd+ c2da+ d2ab ≤ 4.
(Middle Europe 2007)
Solution: We have
a2bc+ b2cd+ c2da+ d2ab− (ac+ bd)(ab+ cd) = −bd(a− c)(b− d),
and
a2bc+ b2cd+ c2da+ d2ab− (bc+ ad)(bd+ ac) = ac(a− c)(b− d).
444
Therefore
a2bc+ b2cd+ c2da+ d2ab ≤ max {(ac+ bd)(ab+ cd), (bc+ ad)(bd+ ac)} .
On the other hand, using the AM-GM Inequality, we see that
(ac+ bd)(ab+ cd) ≤ 1
4(ac+ bd+ ab+ cd)2 =
1
4(a+ d)2(b+ c)2
≤ 1
43(a+ d+ b+ c)4 = 4,
and
(bc+ ad)(bd+ ac) ≤ 1
4(bc+ ad+ bd+ ac)2 =
1
4(a+ b)2(c+ d)2
≤ 1
43(a+ b+ c+ d)4 = 4.
This means that
max{(ac+ bd)(ab+ cd), (bc+ ad)(bd+ ac)} ≤ 4,
and hence, we conclude that
a2bc+ b2cd+ c2da+ d2ab ≤ 4,
as desired. The equality holds if and only if (a, b, c, d) equals (1, 1, 1, 1), or(2, 1, 1, 0) or (1, 1, 0, 2), or (0, 2, 1, 1), or (1, 0, 2, 1).
?F?
07.23. Let a, b, c, d be positive real numbers in the interval
[1
2, 2
]and abcd =
1. Find the maximum value of(a+
1
b
)(b+
1
c
)(c+
1
d
)(d+
1
a
).
(Middle Europe 2007)
Solution: Since1
2≤ a, b, c, d ≤ 2, we get
1
4≤ a
c,b
d≤ 4. This implies
(4a−c)(a−4c) ≤ 0 and (4b−d)(b−4c) ≤ 0, from which deduce that (a+c)2 ≤25
4ac and (b+ d)2 ≤ 25
4bd. Using these inequalities, we obtain
(ab+ 1)(bc+ 1) = b2ac+ b(a+ c) + 1 ≤ b2ac+5
2b√ac+ 1
=1
2
(2b√ac+ 1
) (b√ac+ 2
).
Similarly, we have
(cd+ 1)(da+ 1) ≤ 1
2
(2d√ac+ 1
) (d√ac+ 2
).
445
Multilplying these two inequality and using the remark that(a+
1
b
)(b+
1
c
)(c+
1
d
)(d+
1
a
)= (ab+ 1)(bc+ 1)(cd+ 1)(da+ 1),
we obtain(a+
1
b
)(b+
1
c
)(c+
1
d
)(d+
1
a
)≤
≤ 1
4
(2b√ac+ 1
) (2d√ac+ 1
) (b√ac+ 2
) (d√ac+ 2
).
Now, proceeding the same way as above, we have(2b√ac+ 1
) (2d√ac+ 1
)= 4abcd+ 2
√ac(b+ d) + 1
≤ 4abcd+ 5√abcd+ 1 = 10,
and (b√ac+ 2
) (d√ac+ 2
)= abcd+ 2
√ac(b+ d) + 4
≤ abcd+ 5√abcd+ 4 = 10.
Therefore (a+
1
b
)(b+
1
c
)(c+
1
d
)(d+
1
a
)≤ 25.
On the other hand, we can see that the inequality holds for a = b = 2, c = d =1
2. Thus, we conclude that the searched maximum is 25.
?F?
07.24. Let a1, a2, . . . , an be positive real numbers such that ai ≥1
ifor all
i = 1, 2, . . . , n. Prove the inequality
(a1 + 1)
(a2 +
1
2
)· · ·(an +
1
n
)≥ 2n
(n+ 1)!(1 + a1 + 2a2 + · · ·+ nan).
(Moldova 2007)
Solution: Notice for all x1, x2, . . . , xn ≥ 0, we have
(1 + x1)(1 + x2) · · · (1 + xn) ≥ 1 + x1 + x2 + · · ·+ xn
≥ 1 +2
n+ 1(x1 + x2 + · · ·+ xn).
In this inequality, we replace xi byiai − 1
2≥ 0 to get
(1 +
a1 − 1
2
)(1 +
2a2 − 1
2
)· · ·(
1 +nan − 1
2
)≥
≥ 1 +2
n+ 1
(a1 − 1
2+
2a2 − 1
2+ · · ·+ nan − 1
2
),
446
or equivalently,
n!
2n(a1 + 1)
(a2 +
1
2
)· · ·(an +
1
n
)≥ 1
n+ 1(1 + a1 + 2a2 + · · ·+ nan) .
Dividing each side of the last inequality byn!
2n, we get the desired result. It is
easy to see that the equality holds if and only if ai =1
ifor all i = 1, 2, . . . , n.
?F?
07.25. Let a1, a2, . . . , an ∈ [0, 1]. If S = a31 + a32 + · · ·+ a3n, then prove that
a12n+ 1 + S − a31
+a2
2n+ 1 + S − a32+ · · ·+ an
2n+ 1 + S − a3n≤ 1
3.
(Moldova 2007)
Solution: For every 1 ≤ i ≤ n, the AM-GM Inequality implies
2n+ 1 + S − a3i =∑j 6=i
(a3j + 2) + 3 ≥ 3∑j 6=i
aj + 3 ≥ 3(a1 + a2 + · · ·+ an).
Thereforeai
2n+ 1 + S − a3i≤ 1
3· aia1 + a2 + · · ·+ an
.
Now, adding up the n inequalities obtained from taking i = 1, 2, . . . , n, we getthe result. The equality holds if and only if a1 = a2 = · · · = an = 1.
?F?
07.26. Let a, b, c be positive real numbers such that
a+ b+ c ≥ 1
a+
1
b+
1
c.
Prove that
a+ b+ c ≥ 3
a+ b+ c+
2
abc.
(Peru 2007)
Solution: By the Cauchy Schwarz Inequality, we have
a+ b+ c ≥ 1
a+
1
b+
1
c≥ 9
a+ b+ c,
and thus
a+ b+ c ≥ 3.
Returning to the inequality in question, we see that it can be rewritten as
(a+ b+ c)2 ≥ 3 + 2
(1
ab+
1
bc+
1
ca
).
447
By the AM-GM Inequality, we get
1
ab+
1
bc+
1
ca≤ 1
3
(1
a+
1
b+
1
c
)2
≤ 1
3(a+ b+ c)2,
and so it is sufficient to show that
(a+ b+ c)2 ≥ 3 +2
3(a+ b+ c)2,
which is obviously true according to the fact that a+ b+ c ≥ 3. The equalityholds if and only if a = b = c = 1.
?F?
07.27. a, b, c, d are positive real numbers satisfying the following condition
1
a+
1
b+
1
c+
1
d= 4.
Prove that
3
√a3 + b3
2+
3
√b3 + c3
2+
3
√c3 + d3
2+
3
√d3 + a3
2≤ 2(a+ b+ c+ d)− 4.
(Poland 2007)
Solution: The key to solve this problem is to cancel the cube root, and wecan proceed as follows: Applying the AM-GM Inequality, we have
3
√a3 + b3
2=
4
(a+ b)2· a+ b
2· a+ b
2· 3
√a3 + b3
2
≤ 4
(a+ b)2·
(a+ b)3
8+
(a+ b)3
8+a3 + b3
23
=4
a+ b·
(a+ b)2
4+a2 − ab+ b2
23
=a2 + b2
a+ b.
Therefore, it suffices to prove that
a2 + b2
a+ b+b2 + c2
b+ c+c2 + d2
c+ d+d2 + a2
d+ a≤ 2(a+ b+ c+ d)− 4.
Because 2(a+b+c+d) = (a+b)+(b+c)+(c+d)+(d+a) and a+b− a2 + b2
a+ b=
2ab
a+ b=
21
a+
1
b
, the last inequality can be written as
11
a+
1
b
+1
1
b+
1
c
+1
1
c+
1
d
+1
1
d+
1
a
≥ 2.
448
By the Cauchy Schwarz Inequality, we have
11
a+
1
b
+1
1
b+
1
c
+1
1
c+
1
d
+1
1
d+
1
a
≥
≥ 16(1
a+
1
b
)+
(1
b+
1
c
)+
(1
c+
1
d
)+
(1
d+
1
a
)=
81
a+
1
b+
1
c+
1
d
= 2.
The equality attains if and only if a = b = c = d = 1.?F?
07.28. Let x, y, z be nonnegative real numbers. Prove that the followinginequality holds
x3 + y3 + z3
3≥ xyz +
3
4|(x− y)(y − z)(z − x)| .
(Romania 2007)
Solution: With noting that
x3 + y3 + z3 − 3xyz =(x+ y + z)[(x− y)2 + (y − z)2 + (z − x)2]
3,
we can rewrite the original inequality as
(x+ y + z)[(x− y)2 + (y − z)2 + (z − x)2]
6≥ 3
4|(x− y)(y − z)(z − x)| .
Now, by the AM-GM Inequality, we have
2(x+ y + z) = (x+ y) + (y + z) + (z + x) ≥ |x− y|+ |y − z|+ |z − x|
≥ 3 3√|(x− y)(y − z)(z − x)|,
and
(x− y)2 + (y − z)2 + (z − x)2 ≥ 3 3√|(x− y)2(y − z)2(z − x)2|.
Multiplying these two inequalities and dividing each side of the resulting in-equality by 12, we get the result. It is easy to see that the equality holds ifand only if x = y = z.
?F?
07.29. For n ∈ N, n ≥ 2, determine
maxn∏i=1
(1− xi), for xi ∈ R+, 1 ≤ i ≤ n,n∑i=1
x2i = 1.
(Romania 2007)
449
First solution: Let us analyze E(x, y) = (1 − x)(1 − y), x, y ≥ 0, x2 +
y2 = k2, 0 < k ≤ 1. Take x = k sin θ, y = k cos θ, θ ∈[0,π
2
]. Now,
x + y = k(sin θ + cos θ) = k√
2 sin(θ +
π
4
)= k√
2 cos(θ − π
4
), and xy =
k2 sin θ cos θ =k2
2sin 2θ =
k2
2cos(
2θ − π
2
)=k2
2cos 2
(θ − π
4
), and hence,
xy = k2 cos2(θ − π
4
)− k2
2.
Take u = cos(θ − π
4
), so u ∈
[1√2, 1
], and then E(x, y) = E(u) = k2z2 −
k√
2z + 1 − k2
2. Its minimum value is reached for u0 =
1
k√
2≥ 1√
2, and
therefore
• for 1 <1
k√
2+
(1
k√
2− 1√
2
)=
2− kk√
2, i.e. k < 2
(√2− 1
), the maximum
value for E(u) is reached for u =1√2
, i.e. when θ ∈ {0, π}, which means x or
y being zero;
• for k ≥ 2(√
2− 1), the maximum value for E(u) is reached for u = 1, i.e.
when θ =π
4, which means x = y.
Now, for x2 + y2 + z2 = 1, there will be two, for which we have that x2 + y2 ≤2
3<(2(√
2− 1))2
(we have assumed them to be x and y), so first case applies
for x2+y2 = k2, k < 2(√
2− 1)
(case k = 0 is trivial), therefore the maximumvalue is reached when one of the three variables is zero.
So, when
n∑i=1
x2i = 1, E =n∏i=1
(1− xi)
takes maximum value when all but two variables (be them x, y) are zero, thenE = E(x, y) = (1 − x)(1 − y), with x2 + y2 = 1, and second case applies,
yielding maxE =
(1− 1√
2
)2
, for x = y =1√2
.
Second solution: We will prove by induction that for any x1, x2, . . . , xn, x21+
x22 + · · ·+ x2n = 1,
(1− x1)(1− x2) · · · (1− xn) ≤(
1− 1√2
)2
,
with equality (for example) when x1 = x2 =1√2, x3 = · · · = xn = 0. Indeed,
for n = 2, the inequality becomes
(1− x1)(1− x2) ≤(
1− 1√2
)2
,
450
where x21 + x22 = 1. By the AM-GM Inequality, we have
(1− x1)(1− x2) =(1− x21)(1− x22)(1 + x1)(1 + x2)
=x21x
22
(1 + x1)(1 + x2)≤ x21x
22(
1 +√x1x2
)2=
x1x2(1
√x1x2
+ 1
)2 ≤
x21 + x222(√
2
x21 + x22+ 1
)2 =
(1− 1√
2
)2
.
This proves our claim for n = 2. Now, suppose that the inequality holds forn ≥ 2 and let us prove it for n + 1. Due to symmetry, we may assume that
x1 ≤ x2 ≤ · · · ≤ xn+1, then x21+x22 ≤2
n+ 1. Note that
√x21 + x22, x3, . . . , xn+1
are n nonnegative numbers and(√
x21 + x22
)2+ x23 + · · ·+ x2n+1 = 1, so from
the inductive hypothesis, we find that(1−
√x21 + x22
)(1− x3) · · · (1− xn+1) ≤
(1− 1√
2
)2
.
According to this inequality, we see that it is enough to prove that
(1− x1)(1− x2) ≤ 1−√x21 + x22,
or equivalently,
x1x2 ≤ x1 + x2 −√x21 + x22.
Because x1 + x2 −√x21 + x22 =
2x1x2
x1 + x2 +√x21 + x22
, we can write the above
inequality as
x1x2
(x1 + x2 +
√x21 + x22 − 2
)≤ 0,
which is true because x1x2 ≥ 0 and
x1 + x2 +√x21 + x22 ≤
(√2 + 1
)√x21 + x22 ≤
(√2 + 1
)√ 2
n+ 1
≤(√
2 + 1)√2
3< 2.
This completes our proof.?F?
07.30. Let a, b, c be positive real numbers such that
1
a+ b+ 1+
1
b+ c+ 1+
1
c+ a+ 1≥ 1.
Show thata+ b+ c ≥ ab+ bc+ ca.
(Romania 2007)
451
First solution: By applying the Cauchy Schwarz Inequality, we obtain
(a+ b+ 1)(a+ b+ c2) ≥ (a+ b+ c)2,
and thus1
a+ b+ 1≤ c2 + a+ b
(a+ b+ c)2.
Now by summing cyclically, we obtain
1
a+ b+ 1+
1
b+ c+ 1+
1
c+ a+ 1≤ a2 + b2 + c2 + 2(a+ b+ c)
(a+ b+ c)2.
But from the condition, we can see that
a2 + b2 + c2 + 2(a+ b+ c) ≥ (a+ b+ c)2,
and thereforea+ b+ c ≥ ab+ bc+ ca.
We see that the equality occurs if and only if a = b = c = 1.
Second solution: We first observe that
2 ≥∑(
1− 1
a+ b+ 1
)=∑ a+ b
a+ b+ 1=∑ (a+ b)2
(a+ b)2 + a+ b.
Apply the Cauchy Schwarz Inequality to get
2 ≥∑ (a+ b)2
(a+ b)2 + a+ b≥
[∑(a+ b)
]2∑
[(a+ b)2 + a+ b]
=4∑
a2 + 8∑
ab
2∑
a2 + 2∑
ab+ 2∑
a,
and thus
2(
2∑
a2 + 2∑
ab+ 2∑
a)≥ 4
∑a2 + 8
∑ab.
From this inequality, we deduce that
a+ b+ c ≥ ab+ bc+ ca,
as claimed.?F?
07.31. For n ∈ N, n ≥ 2, ai, bi ∈ R, 1 ≤ i ≤ n, such that
n∑i=1
a2i = 1,n∑i=1
b2i = 1, andn∑i=1
aibi = 0,
prove that (n∑i=1
ai
)2
+
(n∑i=1
bi
)2
≤ n.
452
(C. Lupu and T. Lupu, Romania 2007)
First solution: Denote A =n∑i=1
ai and B =n∑i=1
bi. Applying the Cauchy
Schwarz Inequality, we have
(A2 +B2)2 =
[n∑i=1
(aiA+ biB)
]2≤ n
n∑i=1
(aiA+ biB)2
= n
n∑i=1
(a2iA2 + b2iB
2 + 2aibiAB)
= n
(A2
n∑i=1
a2i +B2n∑i=1
b2i + 2ABn∑i=1
aibi
)= n(A2 +B2).
Hence (A2 + B2)2 ≤ n(A2 + B2), and it follows that A2 + B2 ≤ n, so theinequality is proved.
Second solution: With the notations of the preceding solution, we have
0 ≤n∑i=1
(1− aiA− biB)2 =n∑i=1
(1 + a2iA2 + b2iB
2 − 2aiA− 2biB + 2aibiAB)
= n+A2n∑i=1
a2i +B2n∑i=1
b2i − 2An∑i=1
ai − 2Bn∑i=1
bi + 2ABn∑i=1
aibi
= n+A2 +B2 − 2A2 − 2B2 + 0 = n− (A2 +B2).
and the inequality follows directly from this identity.?F?
07.32. Positive real numbers a, b, c satisfy a+ b+ c = 1. Show that
1
ab+ 2c2 + 2c+
1
bc+ 2a2 + 2a+
1
ca+ 2b2 + 2b≥ 1
ab+ bc+ ca.
(Turkey 2007)
First solution: Applying the AM-GM Inequality, we have
ab+ 2c2 + 2c = ab+ 2c2 + 2c(a+ b+ c) = (2c+ a)(2c+ b)
≤ [b(2c+ a) + a(2c+ b)]2
4ab=
(ab+ bc+ ca)2
ab.
Therefore1
ab+ 2c2 + 2c≥ ab
(ab+ bc+ ca)2,
and it follows that∑ 1
ab+ 2c2 + 2c≥ ab+ bc+ ca
(ab+ bc+ ca)2=
1
ab+ bc+ ca.
453
Note that the equality occurs iff a = b = c =1
3.
Second solution: Similar to the preceding solution, we see that ab+2c2+2c =(2c+ a)(2c+ b), so the original inequality is equivalent to
1
(2a+ b)(2a+ c)+
1
(2b+ c)(2b+ a)+
1
(2c+ a)(2c+ b)≥ 1
ab+ bc+ ca.
Now, setting a =1
x, b =
1
y, z =
1
z, then we have
1
(2a+ b)(2a+ c)=
x2yz
(x+ 2y)(x+ 2z), and
1
ab+ bc+ ca=
xyz
x+ y + z,
and thus, the above inequality is equivalent to
x
(x+ 2y)(x+ 2z)+
y
(y + 2z)(y + 2x)+
z
(z + 2x)(z + 2y)≥ 1
x+ y + z.
By the Cauchy Schwarz Inequality, we get∑ x
(x+ 2y)(x+ 2z)=∑ x2
x(x+ 2y)(x+ 2z)≥ (x+ y + z)2∑
x(x+ 2y)(x+ 2z).
Therefore, the last inequality is deduced from
(x+ y + z)3 ≥ x(x+ 2y)(x+ 2z) + y(y + 2z)(y + 2x) + z(z + 2x)(z + 2y).
After some small computations, one can see that it is equivalent to
xy(x+ y) + yz(y + z) + zx(z + x) ≥ 6xyz.
This is obviously true by the AM-GM Inequality, so the inequality is proved.?F?
07.33. Let a, b, c be positive real numbers such that abc ≥ 1. Prove that(a+
1
a+ 1
)(b+
1
b+ 1
)(c+
1
c+ 1
)≥ 27
8,
and
27(a3+a2+a+1)(b3+b2+b+1)(c3+c2+c+1) ≥ 64(a2+a+1)(b2+b+1)(c2+c+1).
(Ukraine 2007)
Solution: For any positive number x, we have
x+1
x+ 1− 3
4(x+ 1) =
(x− 1)2
4(x+ 1)≥ 0.
This implies that(a+
1
a+ 1
)(b+
1
b+ 1
)(c+
1
c+ 1
)≥ 27
64(a+ 1)(b+ 1)(c+ 1).
454
On the other hand, by the Holder’s Inequality, we have
(a+ 1)(b+ 1)(c+ 1) ≥(
3√abc+ 1
)3= 8.
Combining these two inequalities, we get(a+
1
a+ 1
)(b+
1
b+ 1
)(c+
1
c+ 1
)≥ 27
8.
This proves the first inequality in the required question. Now, let us prove thesecond. Notice that a3 + a2 + a + 1 = (a2 + 1)(a + 1), so we can rewrite theinequality as
(a2 + 1)(b2 + 1)(c2 + 1)
(a2 + a+ 1)(b2 + b+ 1)(c2 + c+ 1)· (a+ 1)(b+ 1)(c+ 1) ≥ 64
27.
Now, by the AM-GM Inequality, we have
(a2 + 1)(b2 + 1)(c2 + 1)
(a2 + a+ 1)(b2 + b+ 1)(c2 + c+ 1)≥
≥ (a2 + 1)(b2 + 1)(c2 + 1)(a2 +
a2 + 1
2+ 1
)(b2 +
b2 + 1
2+ 1
)(c2 +
c2 + 1
2+ 1
) =8
27.
So, it is enough to check that
(a+ 1)(b+ 1)(c+ 1) ≥ 8,
which we have been proved above. The proof is completed. It is easy to seethat the equality (in both inequalities) holds iff a = b = c = 1.
?F?
07.34. Show that for any real numbers a, b, c, then
(a2 + b2)2 ≥ (a+ b+ c)(b+ c− a)(c+ a− b)(a+ b− c).
(United Kingdom 2007)
Solution: From the AM-GM Inequality, we get
(a+ b+ c)(b+ c− a)(c+ a− b)(a+ b− c) =[(a+ b)2 − c2
] [c2 − (a− b)2
]≤ 1
4
[(a+ b)2 − c2 + c2 − (a− b)2
]2= 4a2b2 ≤ (a2 + b2)2,
as desired. The equality holds if and only if (a+ b)2 − c2 = c2 − (a− b)2 anda2 = b2, i.e. when 2a2 = 2b2 = c2.
?F?
07.35. Given a triangle ABC. Determine the minimum value of
cos2A
2cos2
B
2
cos2C
2
+cos2
B
2cos2
C
2
cos2A
2
+cos2
C
2cos2
A
2
cos2B
2
.
455
(Vietnam 2007)
Solution: Let x = tanA
2, y = tan
B
2, z = tan
C
2, then x, y, z > 0 and xy +
yz + zx = 1. By this substitution, we have
cos2A
2cos2
B
2
cos2C
2
=
1
x2 + 1· 1
y2 + 11
z2 + 1
=x2 + 1
(y2 + 1)(z2 + 1)
=(x+ y)(x+ z)
(y + z)(y + z)(z + x)(z + y)=
1
(y + z)2.
So, the desired expression is equal to
1
(y + z)2+
1
(z + x)2+
1
(x+ y)2.
On the other hand, it is known that (see the Iran 1996 problem)
1
(y + z)2+
1
(z + x)2+
1
(x+ y)2≥ 9
4(xy + yz + zx),
with equality iff x = y = z. Using this inequality with noting that xy + yz +zx = 1, we get
cos2A
2cos2
B
2
cos2C
2
+cos2
B
2cos2
C
2
cos2A
2
+cos2
C
2cos2
A
2
cos2B
2
≥ 9
4(xy + yz + zx)=
9
4,
with equality iff tanA
2= tan
B
2= tan
C
2, i.e. when ABC is an equilateral
triangle. From this, we conclude that the searched minimum is9
4.
?F?
08.1. Let a, b, c be real numbers such that a2 + b2 + c2 = 3. Prove that
a2
2 + b+ c2+
b2
2 + c+ a2+
c2
2 + a+ b2≥ (a+ b+ c)2
12.
(Baltic Way 2008)
Solution: From the given hypothesis a2+b2+c2 = 3, we find that |a| , |b| , |c| ≤√3 < 2, and hence 2 + b+ c2 > 0, 2 + c+a2 > 0, 2 +a+ b2 > 0. Now, applying
the Cauchy Schwarz Inequality, we have
a2
2 + b+ c2+
b2
2 + c+ a2+
c2
2 + a+ b2≥ (a+ b+ c)2
6 + (a+ b+ c) + (a2 + b2 + c2)
=(a+ b+ c)2
9 + a+ b+ c
≥ (a+ b+ c)2
9 +√
3(a2 + b2 + c2)
=(a+ b+ c)2
12,
456
as desired. Note that the equality holds if and only if a = b = c = 1.?F?
08.2. Suppose that a, b, c are positive real numbers with a2 + b2 + c2 = 1.Prove that
a5 + b5
ab(a+ b)+
b5 + c5
bc(b+ c)+
c5 + a5
ca(a+ b)≥ 3(ab+ bc+ ca)− 2.
(Bosnia 2008)
Solution: For any positive numbers x, y, we have (x3 − y3)(x2 − y2) ≥ 0,
implying x5 + y5 ≥ x2y2(x+ y), orx5 + y5
xy(x+ y)≥ xy. Accordingly, we find that
the left hand side of the original inequality is not smaller than ab+ bc+ ca. Itsuffices to prove that
ab+ bc+ ca ≥ 3(ab+ bc+ ca)− 2, or 2 ≥ 2(ab+ bc+ ca),
which is true according to the well-known inequality a2 +b2 +c2 ≥ ab+bc+caand the hypothesis x2 + y2 + z2 = 1. The equality holds if and only if a = b =
c =1√3.
?F?
08.3. Let x, y, z be real numbers. Show that the following inequality holds
x2 + y2 + z2 − xy − yz − zx ≥ max
{3(x− y)2
4,3(y − z)2
4,3(z − x)2
4
}.
(Bosnia 2008)
Solution: Without loss of generality, we may assume that (x−z)2 = max{(x−y)2, (y − z)2, (z − x)2}. From this assumption, applying the Cauchy SchwarzInequality, we get
2(x2 + y2 + z2 − xy − yz − zx) = (x− y)2 + (y − z)2 + (x− z)2
≥ 1
2[(x− y) + (y − z)]2 + (x− z)2
=3
2(x− z)2,
and hence, we deduce that
x2+y2+z2−xy−yz−zx ≥ 3
4(x−z)2 = max
{3(x− y)2
4,3(y − z)2
4,3(z − x)2
4
},
as desired.?F?
08.4. Let a, b, c be positive real numbers. Prove that(1 +
4a
b+ c
)(1 +
4b
a+ c
)(1 +
4c
a+ b
)> 25.
457
(Bosnia 2008)
Solution: By expanding, we can rewrite our inequality as
4(a3 + b3 + c3) + 28abc > 4a2(b+ c) + 4b2(c+ a) + 4c2(a+ b),
or
4 [a(a− b)(a− c) + b(b− c)(b− a) + c(c− a)(c− b)] + 16abc > 0.
This is true since 16abc > 0 and∑
a(a − b)(a − c) ≥ 0 from the Schur’s
Inequality (applied for third degree). Our proof is completed.?F?
08.5. Let x, y, z be real numbers such that x+y+z = xy+yz+zx. Determinethe least value of the following expression
P =x
x2 + 1+
y
y2 + 1+
z
z2 + 1.
(Brazil 2008)
Solution: We claim that the minimum value of P is −1
2which attains when
(x, y, z) is a permutation of (−1,−1, 1). We will prove this claim by provingthat
2x
x2 + 1+ 1 +
2y
y2 + 1+ 1 ≥ 1− 2z
z2 + 1,
or equivalently,(x+ 1)2
x2 + 1+
(y + 1)2
y2 + 1≥ (z − 1)2
z2 + 1.
Now, from the given hypothesis, we have z(x+y−1) = x+y−xy. If x+y = 1,
then it follows that xy = x+y = 1, which is impossible since xy ≤(x+ y
2
)2
=
1
4, so we must have x+ y 6= 1, and hence
z =x+ y − xyx+ y − 1
.
Replacing this into the above inequality, we can rewrite it as
(x+ 1)2
x2 + 1+
(y + 1)2
y2 + 1≥
(x+ y − xyx+ y − 1
− 1
)2
(x+ y − xyx+ y − 1
)2
+ 1
,
or(x+ 1)2
x2 + 1+
(y + 1)2
y2 + 1≥ (1− xy)2
(x+ y − xy)2 + (x+ y − 1)2.
458
If x = y = 1, this inequality is trivial. Alternatively, if (x− 1)2 + (y− 1)2 > 0,we make use of the Cauchy Schwarz Inequality and get
(x+ 1)2
x2 + 1+
(y + 1)2
y2 + 1≥ [(1 + x)(1− y) + (1− x)(1 + y)]2
(1− y)2(1 + x2) + (1− x)2(1 + y2)
=4(1− xy)2
(1− y)2(1 + x2) + (1− x)2(1 + y2).
Therefore, we can see that the above inequality is deduced from
4(x+ y − xy)2 + 4(x+ y − 1)2 ≥ (1− y)2(1 + x2) + (1− x)2(1 + y2).
By some easy computations, we can rewrite it as
f(x) = (y2 − 3y + 3)x2 − (3y2 − 8y + 3)x+ 3y2 − 3y + 1 ≥ 0.
Note that f(x) is a quadratic polynomial of x with the highest coefficient ispositive. On the other hand, the discrimimant of f(x) is
∆f = (3y2 − 8y + 3)2 − 4(y2 − 3y + 3)(3y2 − 3y + 1) = −3(y2 − 1)2 ≤ 0.
So, it is clear that f(x) ≥ 0, and our claim is proved. Or in the other words,
we have proved that the minimum value of the expression P is −1
2.
?F?
08.6. Let a, b, c be positive real numbers such that a+ b+ c = 1. Prove that
a− bca+ bc
+b− cab+ ca
+c− abc+ ab
≤ 3
2.
(Canada 2008)
Solution: By the Cauchy Schwarz Inequality, we have
a− bca+ bc
+b− cab+ ca
+c− abc+ ab
= 3− 2
(bc
a+ bc+
ca
b+ ca+
ab
c+ ab
)≤ 3− 2(ab+ bc+ ca)2
3abc+ b2c2 + c2a2 + a2b2
= 3− 2(ab+ bc+ ca)2
3abc(a+ b+ c) + b2c2 + c2a2 + a2b2
= 3− 2(ab+ bc+ ca)2
abc(a+ b+ c) + (ab+ bc+ ca)2
≤ 3− 2(ab+ bc+ ca)2
(ab+ bc+ ca)2
3+ (ab+ bc+ ca)2
=3
2.
Note that the equality holds if and only if a = b = c =1
3.
?F?
459
08.7. Let a, b, c be nonnegative real numbers such that a + b + c = 1. Provethat √
a+(b− c)2
4+√b+√c ≤√
3.
(China 2008)
Solution: Set x =√bc, then we have
√b+√c =
√b+ c+ 2
√bc =
√1− a+ 2
√bc =
√1− a+ 2x,
and √a+
(b− c)24
=
√a+
(b+ c)2 − 4bc
4=
1
2
√(1 + a)2 − 4x2.
Therefore, the inequality in question now rewrites as√(1 + a)2 − 4x2 + 2
√1− a+ 2x ≤ 2
√3.
By the Cauchy Schwarz Inequality, we have√(1 + a)2 − 4x2 + 2
√1− a+ 2x ≤
√3 [(1 + a)2 − 4x2 + 2(1− a+ 2x)],
and hence it suffices to prove that
(1 + a)2 − 4x2 + 2(1− a+ 2x) ≤ 4, or equivalently, a2 − 4x2 + 4x− 1 ≤ 0.
This can be also rewritten as
(1− 2x− a)(1 + a− 2x) ≥ 0,
which is obviously true since
1 + a− 2x ≥ 1− 2x− a = 1− 2√bc− a ≥ 1− (b+ c)− a = 0.
It is easy to see that the equality holds if and only if a = b = c =1
3.
?F?
08.8. Let x, y, z be positive numbers. Find the minimal value of
(a)x2 + y2 + z2
xy + yz;
(b)x2 + y2 + 2z2
xy + yz.
(Croatia 2008)
Solution: (a) Using the identity
x2 + y2 + z2
xy + yz−√
2 =
(x−√
2
2y
)2
+
(z −√
2
2y
)2
xy + yz≥ 0,
460
we conclude that the expressionx2 + y2 + z2
xy + yzhas minimal value
√2 when
x = z =
√2
2y.
(b) Similarly, using the identity
x2 + y2 + 2z2
xy + yz−√
8
3=
(x−
√2
3y
)2
+ 2
(z −
√1
6y
)2
xy + yz≥ 0,
we conclude that the expressionx2 + y2 + 2z2
xy + yzhas minimal value
√8
3when
x =
√2
3y and z =
√1
6y.
?F?
08.9. Determine the smallest constant C such that the following inequality
1 + (x+ y)2 ≤ C(1 + x2)(1 + y2)
holds for any real numbers x, y.(Germany 2008)
First solution: Let x = y =
√2
2, we get C ≥ 4
3. We will prove that the
constant k =4
3is our answer, i.e.
4
3(1 + x2)(1 + y2) ≥ (x+ y)2 + 1.
Expanding, we can rewrite this inequality as follows
f(x) = (4y2 + 1)x2 − 6xy + y2 + 1 ≥ 0.
We see that f(x) is a quadratic function of x with the highest coefficient bepositive. In addition, its discriminant is
∆′f = 9y2 − (y2 + 1)(4y2 + 1) = −(1− 2y2)2 ≤ 0.
This means that f(x) is nonnegative for any real number x. So, our proof iscompleted.
Second solution: Similarly, we need to prove the inequality
4
3(1 + x2)(1 + y2) ≥ (x+ y)2 + 1.
By the Cauchy Schwarz Inequality, we get
4(1 + x2)(1 + y2) = 3 + 2(x2 + y2) + (1 + 2x2)(2y2 + 1)
≥ 3 + (x+ y)2 +(√
2x+√
2y)2
= 3[(x+ y)2 + 1].
461
Dividing both sides of this inequality by 3, we get the desired inequality.?F?
08.10. Prove that if a1, a2, . . . , an are positive integers, then the followinginequality (
a21 + a22 + · · ·+ a2na1 + a2 + · · ·+ an
) knt
≥ a1a2 · · · an
holds for k = max {a1, a2, . . . , an} and t = min {a1, a2, . . . , an} . When do wehave the equality?
(Greece 2008)
Solution: By the Cauchy Schwarz Inequality, we have
n(a21 + a22 + · · ·+ a2n) ≥ (a1 + a2 + · · ·+ an)2,
and hencea21 + a22 + · · ·+ a2na1 + a2 + · · ·+ an
≥ a1 + a2 + · · ·+ ann
.
Since ai is positive integer, we have ai ≥ 1 for all i = 1, 2, . . . , n. Therefore
a1 + a2 + · · ·+ ann
≥ 1.
Also, since the function f(x) = ax is increasing for all a ≥ 1 andk
t≥ 1, we
have (a21 + a22 + · · ·+ a2na1 + a2 + · · ·+ an
) knt
≥(a1 + a2 + · · ·+ an
n
) knt
≥(a1 + a2 + · · ·+ an
n
)n≥ a1a2 · · · an,
as desired. Note that the equality holds if and only if a1 = a2 = · · · = an.?F?
08.11. Let x, y, z be positive real numbers such that x2 + y2 + z2 = 3. Provethe inequality
3
2<
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z< 3.
(Greece 2008)
Solution: From the given hypothesis, we have 0 < x, y, z <√
3. This yields
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z>
1 + y2
2 + 2+
1 + z2
2 + 2+
1 + x2
2 + 2
=x2 + y2 + z2 + 3
4=
3
2,
462
and
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z<
1 + y2
2+
1 + z2
2+
1 + x2
2
=x2 + y2 + z2 + 3
2= 3.
Therefore, we have
3
2<
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z< 3,
as claimed.
Remark: Actually, one can prove that the stronger inequality holds with thesame condition
2 ≤ 1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z<
9
2−√
3.
Indeed, by the AM-GM Inequality and Cauchy Schwarz Inequality, we have
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z≥ 1 + y2
2 +x2 + 1
2
+1 + z2
2 + y2+12
+1 + x2
2 +z2 + 1
2
= 2
(y2 + 1
x2 + 5+z2 + 1
y2 + 5+x2 + 1
z2 + 5
)≥ 2(x2 + y2 + z2 + 3)2∑
(x2 + 5)(y2 + 1)
=72
15 + 6(x2 + y2 + z2) + (x2y2 + y2z2 + z2x2)
≥ 72
15 + 18 +(x2 + y2 + z2)2
3
= 2.
This proves the left inequality. For the right inequality, we will make use ofthe following inequality for all t ∈
(0,√
3),
1
t+ 2− 1
2+
2−√
3
2√
3t2 =
2−√
3
2√
3t2 − t
2(t+ 2)
=t
2(t+ 2)
[2−√
3√3
t(t+ 2)− 1
]
<t
2(t+ 2)
[2−√
3√3·√
3(√
3 + 2)− 1
]= 0,
that is1
t+ 2<
1
2− 2−
√3
2√
3t2.
463
According to this inequality, we have
1 + y2
2 + x+
1 + z2
2 + y+
1 + x2
2 + z<∑
(y2 + 1)
(1
2− 2−
√3
2√
3x2
)
=9
2−√
3− 2−√
3
2√
3(x2y2 + y2z2 + z2x2) <
9
2−√
3.
?F?
08.12. Prove that for any positive real numbers a, b, c, d, we have the followinginequality
(a− b)(a− c)a+ b+ c
+(b− c)(b− d)
b+ c+ d+
(c− d)(c− a)
c+ d+ a+
(d− a)(d− b)d+ a+ b
≥ 0.
(Darij Grinberg, IMO Shortlist 2008)
First solution: We denote with P (a, b, c, d) the left hand side of the originalinequality. The key to solve this inequality is to notice that, without loss ofgenerality, we can assume the following property (a− c)(b− d) ≥ 0. Indeed, if(a−c)(b−d) ≤ 0, then we can let a1 = b, b1 = c, c1 = d, d1 = a, and we will haveP (a1, b1, c1, d1) = P (a, b, c, d), and also (a1−c1)(b1−d1) = −(a−c)(b−d) ≥ 0.This is contradiction so what we have assumed is true. Now, we see that
(a− b)(a− c)a+ b+ c
+(c− d)(c− a)
c+ d+ a=
(a− c)2
a+ b+ c− (a+ 2c)(a− c)(b− d)
(a+ b+ c)(a+ c+ d),
and
(b− c)(b− d)
b+ c+ d+
(d− a)(d− b)d+ a+ b
=(b− d)2
b+ c+ d+
(b+ 2d)(a− c)(b− d)
(b+ c+ d)(d+ a+ b)
≥ (b− d)2
b+ c+ d.
Therefore, the original inequality is just a trivial corollary of
(a− c)2
a+ b+ c+
(b− d)2
b+ c+ d≥ (a+ 2c)(a− c)(b− d)
(a+ b+ c)(a+ c+ d).
According to the AM-GM Inequality, we have
(a− c)2
a+ b+ c+
(b− d)2
b+ c+ d≥ 2(a− c)(b− d)√
(a+ b+ c)(b+ c+ d).
Hence, it suffices to prove that
2(a+ c+ d)
√a+ b+ c
b+ c+ d≥ a+ 2c.
If a ≥ d then it is clear that
√a+ b+ c
b+ c+ d≥ 1 and 2(a+ c+ d) ≥ a+ 2c, so the
above inequality is trivial. Alternatively, if d ≥ a, then we can easily check
that
√a+ b+ c
b+ c+ d≥√a+ c
c+ d, so we only need to prove
2(a+ c+ d)√a+ c ≥ (a+ 2c)
√c+ d.
464
This can be proved as follows
2(a+ c+ d)√a+ c = 2
√a+ c+ d
√(a+ c+ d)(a+ c)
≥ 2(a+ c)√a+ c+ d ≥ 2(a+ c)
√c+ d
≥ (a+ 2c)√c+ d.
It is easy to see that the equality holds iff a = c and b = d.
Second solution: Denote
A =(a− b)(a− c)a+ b+ c
, B =(b− c)(b− d)
b+ c+ d, C =
(c− a)(c− d)
c+ d+ a, D =
(d− a)(d− b)d+ a+ b
.
It is easy to see that 2A = A′ +A′′, where
A′ =(a− c)2
a+ b+ cand A′′ =
(a− c)(a+ c− 2b)
a+ b+ c.
Similarly, we have 2B = B′ + B′′, 2C = C ′ + C ′′ and 2D = D′ + D′′. Byapplying the Cauchy Schwarz Inequality, we have
A′ +B′ + C ′ +D′ =(a− c)2
a+ b+ c+
(b− d)2
b+ c+ d+
(c− a)2
c+ d+ a+
(d− b)2
d+ a+ b
≥ (|a− c|+ |b− d|+ |c− a|+ |d− b|)2
(a+ b+ c) + (b+ c+ d) + (c+ d+ a) + (d+ a+ b)
=4(|a− c|+ |b− d|)2
3(a+ b+ c+ d).
On the other hand, according to the identities
A′′+C ′′ =3(a− c)(d− b)(a+ c)
(a+ b+ c+ d)(a+ c) + bd, B′′+C ′′ =
3(a− c)(b− d)(b+ d)
(a+ b+ c+ d)(b+ d) + ac,
we find that
A′′ +B′′ + C ′′ +D′′ =
= 3(a− c)(b− d)
[b+ d
(a+ b+ c+ d)(b+ d) + ac− a+ c
(a+ b+ c+ d)(a+ c) + bd
]=
3(a− c)(b− d)[bd(b+ d)− ac(a+ c)]
[(a+ b+ c+ d)(b+ d) + ac][(a+ b+ c+ d)(a+ c) + bd].
Furthermore,
[(a+ b+ c+ d)(b+ d) + ac][(a+ b+ c+ d)(a+ c) + bd] >
> [ac(a+ c) + bd(b+ d)](a+ b+ c+ d)
> |ac(a+ c)− bd(b+ d)|(a+ b+ c+ d).
And so, we deduce that
A′′ +B′′ + C ′′ +D′′ ≥ −3|a− c||b− d|a+ b+ c+ d
≥ −3(|a− c|+ |b− d|)2
4(a+ b+ c+ d).
465
From this, we obtain
2(A+B + C +D) ≥ 4(|a− c|+ |b− d|)2
3(a+ b+ c+ d)− 3(|a− c|+ |b− d|)2
4(a+ b+ c+ d)
=7(|a− c|+ |b− d|)2
12(a+ b+ c+ d)≥ 0,
and the conclusion follows.?F?
08.13. (i) If x, y, z are three real numbers, all different from 1, such thatxyz = 1, then prove that
x2
(x− 1)2+
y2
(y − 1)2+
z2
(z − 1)2≥ 1.
(ii) Prove that equality is achieved for infinitely many triples of rational num-bers x, y, z.
(IMO 2008)
First solution: (i) Since xyz = 1, there exist three nonnegative real numbers
a, b, c, such that x =a
b, y =
b
c, z =
c
a(clearly, a, b, c are distinct numbers since
x, y, z 6= 1), the inequality becomes
a2
(a− b)2+
b2
(b− c)2+
c2
(c− a)2≥ 1.
By the Cauchy Schwarz Inequality, we get[∑ a2
(a− b)2
] [∑(a− b)2(a− c)2
]≥[∑
a(a− c)]2
=(∑
a2 −∑
ab)2.
Moreover, we have∑(a− b)2(a− c)2 =
∑(a− b)2(a− c)2 + 2
∑(a− b)(a− c) · (b− c)(b− a)
=[∑
(a− b)(a− c)]2
=(∑
a2 −∑
ab)2.
This proves (i).
(ii) As we have seen in the proof of (i), the equality holds if and only if
a
a− b(a− b)(a− c)
=
b
b− c(b− c)(b− a)
=
c
c− a(c− a)(c− b)
.
This can be rewritten as
a
(a− b)2(a− c)=
b
(b− c)2(b− a)=
c
(c− a)2(c− b),
orb
a+c
b+a
c= 3.
466
Returning to the notations with x, y, z, this is equivalent with
1
x+
1
y+
1
z= 3.
Now, note that in the above equation we can choose
(x, y, z) =
(n
(n+ 1)2,−n(n+ 1),−n+ 1
n2
),
where n is an arbitrary rational number. This shows that the equality inthe inequality in question is achieved for infinitely many triples of rationalnumbers x, y, z.
Second solution: We provides the second way to prove (i). Denote a =x
x− 1, b =
y
y − 1, c =
z
z − 1, then we can find that x =
a
a− 1, y =
b
b− 1, z =
z
z − 1. And since xyz = 1, it follows that abc = (a − 1)(b − 1)(c − 1), or
a+ b+ c− ab− bc− ca = 1. According to the substitution, we need to provea2 + b2 + c2 ≥ 1. Observe that 1 = 2− 1 = 2(a+ b+ c− ab− bc− ca)− 1, thisinequality is equivalent to
a2 + b2 + c2 ≥ 2(a+ b+ c− ab− bc− ca)− 1.
The last one can be simplified into a complete square (a + b + c − 1)2 ≥ 1.This proves our inequality.
Third solution: We give the third proof for (i). Let a = 3√x, b = 3
√y, c = 3
√z,
then we havea2
bc=
3√x2
3√yz
=x
3√xyz
= x,
and similarly,b2
ca= y,
c2
ab= z. Therefore, using this substitution, we can write
our inequality in the form
a4
(a2 − bc)2+
b4
(b2 − ca)2+
c4
(c2 − ab)2≥ 1.
Applying the Cauchy Schwarz Inequality, we have
a4
(a2 − bc)2+
b4
(b2 − ca)2+
c4
(c2 − ab)2≥ (a2 + b2 + c2)2
(a2 − bc)2 + (b2 − ca)2 + (c2 − ab)2.
On the other hand,
(a2 + b2 + c2)2 − [(a2 − bc)2 + (b2 − ca)2 + (c2 − ab)2] = (ab+ bc+ ca)2 ≥ 0.
Therefore, from the above estimation, the result follows immediately.?F?
08.14. Let n ≥ 3 is an integer and let x1, x2, . . . , xn be real numbers suchthat xi > 1 for all i. Prove the following inequality
x1x2x3 − 1
+ · · ·+ xn−1xnx1 − 1
+xnx1x2 − 1
≥ 4n.
467
(Indonesia 2008)
Solution: From the AM-GM Inequality, we get
x1x2x3 − 1
+ · · ·+ xn−1xnx1 − 1
+xnx1x2 − 1
≥ x1x2
x23 + 4
4− 1
+ · · ·+ xn−1xn
x21 + 4
4− 1
+xnx1
x22 + 4
4− 1
= 4
(x1x2x23
+ · · ·+ xn−1xnx21
+xnx1x22
)≥ 4n n
√x1x2x23· · · xn−1xn
x21· xnx1x22
= 4n,
as claimed. Note that the equality holds if and only if x1 = x2 = · · · = xn = 2.?F?
08.15. Let a, b, c be nonnegative real numbers, from which at least two arenonzero and satisfying the condition ab+ bc+ ca = 1. Prove that√
a3 + a+√b3 + b+
√c3 + c ≥ 2
√a+ b+ c.
(Iran 2008)
First solution: From the Holder’s Inequality, we have(∑√a3 + a
)2(∑ a2
a2 + 1
)≥ (a+ b+ c)3,
hence it suffices to prove that
(a+ b+ c)2 ≥ 4
(a2
a2 + 1+
b2
b2 + 1+
c2
c2 + 1
),
or equivalently,
(a+ b+ c)2
ab+ bc+ ca≥ 4
[a2
(a+ b)(a+ c)+
b2
(b+ c)(b+ a)+
c2
(c+ a)(c+ b)
].
This can be rewritten into
(a+ b+ c)2
ab+ bc+ ca≥
4[a2(b+ c) + b2(c+ a) + c2(a+ b)
](a+ b)(b+ c)(c+ a)
,
or in other words,
a2 + b2 + c2
ab+ bc+ ca+
8abc
(a+ b)(b+ c)(c+ a)≥ 2.
Because the inequality is symmetric, we can suppose without loss of generality
that c = min{a, b, c}. Then, applying the evident inequalityA
B≥ A+ C
B + C∀C ≥ 0, A ≥ B > 0 with A = a2 + b2 + c2, B = ab + bc + ca and C = c2, weget
a2 + b2 + c2
ab+ bc+ ca≥ a2 + b2 + c2 + c2
ab+ bc+ ca+ c2=a2 + b2 + 2c2
(a+ c)(b+ c).
468
So, it suffices proving that
a2 + b2 + 2c2
(a+ c)(b+ c)+
8abc
(a+ b)(b+ c)(c+ a)≥ 2.
After some easy computations, we can see that this inequality is equivalent to
(a− b)2(a+ b− 2c)
(a+ b)(b+ c)(c+ a)≥ 0,
and of course, it is true since c = min{a, b, c}. Note that the equality holds if
and only if a = b = c =1√3
or (a, b, c) is a cyclic permutation of (1, 1, ).
Second solution: Since a3 + a = a3 + a(ab+ bc+ ca) = a2(a+ b+ c) + abc,the original inequality can be rewritten as∑√(
a√a+ b+ c
)2+(√
abc)2≥ 2√
(a+ b+ c)(ab+ bc+ ca).
By applying the Minkowsky’s Inequality, we have∑√(a√a+ b+ c
)2+(√
abc)2≥√(∑
a√a+ b+ c
)2+(∑√
abc)2
=√
(a+ b+ c)3 + 9abc.
Therefore, the inequality is deduced from
(a+ b+ c)3 + 9abc ≥ 4(a+ b+ c)(ab+ bc+ ca),
and of course, this is true since it is the Schur’s Inequality applied for thirddegree.
?F?
08.16. Let x, y, z be positive real numbers such that x+y+z = 3. Prove that
x3
y3 + 8+
y3
z3 + 8+
z3
x3 + 8≥ 1
9+
2
27(xy + xz + yz).
(Iran 2008)
First solution: Applying the Cauchy Schwarz Inequality, the well-knowninequality (a+b+c)2 ≥ 3(ab+bc+ca) and the AM-GM Inequality, respectively,we get
x3
y3 + 8+
y3
z3 + 8+
z3
x3 + 8≥ (x3 + y3 + z3)2
x3y3 + y3z3 + z3x3 + 8(x3 + y3 + z3)
≥ (x3 + y3 + z3)2
(x3 + y3 + z3)2
3+ 8(x3 + y3 + z3)
=3(x3 + y3 + z3)
x3 + y3 + z3 + 24= 3− 72
x3 + y3 + z3 + 24
≥ 3− 72
3x+ 3y + 3z + 18=
1
3.
469
On the other hand, from the inequality 3(xy + yz + zx) ≤ (x + y + z)2, wehave
1
9+
2
27(xy + xz + yz) ≤ 1
9+
2
81(x+ y + z)2 =
1
3.
From these two inequalities, we conclude that
x3
y3 + 8+
y3
z3 + 8+
z3
x3 + 8≥ 1
9+
2
27(xy + xz + yz),
as desired. The equality holds if and only if x = y = z = 1.
Second solution: From the AM-GM Inequality, we have
x3
y3 + 8+y + 2
27+y2 − 2y + 4
27≥ x
3,
y3
z3 + 8+z + 2
27+z2 − 2z + 4
27≥ y
3,
z3
x3 + 8+x+ 2
27+x2 − 2x+ 4
27≥ z
3.
Adding up these three inequalities, we deduce that
x3
y3 + 8+
y3
z3 + 8+
z3
x3 + 8≥
≥ x+ y + z
3− x+ y + z + 6
27− x2 + y2 + z2 − 2(x+ y + z) + 12
27
=12− x2 − y2 − z2
27=
12−[(x+ y + z)2 − 2(xy + yz + zx)
]27
=1
9+
2
27(xy + xz + yz),
as desired.?F?
08.17. Find the smallest real k such that for each x, y, z > 0, we have theinequality
x√y + y
√z + z
√x ≤ k
√(x+ y)(y + z)(z + x).
(Iran 2008)
Solution: The answer is3
2√
2. Let x = y = z = 1, we get k ≥ 3
2√
2, so the
number k must be at least3
2√
2. To show that it is our answer, it suffices to
prove that
x√y + y
√z + z
√x ≤ 3
2√
2
√(x+ y)(y + z)(z + x).
470
Indeed, from the Cauchy Schwarz Inequality and the AM-GM Inequality, wehave(
x√y + y
√z + z
√x)2 ≤ (xy + yz + zx)(x+ y + z)
= (x+ y)(y + z)(z + x) + xyz
≤ (x+ y)(y + z)(z + x) +(x+ y)(y + z)(z + x)
8
=9
8(x+ y)(y + z)(z + x).
From this, we conclude that
x√y + y
√z + z
√x ≤ 3
2√
2
√(x+ y)(y + z)(z + x),
as desired. Our proof is completed.?F?
08.18. For any positive real numbers a, b, c, d such that a2 + b2 + c2 + d2 = 1,prove the inequality
a2b2cd+ b2c2da+ c2d2ab+ d2a2bc+ c2a2db+ d2b2ac ≤ 3
32.
(Ireland 2008)
Solution: Firstly, we rewrite the inequality as
abcd(ab+ ac+ ad+ bc+ bd+ cd) ≤ 3
32.
According to the AM-GM Inequality, we have
ab+ ac+ ad+ bc+ bd+ cd =
≤ a2 + b2
2+a2 + c2
2+a2 + d2
2+b2 + c2
2+b2 + d2
2+c2 + d2
2
=3
2(a2 + b2 + c2 + d2) =
3
2,
and
abcd =√a2b2c2d2 ≤
√(a2 + b2 + c2 + d2
4
)4
=1
16.
Multiplying both two inequalities, we can get the result. Note that the equality
holds if and ony if a = b = c = d =1
2.
?F?
08.19. Let a, b, c be positive real numbers satisfying abc = 1. Prove that
1
b(a+ b)+
1
c(b+ c)+
1
a(c+ a)≥ 3
2.
(Kazakhstan 2008)
471
Solution: Firstly, we note that
1
b(a+ b)=
abc
b(a+ b)=
c
b
(1
a+
1
b
) .Therefore, the Cauchy Schwarz Inequality implies that[∑ 1
b(a+ b)
] [∑(1
a+
1
b
)]≥(∑√
c
b
)2
.
The inequality is reduced to(√b
a+
√c
b+
√a
c
)2
≥ 3
(1
a+
1
b+
1
c
),
orb
a+c
b+a
c+ 2
√a
b+ 2
√b
c+ 2
√c
a≥ 3
(1
a+
1
b+
1
c
).
The last inequality is trivial, since by the AM-GM Inequality, we have
b
a+
√c
a+
√c
a≥ 3
3
√b
a· ca
=3
a.
Remark: The inequality in question is sharpening a problem from the JuniorBalkan Mathemathetical Olympiads 2002, which stated that
1
b(a+ b)+
1
c(b+ c)+
1
a(c+ a)≥ 27
2(a+ b+ c)2.
This is a nice problem, and actually, we can show that the more general resultholds: If a, b, c are positive real numbers such that abc = 1, then for any k > 0,we have
1
b(a+ kb)+
1
c(b+ kc)+
1
a(c+ ka)≥ 3
k + 1.
?F?
08.20. Let a, b, c be positive real numbers such that (a+ b) (b+ c) (c+ a) = 8.Prove the inequality
a+ b+ c
3≥ 27
√a3 + b3 + c3
3.
(Macedonia 2008)
Solution: By the AM-GM Inequality, we have
(a+ b+ c)3 = a3 + b3 + c3 + 8(a+ b)(b+ c)(c+ a)
= a3 + b3 + c3 + 24 = 1 · (a3 + b3 + c3) + 8 · 3
≥ 9 9√
38(a3 + b3 + c3).
472
From this, we deduce that
a+ b+ c
3≥ 27
√a3 + b3 + c3
3,
as desired. The equality holds if and only if a = b = c = 1.?F?
08.21. Let a, b, c, d be positive real numbers such that a ≥ b ≥ c ≥ d andabcd = 1. Prove that
1
a3 + 1+
1
b3 + 1+
1
c3 + 1≥ 3
abc+ 1.
(MathLinks Contest 2008)
Solution: Firstly, note that if x, y are positive real numbers satisfying xy ≥ 1,then
1
x2 + 1+
1
y2 + 1− 2
xy + 1=
(x− y)2(xy − 1)
(xy + 1)(x2 + 1)(y2 + 1)≥ 0,
and thus1
x2 + 1+
1
y2 + 1≥ 2
xy + 1.
Now, using it for t2 = ab ≥√abcd = 1, we get
1
a3 + 1+
1
b3 + 1≥ 2
t3 + 1,
hence it suffices to prove that
2
t3 + 1+
1
c3 + 1≥ 3
t2c+ 1,
or in other words,
(t3c+ 2t2c2 − 2t− c)(t− c)2 ≥ 0.
For this, we proceed as follows
t3c+ 2t2c2 − 2t− c = t3c+ 2t2c2 − (2t+ c)(t2cd)3/4
≥ t3c+ 2t2c2 − (2t+ c)(t2c2)3/4
= t3c+ 2t2c2 − tc(2t+ c)√tc
≥ t3c+ 2t2c2 − 1
2tc(t+ c)(2t+ c)
=1
2tc2(t− c) ≥ 0.
Notice that the equality occurs if and only if a = b = c = d = 1.?F?
08.22. Let a, b, c be nonnegative real numbers satisfying ab + bc + ca = 3.Prove that
1
1 + a2(b+ c)+
1
1 + b2(c+ a)+
1
1 + c2(a+ b)≤ 3
1 + 2abc.
473
(MathLinks Contest 2008)
Solution: Note that the inequality is equivalent to∑ a2(b+ c)
1 + a2(b+ c)+
3
1 + 2abc≥ 3.
By the Cauchy Schwarz Inequality, we have∑ a2(b+ c)
1 + a2(b+ c)≥ 36∑
(b+ c)[1 + a2(b+ c)]=
18
9 +(∑
a)
(1− abc),
and hence it suffices to prove that
6
9 +(∑
a)
(1− abc)+
1
1 + 2abc≥ 1.
For this, we proceed by denoting with p, r the terms a + b + c, and abc,respectively. In this case, the last inequality becomes
6
9 + p(1− r)+
1
1 + 2r≥ 1,
which now rewrites as(1− r)(3− pr) ≥ 0.
This one is obviously valid, since by the AM-GM Inequality we have
r = abc ≤(ab+ bc+ ca
3
)3/2
= 1,
and
pr = abc(a+ b+ c) ≤ (ab+ bc+ ca)2
3= 3.
It is easy to see that the equality holds if and only if a = b = c = 1.?F?
08.23. Determine the least value of the expression
P = abc+1
abc
where a, b, c are positive real numbers satisfying a+ b+ c ≤ 3
2.
(Moldova 2008)
Solution: From the given hypothesis and the AM-GM Inequality, we find
that3
2≥ a + b + c ≥ 3 3
√abc, and hence abc ≤ 1
8. From now, we apply the
AM-GM Inequality again to obtain
P = abc+1
abc=
(abc+
1
64abc
)+
63
64abc≥ 1
4+
63
64 · 1
8
=65
8.
474
Note that the equality holds for a = b = c =1
2. This means that
65
8is the
minimum value of P. The problem is solved.?F?
08.24. Let a1, a2, . . . , an be positive real numbers such that a1+a2+· · ·+an ≤n
2. Determine the smallest value of the following expression
A =
√a21 +
1
a22+
√a22 +
1
a23+ · · ·+
√a2n +
1
a21.
(Moldova 2008)
First solution: The answer is
√17
2n, which can be obtained by setting a1 =
a2 = · · · = an =1
2. To prove this claim, we need to prove that A ≥
√17
2n.
Indeed, by the Cauchy Schwarz Inequality, for all x, y > 0, we have√(x2 +
1
y2
)(1
4+ 4
)≥ x
2+
2
y,
and thus √x2 +
1
y2≥ 1√
17
(x+
4
y
).
According to this inequality, we have
A ≥ 1√17
n∑i=1
(ai +
4
ai+1
)=
1√17
(n∑i=1
ai + 4n∑i=1
1
ai
)
≥ 1√17
n∑i=1
ai +4n2
n∑i=1
ai
=1√17
n∑i=1
ai +n2
4n∑i=1
ai
+15n2
4n∑i=1
ai
≥ 1√
17
n+15n2
4 · n2
=
√17
2n.
Second solution: We give another way to prove that A ≥√
17
2n. From the
Minkowsky’s Inequality and Cauchy Schwarz Inequality, we have
A ≥
√(a1 + a2 + · · ·+ an)2 +
(1
a1+
1
a2+ · · ·+ 1
an
)2
≥
√(a1 + a2 + · · ·+ an)2 +
n4
(a1 + a2 + · · ·+ an)2.
475
On the other hand, the AM-GM Inequality implies that
(a1 + a2 + · · ·+ an)2 +n4
16(a1 + a2 + · · ·+ an)2≥ n2
2,
and from the given hypothesis, we have
15n4
16(a1 + a2 + · · ·+ an)2≥ 15n4
16 ·(n
2
)2 =15n2
4.
Adding up these two inequalities, we get
(a1 + a2 + · · ·+ an)2 +n4
(a1 + a2 + · · ·+ an)2≥ 17n2
4,
and thus, we deduce that
A ≥√
17
2n,
as claimed.?F?
08.25. Find the maximum value of constant C such that the inequality
x3 + y3 + z3 + C(xy2 + yz2 + zx2) ≥ (C + 1)(x2y + y2z + z2x)
holds for any nonnegative real numbers x, y, z.(Mongolia 2008)
Solution: Let z = 0, x = 1, y = t (0 < t < 1). Then from the hypothesis thatthe inequality is true, we must have
t3 + 1 + Ct2 ≥ Ct, or C ≤ t3 + 1
t− t2
for all t ∈ (0, 1). This means that the number C satisfies the required question
cannot be greater than the local minimum of the function f(t) =t3 + 1
t− t2on
(0, 1). By some simple computions, we find that
mint∈(0,1)
f(t) = C0 =1 +√
2 +√
2√
2− 1
2+
1√√2 +
√2√
2− 1
≈ 2.4844 . . .
So, we must have C ≤ C0. Now, we will prove that C0 is our answer, i.e.
x3 + y3 + z3 + C0(xy2 + yz2 + zx2) ≥ (C0 + 1)(x2y + y2z + z2x).
Due to the cyclicity, we may assume that z = min {x, y, z} and let x − z =p, y − z = q. By this assumption, we have p, q ≥ 0. Now, after some easycomputions, we can write that above inequality in the following equivalentform
2(p2 − pq + q2)z + p3 − (C0 + 1)p2q + C0pq2 + q3 ≥ 0.
476
Because 2(p2 − pq + q2)z ≥ 0, it suffices to show that
p3 − (C0 + 1)p2q + C0pq2 + q3 ≥ 0.
Let q = tp where t ≥ 0, then this inequality is equivalent to
p3 − (C0 + 1)tp3 + C0p3t2 + t3p3 ≥ 0,
or equivalently,1− (C0 + 1)t+ C0t
2 + t3 ≥ 0.
For t = 0, it is clear. For t ≥ 1, we have C0t2 + t3 ≥ C0t + t = (C0 + 1)t, so
the above inequality also holds. Suppose that 0 < t < 1, then we may writethe inequality as
t3 + 1 ≥ C0(t− t2) or C0 ≤t3 + 1
t− t2= f(t),
which is true because C0 = mint∈(0,1)
f(t). This completes our proof.
?F?
08.26. If a, b, c are nonnegative real numbers, then the inequality holds
4(√
a3b3 +√b3c3 +
√c3a3
)≤ 4c3 + (a+ b)3.
(Poland 2008)
Solution: Put x =√a, y =
√b, z =
√c, then our inequality becomes
(x2 + y2)3 + 4z6 ≥ 4x3y3 + 4z3(x3 + y3).
By the AM-GM Inequality, we have 4z3(x3 + y3) ≤ 4z6 + (x3 + y3)2. Conse-quently, we can see that the above inequality is deduced from
(x2 + y2)3 ≥ 4x3y3 + (x3 + y3)2.
We have (x2 + y2)3 − 4x3y3 − (x3 + y3)2 = 3x2y2(x− y)2, which is obviouslynonnegative. And this finishes our proof. Note that the equality holds if andonly if a = b = c.
?F?
08.27. For real numbers xi > 1, 1 ≤ i ≤ n, n ≥ 2, such that
x2ixi − 1
≥ S =
n∑j=1
xj , for all i = 1, 2, . . . , n,
find, with proof, supS.(Romania 2008)
Solution: For n = 2, S is unbounded to the right, since for the pairs(v,
v
v − 1
), where v > 1, we have S =
v2
v − 1, and then lim
v→1
v2
v − 1=∞.
477
For n > 2, we will prove the tight upper bound S ≤ n2
n− 1, with equality
if and only if xi =n
n− 1, for all i = 1, 2, . . . , n. Consider the function f :
(1, +∞)→ [4, +∞), defined by f(x) =x2
x− 1. It is straightforward to prove
that f is strictly decreasing on (1, 2]. Let now M = max{x1, x2, . . . , xn}, then
M2
M − 1≥ S > M + (n− 1),
and therefore M <n− 1
n− 2≤ 2, which implies that xi ∈ (1, 2) for all i. Now, for
m > 1, either xi ≥ m for some i, and then S ≤ x2ixi − 1
≤ m2
m−1 (according to
the monotonicity of f), or xi ≤ m for all i, and then clearly S ≤ nm. Solving
the equation m2
m−1 = nm, obtained by equaling the two possible upper bounds
for S, yields as unique solution m0 =n
n− 1, and therefore, in all cases,
S ≤ m20
m0 − 1= nm0 =
n2
n− 1,
with equality if and only if all xi’s are equal withn
n− 1.
Remark: Note that from this inequality, we can deduce the following problemof S. Berlov, which appeared in 2005 at the Russian MO:
Real numbers a1, a2, a3 > 1 satisfy a1 + a2 + a3 = S anda2i
ai − 1> S for
i = 1, 2, 3. Prove that
1
a1 + a2+
1
a2 + a3+
1
a3 + a3> 1.
?F?
08.28. Show that for all integers n ≥ 1, we have
n
(1 +
1
2+ · · ·+ 1
n
)≥ (n+ 1)
(1
2+
1
3+ · · ·+ 1
n+ 1
).
(Romania 2008)
Solution: The inequality is equivalent to
n∑k=1
(n
k− n+ 1
k + 1
)≥ 0,
orn∑k=1
n− kk(k + 1)
≥ 0,
which is obviously true since n ≥ k for all k = 1, 2, . . . , n.
478
?F?
08.29. Let a, b, c be positive real numbers with ab+ bc+ ca = 3. Prove that
1
1 + a2(b+ c)+
1
1 + b2(c+ a)+
1
1 + c2(a+ b)≤ 1
abc.
(Romania 2008)
Solution: Using the AM-GM Inequality, we deriveab+ bc+ ca
3≥ 3√a2b2c2.
As ab+ bc+ ca = 3, then abc ≤ 1. Now∑ 1
1 + a2(b+ c)=∑ 1
1 + a(ab+ ac)=∑ 1
1 + a(3− bc)
=∑ 1
3a+ (1− abc)≤∑ 1
3a=ab+ bc+ ca
3abc=
1
abc,
as required. Note that the equality holds if and only if a = b = c = 1.?F?
08.30. Determine the maximum value of real number k such that
(a+ b+ c)
(1
a+ b+
1
b+ c+
1
c+ a− k)≥ k
for all real numbers a, b, c ≥ 0 with a+ b+ c = ab+ bc+ ca.(Romania 2008)
First solution: Observe that the numbers a = b = 2, c = 0 fulfill the condi-tion a+ b+ c = ab+ bc+ ca. Plugging into the given inequality, we derive thatk ≤ 1. We claim that the inequality holds for k = 1, proving the maximumvalue of k is 1. To this end, rewrite the inequality as follows
(ab+ bc+ ca)
(1
a+ b+
1
b+ c+
1
c+ a− 1
)≥ 1,
or equivalently, ∑ ab+ bc+ ca
a+ b≥ ab+ bc+ ca+ 1.
Since∑ ab+ bc+ ca
a+ b=∑(
ab
a+ b+ c
)=∑ ab
a+ b+a+b+c and a+b+c =
ab+ bc+ ca, the last one can be rewritten as
ab
a+ b+
bc
b+ c+
ca
c+ a≥ 1.
Notice thatab
a+ b≥ ab
a+ b+ c, since a, b, c ≥ 0. Suming over a cyclic permu-
tation of a, b, c, we get∑ ab
a+ b≥∑ ab
a+ b+ c=ab+ bc+ ca
a+ b+ c= 1,
479
as needed.
Second solution: The inequality is equivalent to the following
S =a+ b+ c
a+ b+ c+ 1
(1
a+ b+
1
b+ c+
1
c+ a
)≥ k.
Under the given condition, we get
1
a+ b+
1
b+ c+
1
c+ a=a2 + b2 + c2 + 3(ab+ bc+ ca)
(a+ b)(b+ c)(c+ a)
=a2 + b2 + c2 + 2(ab+ bc+ ca) + (a+ b+ c)
(a+ b+ c)(ab+ bc+ ca)− abc
=(a+ b+ c)(a+ b+ c+ 1)
(a+ b+ c)2 − abc,
hence
S =(a+ b+ c)2
(a+ b+ c)2 − abc.
It is now clear that S ≥ 1, and the equality holds iff abc = 0. Consequently,k = 1 is the maximum value.
?F?
08.31. Let a, b ∈ [0, 1]. Prove the inequality
1
1 + a+ b≤ 1− a+ b
2+ab
3.
(Romania 2008)
First solution: By brute force, the given inequality is equivalent to
2a2b+ 2ab2 − 3a2 − 3b2 − 4ab+ 3a+ 3b ≥ 0.
Rearrange terms, we can rewrite it as
2(a+ b)(a− 1)(b− 1) + a(1− a) + b(1− b) ≥ 0,
which holds true since all terms are nonnegative by the given condition. Notethat the equality holds if and only if a = b = 0 or a = 1, b = 0 or a = 0, b = 1.
Second solution: We have two casesCase 1. a+ b ≥ 1. Since a, b ∈ [0, 1] , we have (a− 1)(b− 1) ≥ 0 from which itfollows that ab ≥ a+ b− 1. Therefore
1− a+ b
2+ab
3− 1
1 + a+ b≥ 1− a+ b
2+a+ b− 1
3− 1
1 + a+ b
=4− a− b
6− 1
1 + a+ b
=(a+ b− 1)(2− a− b)
6(a+ b+ 1)≥ 0.
480
Case 2. 1 > a+ b ≥ 0. In this case, we have
1− a+ b
2+ab
3− 1
1 + a+ b≥ 1− a+ b
2− 1
1 + a+ b
=(a+ b)(1− a− b)
2(a+ b+ 1)≥ 0.
These two cases allow us to conclude that the following inequality holds forany a, b ∈ [0, 1]
1
1 + a+ b≤ 1− a+ b
2+ab
3,
as desired.?F?
08.32. Let n ≥ 3 is an odd integer. Determine the maximum value of thecyclic sum, for 0 ≤ xi ≤ 1, i = 1, 2, . . . , n,
E =√|x1 − x2|+
√|x2 − x3|+ · · ·+
√|xn − x1|.
(Romania 2008)
Solution: The expression E is cyclic, so we may assume without loss ofgenerality that x1 is the maximum number among x1, x2, . . . , xn. Denotingxn+1 = x1, we will prove that exists i ∈ {1, 2, . . . , n} such that (xi+1 −xi)(xi+1 − xi+2) ≤ 0. Indeed, if it does not exist a number i satisfying thisproperty, then for all i = 1, 2, . . . , n, we have
(xi+1 − xi)(xi+1 − xi+2) > 0.
If i = 1, we get (x2 − x1)(x2 − x3) > 0. Because x1 = maxxi, we obtainx2 < x3. Similarly, if i = 2, we get (x3 − x2)(x3 − x4) > 0, and thereforex3 > x4. We can proceed this way, step by step, to get easily as conclusion
that x2k+1 > max{x2k, x2k+2} for all 1 ≤ k ≤ n− 1
2(note that n is an odd
number). From this conclusion, we have
x2·n−12
+1 > max{x2·n−1
2, x2·n−1
2+2
},
or xn > max{xn−1, xn+1} = max{xn−1, x1} = x1, which is contradiction aswe have x1 ≥ xi for all i = 1, 2, . . . , n. Therefore, it must exist a number isuch that (xi+1 − xi)(xi+1 − xi+2) ≤ 0. Now, the Cauchy Schwarz Inequalityimplies that√
|xi − xi+1|+√|xi+1 − xi+2| ≤
√2(|xi − xi+1|+ |xi+1 − xi+2|).
It is easy to verify this fact: If ab ≤ 0 then |a| + |b| = |a − b|. Applying thispropriety for a = xi+1 − xi, b = xi+1 − xi+2, we have
|xi − xi+1|+ |xi+1 − xi+2| = |(xi+1 − xi)− (xi+1 − xi+2)| = |xi − xi+2| ≤ 1.
481
Therefore√|xi − xi+1|+
√|xi+1 − xi+2| ≤
√2. Moreover, because xi ∈ [0, 1],
the following inequality is trivial
∑j 6=i,j 6=i+1
√|xj − xj+1| ≤ n− 2.
So we can easily obtain that E ≤ n−2 +√
2. This is the maximum value of E
because the equality can occur for x1 =1
2, x2 = 1, x3 = 0, . . . , xn−1 = 1, xn =
0.?F?
08.33. Let a, b, c be three positive real numbers satisfying abc = 8. Provethat
a− 2
a+ 1+b− 2
b+ 1+c− 2
c+ 1≤ 0.
(Romania 2008)
Solution: The inequality in question rewrites as
1 ≤ 1
a+ 1+
1
b+ 1+
1
c+ 1.
Consider the substitutions a =2x
y, b =
2y
z, c =
2z
x, we have
1
a+ 1+
1
b+ 1+
1
c+ 1=
y2
2xy + y2+
z2
2yz + z2+
x2
2zx+ x2
≥ (x+ y + z)2
x2 + y2 + z2 + 2xy + 2yz + 2zx= 1.
?F?
08.34. Let a1, a2, . . . , an be positive real numbers satisfying the conditionthat a1 + a2 + · · ·+ an = 1. Prove that
n∑j=1
aj1 + a1 + · · ·+ aj
<1√2.
(Romania 2008)
Solution: For convenience, we take a0 = 0 and denote P as the left hand sideof the orginal inequality. Then, applying the Cauchy Schwarz Inequality, we
482
have that
P =
n∑j=1
aj1 + a0 + · · ·+ aj
=
n∑j=1
( √aj
1 + a0 + · · ·+ aj· √aj
)
≤
√√√√√ n∑j=1
aj(1 + a0 + · · ·+ aj)2
n∑j=1
aj
<
√√√√ n∑j=1
aj(1 + a0 + · · ·+ aj−1)(1 + a0 + · · ·+ aj)
=
√√√√ n∑j=1
(1
1 + a0 + · · ·+ aj−1− 1
1 + a0 + · · ·+ aj
)
=
√1− 1
1 + a0 + · · ·+ an=
1√2.
?F?
08.35. Let x, y, z be positive real numbers such that x+ y+ z = 1. Prove theinequality
1
yz + x+1
x
+1
xz + y +1
y
+1
xy + z +1
z
≤ 27
31.
(Serbia 2008)
First solution: Put a = 3x, b = 3y, c = 3z, then we have a + b + c = 3 andthe desired inequality becomes
a
3a2 + abc+ 27+
b
3b2 + abc+ 27+
c
3c2 + abc+ 27≤ 3
31.
According to the Schur’s Inequality (applied for third degree), we have
(a+ b+ c)3 + 9abc ≥ 4(a+ b+ c)(ab+ bc+ ca),
and thus abc ≥ 4(ab+ bc+ ca)− 9
3. Using this inequality, it suffices to prove
that ∑ a
3a2 +4(ab+ bc+ ca)− 9
3+ 27
≤ 3
31,
or equivalently, ∑ 3a
9a2 + 4(ab+ bc+ ca) + 72≤ 3
31.
We rewrite the latter inequality as
∑[1− 31a(a+ b+ c)
9a2 + 4(ab+ bc+ ca) + 72
]≥ 0,
483
that is ∑ 9a2 + 4(ab+ bc+ ca) + 8(a+ b+ c)2 − 31a(a+ b+ c)
a2 + s≥ 0,
where s =4(ab+ bc+ ca) + 72
9. Because 9a2 +4(ab+ bc+ ca)+8(a+ b+ c)2−
31a(a + b + c) = (7a + 8c + 10b)(c − a) − (7a + 8b + 10c)(a − b), the aboveinequality is equivalent to∑ (7a+ 8c+ 10b)(c− a)− (7a+ 8b+ 10c)(a− b)
a2 + s≥ 0,
that is ∑(a− b)
(8a+ 7b+ 10c
b2 + s− 7a+ 8b+ 10c
a2 + s
)≥ 0.
By some easy computations, we can rewrite it as∑(a− b)2 · 8a2 + 8b2 + 15ab+ 10c(a+ b) + s
(a2 + s)(b2 + s)≥ 0.
The last one is obviously true, therefore our proof is completed. Note that the
equality holds if and only if x = y = z =1
3.
?F?
08.36. Determine the least value of the expression x2 + y2 + z2, where x, y, zare real numbers such that x3 + y3 + z3 − 3xyz = 1.
(United Kingdom 2008)
Solution: Using the well-known identity x3+y3+z3−3xyz = (x+y+z)(x2+y2 + z2 − xy − yz − zx) and the Cauchy Schwarz Inequality, we find that
1 = (x3 + y3 + z3 − 3xyz)2
= (x+ y + z)2(x2 + y2 + z2 − xy − yz − zx)2
≤[
(x+ y + z)2 + 2(x2 + y2 + z2 − xy − yz − zx)
3
]3= (x2 + y2 + z2)3.
And since x2 + y2 + z2 ≥ 0, the above inequality implies that
x2 + y2 + z2 ≥ 1.
On the other hand, it is clear that for x = 1, y = z = 0, we have x3 + y3 +z3 − 3xyz = 1 and x2 + y2 + z2 = 1. This allows us to conclude that 1 is theminimum of x2 + y2 + z2, and the problem is solved.
?F?
08.37. If a, b, c, d are positive real numbers, then
(a+ b)(b+ c)(c+ d)(d+ a)(
1 +4√abcd
)4≥ 16abcd(1 + a)(1 + b)(1 + c)(1 + d).
484
(Ukraine 2008)
First solution: Firstly, we claim that for all x, y > 0, the following inequalityholds
x+ y
(1 + x)(1 + y)≥
2√xy(
1 +√xy)2 .
Indeed, we have
x+ y
2√xy− (1 + x)(1 + y)(
1 +√xy)2 =
(x+ y
2√xy− 1
)−
[(1 + x)(1 + y)(
1 +√xy)2 − 1
]
=
(√x−√y
)22√xy
−(√x−√y
)2(1 +√xy)2
=
(√x−√y
)2(xy + 1)
2√xy(1 +√xy)2 ≥ 0.
This proves our claim. Now, by using the claim and the Cauchy SchwarzInequality, we have
(a+ b)(b+ c)(c+ d)(d+ a)
(1 + a)(1 + b)(1 + c)(1 + d)≥ 4
√abcd(b+ c)(d+ a)(
1 +√ab)2 (
1 +√cd)2 ≥ 4
√abcd
(√ab+
√cd)2
(1 +√ab)2 (
1 +√cd)2
≥ 4√abcd
2√√
ab ·√cd(
1 +√√
ab ·√cd)22
=16abcd(
1 + 4√abcd
)4 .Multiplying both sides of this inequality by (1+a)(1+b)(1+c)(1+d)
(1 + 4√abcd
)4,
we get
(a+ b)(b+ c)(c+ d)(d+ a)(
1 +4√abcd
)4≥ 16abcd(1 + a)(1 + b)(1 + c)(1 + d),
as desired. Note that the equality holds if and only if a = b = c = d.
Second solution: Denote A = (a+ b+ c+ d)(abc+ bcd+ cda+ dab). Thenwe have
(a+ b)(b+ c)(c+ d)(d+ a)−A = (ac− bd)2 ≥ 0.
Hence, it suffices to prove that
A(
1 +4√abcd
)4≥ 16abcd(1 + a)(1 + b)(1 + c)(1 + d).
By expanding, we can rewrite the latter inequality as S1+S2+S3+S4+S5 ≥ 0,where
S1 = A− 16abcd,
S2 = 4A4√abcd− 16abcd(a+ b+ c+ d),
S3 = 6A√abcd− 16abcd(ab+ ac+ ad+ bc+ bd+ cd),
S4 = 4A4√a3b3c3d3 − 16abcd(abc+ bcd+ cda+ dab),
S5 = Aabcd− 16a2b2c2d2.
485
To finish the proof, we only need to prove that each term of the above sum isnonnegative. Actually, we will prove that S1 ≥ 0, S2 ≥ 0, S3 ≥ 0, S4 ≥ 0 andS5 ≥ 0. The inequalities S1 ≥ 0, S2 ≥ 0, S4 ≥ 0 and S5 ≥ 0 are obvious fromthe AM-GM Inequality (Actually, we use the inequalities a+b+c+d ≥ 4 4
√abcd
and abc+ bcd+ cda+ dab ≥ 44√a3b3c3d3 to prove S1 ≥ 0 and S5 ≥ 0. We use
abc+bcd+cda+dab ≥ 44√a3b3c3d3 to prove S2 ≥ 0 and a+b+c+d ≥ 4 4
√abcd to
prove that S4 ≥ 0). Now we will prove that S3 ≥ 0. Applying the Maclaurin’sInequality, we get
a+ b+ c+ d
4≥√ab+ ac+ ad+ bc+ bd+ cd
6,
and1
a+
1
b+
1
c+
1
d4
≥
√√√√ 1
ab+
1
ac+
1
ad+
1
bc+
1
bd+
1
cd6
.
Multiplying both two inequalities, we deduce that
(a+ b+ c+ d)(abc+ bcd+ cda+ dab)
16abcd≥ ab+ ac+ ad+ bc+ bd+ cd
6√abcd
,
and hence
A ≥ 8
3
√abcd(ab+ ac+ ad+ bc+ bd+ cd).
Using this inequality, we have
S3 ≥ 6√abcd·8
3
√abcd(ab+ac+ad+bc+bd+cd)−16abcd(ab+ac+ad+bc+bd+cd) = 0.
Thus, we have proved that S1, S2, S3, S4, S5 are all nonnegative and the resultfollows immediately.
?F?
08.38. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Provethe inequality√
a2
a2 + b+ c+
√b2
b2 + c+ a+
√c2
c2 + a+ b≤√
3.
(Ukraine 2008)
First solution: By the AM-GM Inequality, we can conclude that
a+ b+ c ≤ a2 + 1
2+b2 + 1
2+c2 + 1
2= 3.
Now, denote with P the left hand side of the original inequality, then theCauchy Schwarz Inequality implies
P 2 ≤ (a+ b+ c)
(a
a2 + b+ c+
b
b2 + c+ a+
c
c2 + a+ b
)
≤ (a+ b+ c)
∑ a
a2 +1
3(b+ c)(a+ b+ c)
.486
Therefore, it suffices to prove that
(a+ b+ c)
∑ a
a2 +1
3(b+ c)(a+ b+ c)
≤ 3.
This is a homogeneous inequality of a, b, c, so we may give up the relationa2 + b2 + c2 = 3 to normalize a + b + c = 1. By this normalization, we canrewrite the last inequality as
a
3a2 − a+ 1+
b
3b2 − b+ 1+
c
3c2 − c+ 1≤ 1.
This is true and we can prove it as follows∑ a
3a2 − a+ 1=∑ 3a
9a2 − 3a+ 3=∑ 3a
(3a− 1)2 + 3a+ 2≤∑ 3a
3a+ 2
=∑(
1− 2
3a+ 2
)= 3− 2
(1
3a+ 2+
1
3b+ 2+
1
3c+ 2
)≤ 3− 18
3(a+ b+ c) + 6= 1.
Note that the equality holds iff a = b = c = 1.
Second solution: According to the Cauchy Schwarz Inequality, we have
(a2 + b+ c)(1 + b+ c) ≥ (a+ b+ c)2.
This implies√a2
a2 + b+ c=
a√
1 + b+ c√(a2 + b+ c)(1 + b+ c)
≤ a√
1 + b+ c
a+ b+ c.
Therefore, it suffices to prove that
a√
1 + b+ c+ b√
1 + c+ a+ c√
1 + a+ b
a+ b+ c≤√
3.
We apply the Cauchy Schwarz Inequality again and obtain
a√
1 + b+ c+ b√
1 + c+ a+ c√
1 + a+ b ≤
≤√
(a+ b+ c)[a(1 + b+ c) + b(1 + c+ a) + c(1 + a+ b)]
=√
(a+ b+ c)[(a+ b+ c) + 2(ab+ bc+ ca)].
This inequality gives us an upper bound of the left hand side of the aboveinequality, that is√
(a+ b+ c)[(a+ b+ c) + 2(ab+ bc+ ca)]
a+ b+ c=
√1 +
2(ab+ bc+ ca)
a+ b+ c.
And so, it suffices to check
ab+ bc+ ca
a+ b+ c≤ 1.
487
However, this is true since from the AM-GM Inequality, we have
ab+ bc+ ca
a+ b+ c≤ a+ b+ c
3≤
a2 + 1
2+b2 + 1
2+c2 + 1
23
= 1.
?F?
08.39. Let x, y, z be distinct nonnegative real numbers. Prove that
1
(x− y)2+
1
(y − z)2+
1
(z − x)2≥ 4
xy + yz + zx.
(Vietnam 2008)
Solution: Without loss of generality, we may assume that z = min {x, y, z} .Now, observe that
(x− z)2 + (y − z)2 = (x− y)2 + 2(x− z)(y − z).
Therefore, by the AM-GM Inequality, we get
1
(x− y)2+
1
(y − z)2+
1
(z − x)2=
1
(x− y)2+
(x− y)2
(x− z)2(y − z)2+
2
(x− z)(y − z)
≥ 2
(x− z)(y − z)+
2
(x− z)(y − z)
=4
(x− z)(y − z)≥ 4
xy≥ 4
xy + yz + zx,
as desired.?F?
488
Chapter 4
Anexa 1
Tehnici de baza ın demonstrarea inegalitatilor
In compararea cantitatilor observate ın viata reala, relatia de egalitate estelocala si relativa, ın timp ce relatia de inegalitate este universala si absoluta.Esenta inegalitatii este relatia ıntre cantitati inegale.Putem compara ıntotdeauna doua cantitati de acelasi fel, ıncercand sa aratamca una este mai mare decat cealalta: cu alte cuvinte, sa demonstram o inegal-itate. Tehnicile care pot fi utilizate ın demonstrarea unei inegalitati depindde forma si natura inegalitatii, dar si de experienta si abilitatea rezolvitoru-lui, bazandu-se de multe ori pe anumite inegalitati fundamentale, cum ar fiinegalitatea mediilor (ıntre media aritmetica si media geometrica), inegali-tatea Cauchy-Schwarz, inegalitatea Schur etc.; alte tehnici implica anumiterearanjari algebrice mai avansate.In acest capitol introducem cele mai simple tehnici de baza pentru demon-strarea inegalitatilor. Subliniem faptul ca pentru demonstrarea unor ine-galitati mai dificile, adesea trebuie sa aplicam doua sau mai multe tehnici(de baza si/sau avansate). De regula, o demonstratie este frumoasa atuncicand se bazeaza pe un numar redus de tehnici simple.
4.1 Metoda compararii
In mod natural, avem doua cai pentru a compara doua cantitati:(1) Comparare prin scadere: pentru a arata ca A ≥ B, este suficient sa aratamca A−B ≥ 0;(2) Comparare prin ımpartire: presupunand ca B > 0, pentru a arata caA ≥ B, este suficient sa aratam ca A
B ≥ 1.In cadrul metodei compararii, inegalitatea de demonstrat trebuie rearanjataıntr-o forma convenabila, evidenta sau demonstrabila printr-o metoda cunos-cuta, utilizand operatii de reducere, combinare, factorizare, spargere si com-pletare cu patrate, care necesita abilitate, experienta si buna observatie dinpartea rezolvitorului.
Ex. 1 Pentru numerele reale x, y, z care satisfac xy + yz + zx = −10, aratatica:
x2 + 5y2 + 8z2 ≥ 40.
489
Proof. Deoarece
x2 + 5y2 + 8z2 − 40 = x2 + 5y2 + 8z2 + 4(xy + yz + zx)
= (x+ 2y + 2z)2 + (y − 2z)2 ≥ 0,
rezulta
x2 + 5y2 + 8z2 ≥ 40.
Egalitatea are loc atunci cand xy + yz + zx = −10, x + 2y + 2z = 0si y − 2z = 0, deci cand x = 4, y = −2, z = −1 sau x = −4, y = 2,z = 1.
Ex. 2 Fie a, b, c ∈ R+. Demonstrati ca pentru orice numere reale x, y, z, avem:
x2+y2+z2 ≥ 2
√abc
(a+ b)(b+ c)(c+ a)
(a+ b
cxy +
b+ c
ayz +
c+ a
bzx
).
[Observatie] Deoarece pentru a = b = c se obtine cunoscuta inegalitatex2+y2+z2 ≥ xy+yz+zx, echivalenta cu (x−y)2+(y−z)2+(z−x)2 ≥0, putem apela la o metoda similara pentru a demonstra inegalitateaceruta.
Proof. Scriem inegalitatea sub forma[b
b+ cx2 +
a
c+ ay2 − 2
√ab
(b+ c)(c+ a)xy
]
+
[c
c+ ay2 +
b
a+ bz2 − 2
√bc
(c+ a)(a+ b)yz
]
+
[a
a+ bz2 +
c
b+ cx2 − 2
√ca
(b+ c)(a+ b)zx
]≥ 0,
sau ∑ab
[x√
a(b+ c)− y√
b(c+ a)
]2≥ 0,
unde∑
este simbolul ınsumarii ciclice. Ultima inegalitate este evidentadevarata. Egalitatea are loc atunci cand
x√a(b+ c)
=y√
b(c+ a)=
z√c(a+ b)
.
Ex.3 Fie a, b, c ∈ R+. Aratati ca:
a2ab2bc2c ≥ ab+cbc+aca+b.
490
Proof. Impartind ambii membri prin ab+cbc+aca+b, inegalitatea poate fiscrisa astfel (a
b
)a−b(bc
)b−c ( ca
)c−a≥ 1.
Luand pe rand a ≥ b si a ≤ b, constatam ca(ab
)a−b≥ 1.
Din aceasta inegalitate si din celelelte doua similare rezulta imediat ine-galitatea ceruta. Egalitatea are loc pentru a = b = c.
[Nota] Putem scrie inegalitatea de mai sus sub forma
a3ab3bc3c ≥ aa+b+cba+b+cca+b+c,
sau
aabbcc ≥ (abc)a+b+c
3 .
In general, daca xi ∈ R+, i = 1, 2, · · · , n, avem
xx11 · xx22 · · · · · x
xnn ≥ (x1x2 · · ·xn)
x1+x2+···+xnn .
Demonstratia este similara, scriind inegalitatea sub forma
∏i<j
(xixj
)xi−xj≥ 1.
Ex. 4 Daca a, b, c sunt numere reale nenegative, atunci
a2 + b2 + c2 + 2abc+ 1 ≥ 2(ab+ bc+ ca).
Proof. Printre numerele 1−a, 1− b si 1− c exista doua cu acelasi semn;fie acestea 1− b si 1− c; deci
(1− b)(1− c) ≥ 0.
Avem
a2 + b2 + c2 + 2abc+ 1− 2(ab+ bc+ ca)
= (a− 1)2 + (b− c)2 + 2a+ 2abc− 2(ab+ ca)
= (a− 1)2 + (b− c)2 + 2a(1− b)(1− c) ≥ 0.
Egalitatea are loc daca si numai daca a = b = c = 1.
Ex. 5 Daca a, b, c sunt numere reale nenegative, aratati ca
a3 + b3 + c3 − 3abc ≥ 2
(b+ c
2− a)3
.
491
Proof. Scriem inegalitatea sub forma
2(a+ b+ c)[(a− b)2 + (b− c)2 + (c− a)2] ≥ (b+ c− 2a)3.
In cazul netrivial b + c − 2a > 0, deoarece a + b + c ≥ b + c − 2a si(b− c)2 ≥ 0, este suficient sa demonstram ca
2[(a− b)2 + (c− a)2] ≥ (b+ c− 2a)2.
Intr-adevar, cu notatiile x = b− a si y = c− a, avem
2[(a− b)2 + (c− a)2]− (b+ c− 2a)2 = 2(x2 + y2)− (x+ y)2
= (x− y)2 = (b− c)2 ≥ 0.
Egalitatea are loc atunci cand a = b = c, precum si atunci cand unuldintre numerele a, b, c este 0, iar ceilelalte sunt egale.
Ex. 6 Fie a, b, c numere reale nenegative astfel ıncat a2 + b2 + c2 = 3. Aratatica
ab2 + bc2 + ca2 ≤ abc+ 2.
Proof. Fara a pierde din generalitate, presupunem ca b este cuprins ıntrea si c, adica
(b− a)(b− c) ≤ 0.
Avem
2− ab2 − bc2 − ca2 + abc = 2− ab2 − b(3− a2 − b2)− ca2 + abc
= b3−3b+ 2−a(b2−ab+ ca− bc) = (b−1)2(b+ 2)−a(b−a)(b− c) ≥ 0.
Egalitatea are loc atunci cand a = b = c = 1, precum si atunci cand(a, b, c) este o permutatre ciclica a tripletului (0, 1,
√2) .
Ex. 7 Daca a, b, c sunt numere reale pozitive, atunci
a
b+b
c+c
a− 2 ≥ a2 + b2 + c2
ab+ bc+ ca.
Proof. Deoarece inegalitatea este ciclica, fara a pierde din generalitatepresupunem ca c = min{a, b, c}. Avem
a
b+b
c+c
a−3 =
(a
b+b
a− 2
)+
(b
c+c
a− b
a− 1
)=
(a− b)2
ab+
(a− c)(b− c)ac
sia2 + b2 + c2
ab+ bc+ ca− 1 =
(a− b)2 + (a− c)(b− c)ab+ bc+ ca
.
Rezultaa
b+b
c+c
a− 2− a2 + b2 + c2
ab+ bc+ ca
492
=
(1
ab− 1
ab+ bc+ ca
)(a− b)2 +
(1
ac− 1
ab+ bc+ ca
)(a− c)(b− c)
=c2(a+ b)(a− b)2 + b2(a+ c)(a− c)(b− c)
abc(ab+ bc+ ca)≥ 0.
Egalitatea are loc atunci cand a = b = c.
Ex. 8 Fie a, b, c ∈ R+ astfel ıncat a2 + b2 + c2 = 1. Gasiti minimul pentru
S =1
a2+
1
b2+
1
c2− 2(a3 + b3 + c3)
abc.
[Observatie] Nu ezitati sa ghiciti raspunsul corect al unei probleme deextremum, deoarece acest lucru va poate ajuta la rezolvarea problemei!Operatia de ”ghicire” a valorii extreme este strans legata de valorilevariabilelor pentru care inegalitatea devine egalitate, de care trebuie satinem seama si ın etapa de demonstrare a inegalitatii.
Proof. Pentru a = b = c = 1√3, rezulta S = 3. Vom ıncerca sa aratam
ca S ≥ 3. Intr-adevar,
S − 3 =1
a2+
1
b2+
1
c2− 3− 2(a3 + b3 + c3)
abc
=a2 + b2 + c2
a2+a2 + b2 + c2
b2+a2 + b2 + c2
c2− 3− 2
(a2
bc+b2
ca+c2
ab
)
= a2(
1
b2+
1
c2
)+ b2
(1
a2+
1
c2
)+ c2
(1
a2+
1
b2
)− 2
(a2
bc+b2
ca+c2
ab
)
= a2(
1
b− 1
c
)2
+ b2(
1
c− 1
a
)2
+ c2(
1
a− 1
b
)2
≥ 0.
Prin urmare, S are valoarea minima 3, care este atinsa atunci canda = b = c = 1√
3.
Ex. 9 Daca x ∈ R, y ≥ 0 si y(y + 1) ≤ (x+ 1)2, atunci
y(y − 1) ≤ x2.
Proof. In cazul 0 ≤ y ≤ 1, avem evident y(y − 1) ≤ 0 ≤ x2. In cazuly > 1, scriem ipoteza y(y + 1) ≤ (x+ 1)2 sub forma
y ≤√
(x+ 1)2 +1
4− 1
2,
iar inegalitatea ceruta y(y − 1) ≤ 0 ≤ x2 sub forma echivalenta
y ≤√x2 +
1
4+
1
2.
493
Prin urmare, este suficient sa aratam ca√(x+ 1)2 +
1
4− 1
2≤√x2 +
1
4+
1
2,
adica √(x+ 1)2 +
1
4≤√x2 +
1
4+ 1.
Prin ridicare la patrat, obtinem inegalitatea
2x ≤ 2
√x2 +
1
4,
care este ın mod clar adevarata. Cu aceasta, demonstratia este ıncheiata.Avem egalitate pentru x = 0 si y ∈ {0, 1}.
Ex. 10 Fie a, b, c numere reale pozitive. Aratati ca
a
b+b
c+c
a≥ 3 +
(a− c)2
ab+ bc+ ca.
Proof. Prin dezvoltare si reducere, inegalitatea devine astfel
b2 +ab2
c+a2c
b+bc2
a≥ 2ab+ 2bc.
Scriem aceasta inegalitate sub forma echivalenta(bc2
a+ ab− 2bc
)+
(ab2
c+a2c
b+ b2 − 3ab
)≥ 0.
Din inegalitatea mediilor a doua si respectiv trei numere , avem
bc2
a+ ab ≥ 2
√bc2
a· ab = 2bc
si
ab2
c+a2c
b+ b2 ≥ 3
3
√ab2
c· a
2c
b· b2 = 3ab,
iar prin ınsumarea acestor doua inegalitati obtinem inegalitatea ceruta.Avem egalitate daca si numai daca a = b = c.
Ex. 11 Daca a, b, c sunt numere reale pozitive, atunci
a(a− b)b+ c
+b(b− c)c+ a
+c(c− a)
a+ b≥ 0.
Proof. Deoarece
a(a− b)b+ c
=
(a
b+ c+ 1
)(a− b)− a+ b = (a+ b+ c)
a− bb+ c
− a+ b,
494
rezulta∑ a(a− b)b+ c
= (a+ b+ c)∑ a− b
b+ c= (a+ b+ c)
∑(a+ c
b+ c− 1
).
Prin urmare, inegalitatea dorita este echivalenta cu
(a+ b+ c)
(∑ a+ c
b+ c− 3
)≥ 0.
Aceasta inegalitate rezulta imediat din inegalitatea mediilor
∑ a+ c
b+ c≥ 3 3
√a+ c
b+ c· b+ a
c+ a· c+ b
a+ b= 3.
Avem egalitate daca si numai daca a = b = c.
Ex. 12 Pentru n ∈ N+, demonstrati ca
1
n+ 1(1 +
1
3+ · · ·+ 1
2n− 1) ≥ 1
n(1
2+
1
4+ · · ·+ 1
2n).
Proof. Scriem inegalitatea sub forma echivalenta
n(1 +1
3+ · · ·+ 1
2n− 1) ≥ (n+ 1)(
1
2+
1
4+ · · ·+ 1
2n),
saun
2+ n(
1
3+
1
5+ · · ·+ 1
2n− 1)
≥ n(1
4+
1
6+ · · ·+ 1
2n) + (
1
2+
1
4+ · · ·+ 1
2n).
Ultima inegalitate este adevarata, deoarece
n
2≥ 1
2+
1
4+ · · ·+ 1
2n
si1
3+
1
5+ · · ·+ 1
2n− 1≥ 1
4+
1
6· · ·+ 1
2n.
Avem egalitate numai pentru n = 1.
Ex. 13 Fie a, b, c ∈ R+ astfel ıncat abc = 1. Aratati ca
(a+ b)(b+ c)(c+ a) ≥ 4(a+ b+ c− 1).
Proof. Fara a pierde din generalitate, presupunem ca a ≥ 1. Inegalitateadorita este echivalenta cu
a2(b+ c) + b2(c+ a) + c2(a+ b) + 6 ≥ 4(a+ b+ c),
sau
(a2 − 1)(b+ c) + b2(c+ a) + c2(a+ b) + 6 ≥ 4a+ 3(b+ c).
495
Deoarece (a+ 1)(b+ c) ≥ 2√a · 2√bc = 4, este suficient sa aratam ca
4(a− 1) + b2(c+ a) + c2(a+ b) + 6 ≥ 4a+ 3(b+ c),
adicaa(b2 + c2) + bc(b+ c)− 3(b+ c) + 2 ≥ 0.
Aplicand inegalitatea mediilor si inegalitatea b2 + c2 ≥ 1
2(b+ c)2, avem
a(b2 + c2) + bc(b+ c) ≥ 2√abc(b2 + c2)(b+ c) = 2
√(b2 + c2)(b+ c)
≥ (b+ c)√
2(b+ c).
Prin urmare, este suficient sa demonstram ca
(b+ c)√
2(b+ c)− 3(b+ c) + 2 ≥ 0.
Cu notatia x =b+ c
2, inegalitatea poate fi scrisa astfel
2x3 − 3x2 + 1 ≥ 0.
Deoarece2x3 − 3x2 + 1 = (x− 1)2(2x+ 1) ≥ 0,
demonstratia este completa. Egalitatea se atinge atunci cand a = b =c = 1.
Ex. 14 Fie a, b, c numere pozitive astfel ıncat abc = 1. Aratati ca
a
b+b
c+
1
a≥ a+ b+ 1.
Proof. Scriem inegalitatea sub forma echivalenta(2 · a
b+b
c
)+
(b
c+
1
a
)+
(1
a+ a
)≥ 3a+ 2b+ 2.
Aplicand inegalitatea mediilor, rezulta(2 · a
b+b
c
)+
(b
c+
1
a
)+
(1
a+ a
)≥ 3
3
√a2
bc+ 2
√b
ca+ 2 = 3a+ 2b+ 2,
adica inegalitatea ceruta. Avem egalitate pentru a = b = c = 1.
[Nota] In aceleasi conditii, urmatoarea inegalitate mai ”tare” are loc:
a
b+b
c+
1
a≥√
3(a2 + b2 + 1).
Aceasta inegalitate este echivalenta cu
a
(1
b+ b2
)+
1
a≥√
3(a2 + b2 + 1),
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care, prin ridicare la patrat, devine
a2(b4 + 2b− 3 +
1
b2
)+
1
a2≥ b2 + 3− 2
b.
Deoarece
b4 + 2b− 3 +1
b2> 2b− 3 +
1
b2=
(b− 1)2(2b+ 1)
b2≥ 0,
din inegalitatea mediilor rezulta
a2(b4 + 2b− 3 +
1
b2
)+
1
a2≥ 2
√b4 + 2b− 3 +
1
b2,
ramanand de aratat ca
2
√b4 + 2b− 3 +
1
b2≥ b2 + 3− 2
b.
Efectuand o noua ridicare la patrat, obtinem
b5 − 2b3 + 4b2 − 7b+ 4 ≥ 0,
adicab(b2 − 1)2 + 4(b− 1)2 ≥ 0.
Ex. 15 Daca a, b, c sunt numere reale pozitive, atunci
1
a(a+ b)+
1
b(b+ c)+
1
c(c+ d)+
1
d(d+ a)≥ 8
(a+ c)(b+ d).
Proof. Scriem inegalitatea sub forma∑ a(b+ d) + c(b+ d)
a(a+ b)≥ 8,
echivalenta cu ∑ b+ d
a+ b+∑ c(b+ d)
a(a+ b)≥ 8.
Vom arata ca ∑ b+ d
a+ b≥ 4
si ∑ c(b+ d)
a(a+ b)≥ 4.
Tinand seama de inegalitatea
1
x+
1
y≥ 4
x+ y,
echivalenta cu (x− y)2 ≥ 0 pentru x, y > 0, avem∑ b+ d
a+ b= (b+ d)
(1
a+ b+
1
c+ d
)+ (a+ c)
(1
b+ c+
1
a+ d
)497
≥ 4(b+ d)
(a+ b) + (c+ d)+
4(a+ c)
(b+ c) + (a+ d)= 4.
De asemenea, aplicand de doua ori inegalitatea mediilor x+ y ≥ 2√xy,
echivalenta cu (√x−√y)2 ≥ 0, avem
∑ c(b+ d)
a(a+ b)= (b+d)
[c
a(a+ b)+
a
c(c+ d)
]+(a+c)
[d
b(b+ c)+
b
d(d+ a)
]
≥ 2(b+ d)√(a+ b)(c+ d)
+2(a+ c)√
(b+ c)(d+ a)≥ 4(b+ d)
(a+ b) + (c+ d)+
4(a+ c)
(b+ c) + (d+ a)= 4.
Egalitatea are loc atunci cand a = c si b = d.
Ex. 16 (Cırtoaje) Daca a, b, c sunt numere reale, atunci
a2 − bc4a2 + b2 + 4c2
+b2 − ca
4b2 + c2 + 4a2+
c2 − ab4c2 + a2 + 4b2
≥ 0.
Proof. Deoarece
4(a2 − bc)4a2 + b2 + 4c2
= 1− (b+ 2c)2
4a2 + b2 + 4c2,
putem scrie inegalitatea sub forma echivalenta
(b+ 2c)2
4a2 + b2 + 4c2+
(c+ 2a)2
4b2 + c2 + 4a2+
(a+ 2b)2
4c2 + a2 + 4b2≤ 3.
Aplicand inegalitatea Cauchy-Schwarz (vezi Ex. 38), avem
(b+ 2c)2
4a2 + b2 + 4c2=
(b+ 2c)2
(2a2 + b2) + 2(2c2 + a2)≤ b2
2a2 + b2+
2c2
2c2 + a2.
In mod similar,
(c+ 2a)2
4b2 + c2 + 4a2≤ c2
2b2 + c2+
2a2
2a2 + b2,
(a+ 2b)2
4c2 + a2 + 4b2≤ a2
2c2 + a2+
2b2
2b2 + c2.
Prin adunarea acestor inegalitati rezulta imediat inegalitatea dorita.Avem egalitate pentru a = b = c, precum si pentru a = 2b = 4c, saub = 2c = 4a, sau c = 2a = 4b.
[Nota] In mod similar putem demonstra ca pentru orice k > 0, avem
a2 − bc2ka2 + b2 + k2c2
+b2 − ca
2kb2 + c2 + k2a2+
c2 − ab2kc2 + a2 + k2b2
≥ 0,
cu egalitate pentru a = b = c, precum si pentru a = kb = k2c, saub = kc = k2a, sau c = ka = k2b.
498
4.2 Metoda majorarii si minorarii
Daca ne-am blocat ın demonstrarea directa a inegalitatii A ≤ B, putemıncerca sa gasim o cantitate C care actioneaza ca o punte ıntre A si B:daca avem A ≤ C si C ≤ B, atunci ın mod natural avem A ≤ B. Cualte cuvinte, putem majora pe A cu C, apoi pe C cu B (aceeasi idee seaplica la minorare). Important este de a gasi un nivel corespunzator demajorare sau de minorare.
Ex. 17 Demonstrati ca pentru toate numerele reale pozitive a, b, c, avem
1
a3 + b3 + abc+
1
b3 + c3 + abc+
1
c3 + a3 + abc≤ 1
abc.
[Observatie] Cand demonstram inegalitati cu fractii, adesea este binesa aducem fractiile la un numitor comun. Aceasta operatie se poaterealiza inclusiv prin majorarea sau minorarea numitorilor.
Proof. Intrucat a3 + b3 = (a+ b)(a2 + b2 − ab) ≤ (a+ b)ab, avem
1
a3 + b3 + abc≤ 1
ab(a+ b) + abc=
c
abc(a+ b+ c).
Similar,1
b3 + c3 + abc≤ a
abc(a+ b+ c),
si1
c3 + a3 + abc≤ b
abc(a+ b+ c),
Adunand cele trei inegalitati de mai sus, obtinem
1
a3 + b3 + abc+
1
b3 + c3 + abc+
1
c3 + a3 + abc≤ 1
abc,
cu egalitate daca si numai daca a = b = c.
Ex. 18 Fie a, b, c numere reale pozitive astfel ıncat a+ b+ c = 3. Aratati ca
1 + 8abc ≥ 9 min{a, b, c}.
Proof. Fara a pierde din generalitate, presupunem ca a ≤ b ≤ c. Trebuiesa aratam ca
1 + 8abc ≥ 9a.
Din a + b + c = 3 si a ≤ b ≤ c rezulta a ≤ 1, iar din (a − b)(a − c) ≥ 0obtinem
bc ≥ a(b+ c)− a2 = a(3− a)− a2 = a(3− 2a).
Prin urmare, avem
1 + 8abc− 9a ≥ 1 + 8a2(3− 2a)− 9a = (1− a)(1− 4a)2 ≥ 0.
Egalitatea are loc atunci cand a = b = c = 1, precum si atunci cand
(a, b, c) este o permutare ciclica a tripletului
(1
4,1
4,5
2
).
499
Ex. 19 Fie a, b, c numere reale pozitive astfel ıncat a ≤ b ≤ c si ab+ bc+ ca = 3.Aratati ca
1
2a+ b+
1
2b+ c+
1
2c+ a≥ 1.
Proof. Vom utiliza cunoscuta inegalitate
(x+ y + z)2 ≥ 3(xy + yz + zx),
echivalenta cu
(x− y)2 + (y − z)2 + (z − x)2 ≥ 0.
Astfel, pentru x =1
2a+ b, y =
1
2b+ csi z =
1
2c+ a, avem
(1
2a+ b+
1
2b+ c+
1
2c+ a
)2
≥ 9(a+ b+ c)
(2a+ b)(2b+ c)(2c+ a).
Ramane sa aratam ca
9(a+ b+ c) ≥ (2a+ b)(2b+ c)(2c+ a).
Scriem aceasta inegalitate sub forma omogena
3(a+ b+ c)(ab+ bc+ ca) ≥ (2a+ b)(2b+ c)(2c+ a),
echivalenta cuab2 + bc2 + ca2 ≥ a2b+ b2c+ c2a,
sau(a− b)(b− c)(c− a) ≥ 0.
Pentru a ≤ b ≤ c, ultima inegalitate este evident adevarata. Egalitateaare loc pentru a = b = c.
Ex. 20 (Schur) Daca a, b, c sunt numere reale nenegative, atunci
(1) a(a− b)(a− c) + b(b− c)(b− a) + c(c− a)(c− b) ≥ 0;
(2) a2(a− b)(a− c) + b2(b− c)(b− a) + c2(c− a)(c− b) ≥ 0.
Proof. Deoarece inegalitatea este simetrica ın raport cu a, b, c, fara apierde din generalitate vom considera ca a ≥ b ≥ c.(1) In ipotez a ≥ b ≥ c, este suficient sa aratam ca
a(a− b)(a− c) + b(b− c)(b− a) ≥ 0.
Intr-adevar, avem
a(a− b)(a− c) + b(b− c)(b− a) = (a− b)2(a+ b− c) ≥ 0.
Egalitatea are loc atunci cand a = b = c, precum si atunci cand unuldintre numerele a, b, c este 0, iar celelalte sunt egale.
500
(2) In mod similar, este suficient sa aratam ca
a2(a− b)(a− c) + b2(b− c)(b− a) ≥ 0.
Intr-adevar,
a2(a− b)(a− c) + b2(b− c)(b− a) = (a− b)2(a2 + ab+ b2 − bc− ca) ≥ 0.
Egalitatea are loc atunci cand a = b = c, precum si atunci cand unuldintre numerele a, b, c este 0, iar celelalte sunt egale.
[Nota 1] Inegalitatea Schur de gradul trei are urmatoarele forme echiva-lente, frecvent utilizate ın demonstrarea altor inegalitati:
a3 + b3 + c3 + 3abc ≥ ab(a+ b) + bc(b+ c) + ca(c+ a)
si
(a+ b+ c)3 + 9abc ≥ 4(a+ b+ c)(ab+ bc+ ca).
[Nota 2] Inegalitatea Schur de gradul patru este adevarata pentru oricenumere reale a, b, c si are formele echivalente
a4 + b4 + c4 + abc(a+ b+ c) ≥ ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2)
si
6abcp ≥ (p2 − q)(4q − p2),
unde p = a+ b+ c si q = ab+ bc+ ca.
Ex. 21 Fie a, b, c numere reale pozitive astfel ıncat a ≤ 1 ≤ b ≤ c si
a+ b+ c =1
a+
1
b+
1
c.
Aratati ca1
a2+
1
b2+
1
c2≥ a2 + b2 + c2.
Proof. Scriem inegalitatea sub forma
(a2 + c2)
(1
a2c2− 1
)≥ b2 − 1
b2,
sau
(a2 + c2)
(1
ac− 1
)(1
ac+ 1
)≥(b− 1
b
)(b+
1
b
).
Din conditia a+ b+ c =1
a+
1
b+
1
c, rezulta
(a+ c)
(1
ac− 1
)= b− 1
b≥ 0.
501
Astfel, trebuie sa aratam ca
(a2 + c2)
(1
ac+ 1
)≥ (a+ c)
(b+
1
b
).
Din inegalitatea evidenta (b− c)(
1
bc− 1
)≥ 0, rezulta
c+1
c≥ b+
1
b.
Prin urmare, este suficient sa demonstram ca
(a2 + c2)
(1
ac+ 1
)≥ (a+ c)
(c+
1
c
).
Intr-adevar, avem
(a2 + c2)
(1
ac+ 1
)− (a+ c)
(c+
1
c
)=
(1− a2)(c− a)
a≥ 0.
Egalitatea are loc pentru b = 1 si ac = 1.
Ex. 22 Fie ai ≥ 1 (i = 1, 2, · · · , n). Demonstrati ca
(1 + a1)(1 + a2) · · · (1 + an) ≥ 2n
n+ 1(1 + a1 + a2 + · · ·+ an).
[Observatie] Examinand cele doua parti ale inegalitatii de mai sus, cumputem proceda pentru a obtine 2n si ın membrul stang?
Proof. Avem
(1 + a1)(1 + a2) · · · (1 + an)
= 2n(1 +a1 − 1
2)(1 +
a2 − 1
2) · · · (1 +
an − 1
2).
Intrucat ai − 1 ≥ 0, rezulta
(1 + a1)(1 + a2) · · · (1 + an)
≥ 2n(1 +a1 − 1
2+a2 − 1
2+ · · ·+ an − 1
2)
≥ 2n(1 +a1 − 1
n+ 1+a2 − 1
n+ 1+ · · ·+ an − 1
n+ 1)
=2n
n+ 1[n+ 1 + (a1 − 1) + (a2 − 1) + · · ·+ (an − 1)]
=2n
n+ 1(1 + a1 + a2 + · · ·+ an).
Am obtinut astfel inegalitatea initiala. Pentru n ≥ 2, egalitatea are locatunci cand a1 = a2 = · · · = an = 1.
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Ex. 23 Gasiti cel mai mare numar real α, astfel ıncat
x√y2 + z2
+y√
x2 + z2+
z√x2 + y2
≥ α
pentru orice numere reale nenegative x, y, z, dintre care cel putin douanenule.
Proof. Pentru x = 0 si y = z, membrul stang al inegalitatii este egalcu 2. Prin urmare, α nu poate fi mai mare ca 2. Sa consideram asadarα = 2.
Fara a pierde din generalitate, presupunem ca x ≤ y ≤ z. Vom arata ca
x√y2 + z2
+y√
x2 + z2+
z√x2 + y2
≥√x2 + y2√x2 + z2
+
√x2 + z2√x2 + y2
≥ 2.
Inegalitatea din dreapta rezulta imediat din inegalitatea mediilor. Scriemacum inegalitatea din stanga sub forma
x√y2 + z2
≥√x2 + y2 − y√x2 + z2
+
√x2 + z2 − z√x2 + y2
,
sau
x√y2 + z2
≥ x2√x2 + z2(
√x2 + y2 + y)
+x2√
x2 + y2(√x2 + z2 + z)
.
Deoarece√x2 + y2 + y > 2y si
√x2 + z2 + z > 2z, este suficient sa
aratam ca1√
y2 + z2≥ x
2y√x2 + z2
+x
2z√x2 + y2
.
In cazul netrivial x > 0, avem
x√x2 + z2
=1√
1 + ( zx)2≤ 1√
1 + ( zy )2=
y√y2 + z2
six√
x2 + y2=
1√1 + ( yx)2
≤ 1√2.
Prin urmare, este suficient sa demonstram inegalitatea
1√y2 + z2
≥ 1
2√y2 + z2
+1
2√
2z,
care este echivalenta cu inegalitatea evidenta
√2z ≥
√y2 + z2.
Astfel, inegalitatea ceruta este demonstrata. Prin urmare, αmax = 2. Inipoteza x ≤ y ≤ z, egalitatea are loc pentru x = 0 si y = z.
503
Ex. 24 Daca x, y, z ∈ [1, 4], atunci
8
(x
y+y
z+z
x
)≥ 5
(y
x+z
y+x
z
)+ 9.
Proof. Fie
E(x, y, z) = 8
(x
y+y
z+z
x
)− 5
(y
x+z
y+x
z
)− 9.
Deoarece inegalitatea este ciclica, fara a pierde din generalitate, pre-supunem ca x = max{x, y, z}. Vom arata ca
E(x, y, z) ≥ E(x,√xz, z) ≥ 0.
Avem
E(x, y, z)− E(x,√xz, z) = 8
(x
y+y
z− 2
√x
z
)− 5
(y
x+z
y− 2
√z
x
)
=(y −
√xz)2(8x− 5z)
xyz≥ 0.
De asemenea, cu notatia t =
√x
z, 1 ≤ t ≤ 2, avem
E(x,√xz, z) = 8
(2
√x
z+z
x− 3
)− 5
(2
√z
x+x
z− 3
)
= 8
(2t+
1
t2− 3
)−5
(2
t+ t2 − 3
)=
8
t2(t−1)2(2t+1)− 5
t(t−1)2(t+2)
=(t− 1)2(4 + 5t)(2− t)
t2≥ 0.
Avem egalitate atunci cand x = y = z, precum si atunci cand (x, y, z)este o permutare ciclica a tripletului (4, 2, 1).
4.3 Metoda coeficientilor nedeterminati
Demonstrarea unor inegalitati se poate face prin spargerea acestora ıninegalitıti mai simple, cu coeficienti nedeterminati, dar care urmeaza afi calculati astfel ıncat inegalitatile respective sa fie adevarate, sau sa seajunga la rezultatul dorit. In multe cazuri, succesul metodei depinde deigeniozitatea si experienta rezolvitorului.
Ex. 25 Daca x, y, z ∈ [1, 2], atunci
3
(1
x+ 2y+
1
y + 2z+
1
z + 2x
)≥ 2
(1
x+ y+
1
y + z+
1
z + x
).
504
Proof. Presupunem ca exista o constanta reala α astfel ıncat
3
x+ 2y− 2
x+ y+ α
(1
x− 1
y
)≥ 0
pentru orice x, y ∈ [1, 2]. Atunci, ınsumand aceasta inegalitate cu cele-lalte doua similare, obtinem inegalitatea dorita. Deoarece
3
x+ 2y− 2
x+ y+ α
(1
x− 1
y
)=
(x− y)[xy − α(x+ y)(x+ 2y)]
xy(x+ y)(x+ 2y),
alegand α =1
6, rezulta
3
x+ 2y− 2
x+ y+
1
6
(1
x− 1
y
)=
(x− y)2(2y − x)
6xy(x+ y)(x+ 2y)≥ 0.
Avem egalitate daca si numai daca x = y = z.
Ex. 26 Fie x, y, z numere reale, nu toate nule. Gasiti valoarea maxima a fractiei
xy + 2yz
x2 + y2 + z2.
Proof. Pentru a gasi valoarea maxima ceruta, este suficient sa aratamca exista o constanta c, astfel ıncat
xy + 2yz
x2 + y2 + z2≤ c,
si ca semnul egal se obtine cel putin pentru un triplet x, y, z dat. Deoareceambii termeni ai sumei xy+2yz contin variabila y, vom sparge termenuly2 al sumei x2 + y2 + z2 ın αy2 si (1− α)y2. Din inegalitatea mediilor,avem
x2 + αy2 ≥ 2√α · xy
si(1− α)y2 + z2 ≥ 2
√1− α · yz,
iar prin ınsumarea acestor inegalitati obtinem
x2 + y2 + z2 ≥ 2√α · xy + 2
√1− α · yz,
x2 + y2 + z2 ≥ 2√α
(xy +
√1− αα· yz
).
Pentru a avea
√1− αα
= 2, trebuie sa alegem α =1
5. Rezulta
xy + 2yz
x2 + y2 + z2≤√
5
2,
cu egalitate pentru x =y√5
=z
2. Prin urmare, fractia
xy + 2yz
x2 + y2 + z2are
valoarea maxima
√5
2.
505
Ex. 27 Fie x, y, z numere reale nenegative, dintre care cel putin doua nenule.Demonstrati ca
x√y2 + z2
+y√
x2 + z2+
z√x2 + y2
≥ 2.
Proof. Daca exista o constanta reala α astfel ıncat
x√y2 + z2
≥ 2xα
xα + yα + zα
pentru orice numere reale nenegative x, y, z, atunci obtinem inegali-tatea dorita prin ınsumarea acestei inegalitati cu celelalte doua similare.Deoarece
xα + yα + zα ≥ 2√xα(yα + zα),
inegalitatea de mai sus este adevarata daca
x√y2 + z2
≥ xα√xα(yα + zα)
,
adicayα + zα ≥ xα−2(y2 + z2).
Este usor de observat ca ultima inegalitate este verificata ca identitatepentru α = 2. Cu aceasta, demonstratia este ıncheiata. Inegalitateainitiala devine egalitate cand unul dintre numerele x, y, z este 0, iar cele-lalte doua sunt egale.
Ex. 28 (Cirtoaje) Fie x1, x2, ..., xn numere reale pozitive astfel ıncat x1x2...xn =1. Aratati ca
1
1 + (n− 1)x1+
1
1 + (n− 1)x2+ ...+
1
1 + (n− 1)xn≥ 1.
Proof. Presupunem ca exista o constanta reala k astfel ıncat
1
1 + (n− 1)xi≥ xkixk1 + xk2 + ...+ xkn
pentru i ∈ {1, 2, ..., n}. Atunci, ınsumand toate aceste inegalitati pentrui = 1, 2, ..., n, obtinem inegalitatea dorita. Scriem inegalitatea de maisus sub forma
xk1 + ...+ xki−1 + xri+1 + ...+ xkn ≥ (n− 1)xk+1i .
Din inegalitatea mediilor, avem
xk1 + ...+ xki−1 + xki+1 + ...+ xkn ≥ (n− 1) n−1
√(x1...xi−1xi+1...xn)k
= (n− 1)x−kn−1
i ,
iar pentru k =1
n− 1 obtinem chiar inegalitatea dorita. Cu aceasta,
inegalitatea este demonstrata. Avem egalitate daca si numai daca x1 =x2 = ... = xn = 1.
506
Ex. 29 [Cırtoaje] Fie a, b, c, d numere reale pozitive astfel ıncat abcd = 1. Aratatica
1
a+ b+ 2+
1
b+ c+ 2+
1
c+ d+ 2+
1
d+ a+ 2≤ 1.
Proof. Daca exista o constanta reala k astfel ıncat
2
a+ b+ 2≤ ck + dk
ak + bk + ck + dk,
atunci putem obtine inegalitatea dorita prin ınsumarea acestei inegalitaticu celelalte inegalitati similare. Inegalitatea de mai sus este echivalentacu
(a+ b)(ck + dk) ≥ 2(ak + bk).
Deoarece
ck + dk ≥ 2(cd)k/2 =2
(ab)k/2,
este suficient sa aratam ca
a+ b ≥ (ab)k/2(ak + bk).
Aceasta inegalitate este omogena pentru k =1
2. In acest caz, inegalitatea
devinea+ b ≥ 4
√ab(√a+√b),
sau(
4√a3 − 4
√b3)( 4√a− 4√b) ≥ 0,
ultima forma fiind evident adevarata. Egalitatea are loc pentru a = b =c = d = 1.
[Nota] In general, pentru a1, a2, ..., an numere reale pozitive satisfacanda1a2...an = 1, are loc inegalitatea∑ 1
a1 + a2 + ...+ ak + n− k≤ 1,
unde 1 ≤ k ≤ n− 1.
Ex. 30 Fie a, b, c numere reale pozitive astfel ıncat a+ b+ c = 3. Aratati ca
1
a2+
1
b2+
1
c2≥ a2 + b2 + c2.
Proof. Daca exista o constanta reala k astfel ıncat
1
a2− a2 + k(a− 1) ≥ 0,
atunci putem obtine inegalitatea dorita prin ınsumarea acestei inegalitaticu celelalte inegalitati similare. Deoarece
1
a2− a2 + k(a− 1) =
(1− a)[1 + a+ (1− k)a2 + a3)
a2,
507
alegand k = 4, rezulta
1
a2− a2 + 4(a− 1) =
(1− a)2(1 + 2a− a2)a2
.
Prin urmare, pentru a ≤ 1 +√
2, avem
1
a2− a2 + 4(a− 1) ≥ 0
In continuare, fara a pierde din generalitate, consideram a ≥ b ≥ c.Avem doua cazuri.
Cazul 1. a ≤ 1 +√
2. Prin ınsumarea inegalitatilor
1
a2− a2 + 4(a− 1) ≥ 0,
1
b2− b2 + 4(b− 1) ≥ 0,
1
c2− c2 + 4(c− 1) ≥ 0,
obtinem inegalitatea dorita.
Cazul 2. a > 1 +√
2. Deoarece b+ c = 3− a < 2−√
2 <2
3, avem
bc ≤ 1
4(b+ c)2 <
1
9,
deci
1
a2+
1
b2+
1
c2>
1
a2+
1
b2≥ 2
bc> 18 > (a+ b+ c)2 > a2 + b2 + c2.
Cu aceasta, inegalitatea este demonstrata. Egalitatea are loc numai ıncazul a = b = c = 1.
Ex. 31 Fie a, b, c, d numere reale nenegative astfel ıncat
a2 + b2 + c2 + d4 = a+ b+ c+ d.
Aratati caa4 + b4 + c4 + d4 ≥ a+ b+ c+ d.
Proof. Pentru orice α ∈ R, inegalitatea dorita este echivalenta cu fiecaredin inegalitatile de mai jos:
a4 + b4 + c4 + d4 +α(a+ b+ c+ d) ≥ a+ b+ c+ d+α(a2 + b2 + c2 + d2),∑a(a3 − αa+ α− 1) ≥ 0,∑
a(a− 1)(a2 + a− α+ 1) ≥ 0.
Alegand α = 3, inegalitatea devine astfel∑a(a+ 2)(a− 1)2 ≥ 0,
fiind evident adevarata. Egalitatea are loc atunci cand a, b, c, d ∈ {0, 1}.
508
Ex. 32 (Ostrowski) Fie trei seturi de numere reale a1, a2, · · · , an, b1, b2, · · · , bnsi x1, x2, · · · , xn care satisfac relatiile
a1x1 + a2x2 + · · ·+ anxn = 0,
b1x1 + b2b2 + · · ·+ bnxn = 1.
Demonstrati ca
x21 + x22 + · · ·+ x2n ≥A
AB − C2,
unde
A =
n∑i=1
a2i 6= 0, B =
n∑i=1
b2i , C = (
n∑i=1
aibi)2.
Proof. Oricare ar fi numerele reale α si β, avem
n∑i=1
x2i =n∑i=1
x2i + αn∑i=1
aixi + β(n∑i=1
bixi − 1),
n∑i=1
x2i =
n∑i=1
(xi +αai + βbi
2)2 −
n∑i=1
(αai + βbi)2
4− β,
decin∑i=1
x2i ≥ −n∑i=1
(αai + βbi)2
4− β, (4.1)
cu egalitate daca si numai daca
xi = −αai + βbi2
, i = 1, 2, · · · , n. (4.2)
Inlocuind (4.2) ın relatiile∑n
i=1 aixi = 0 si∑n
i=1 bixi = 1, obtinem
αA+ βC = 0, αC + βB = −1,
deci
α =2C
AB − C2, β = − 2A
AB − C2. (4.3)
Astfel, din (4.1) si (4.3) rezulta
n∑i=1
x2i ≥A
AB − C2,
adica inegalitatea ceruta. Din (4.2) si (4.3) rezulta ca egalitatea are locatunci cand
xi =Abi − CaiAB − C2
, i = 1, 2, · · · , n.
509
[Nota] Inegalitatea de mai sus poate fi mai simplu demonstrata utilizandinegalitatea Cauchy-Schwarz (vezi Ex. 38). Astfel, pentru t ∈ R, avem
[
n∑i=1
(ait+ bi)2] · (
n∑i=1
x2i ) ≥ [
n∑i=1
(ait+ bi)xi]2 = 1,
(At2 + 2Ct+B)(x21 + x22 + · · ·x2n)− 1 ≥ 0,
At2 + 2Ct+B − 1
x21 + x22 + · · ·+ x2n≥ 0.
Deoarece A > 0, ultima inegalitate este adevarata pentru orice t realdaca si numai daca ∆ ≤ 0, unde
∆ = 4C2 − 4A(B − x21 − x22 − · · · − x2n)
este discriminantul trinomului de gradul doi ın t din membrul stang alinegalitatii. Din conditia ∆ ≤ 0 obtinem inegalitatea dorita.
4.4 Metoda normalizarii
Cand o inegalitate f(x1, x2, ..., xn) ≥ 0 este omogena de ordinul k, adicaf(tx1, tx2, ..., txn) = tkf(x1, x2, ..., xn), putem presupune, fara a pierdedin generalitate, ca suma variabilelor este constanta, ın scopul de a aduceinegalitatea la o forma mai simpla, mai usor demonstrabila. In gen-eral, putem considera ca g(x1, x2, ..., xn) = const, unde g(x1, x2, ..., xn)este o expresie omogena de ordin arbitrar, sau chiar f1(x1, x2, ..., xn) =f2(x1, x2, ..., xn), unde f1(x1, x2, ..., xn) si f2(x1, x2, ..., xn) sunt expresiiomogene de ordin diferit.
Ex. 33 Daca a, b, c sunt numere reale pozitive, atunci
(2a+ b+ c)2
2a2 + (b+ c)2+
(2b+ c+ a)2
2b2 + (c+ a)2+
(2c+ a+ b)2
2c2 + (a+ b)2≤ 8.
Proof. Deoarece inegalitatea este omogena, presupunem ca a+b+c = 3.Trebuie sa aratam ca
(a+ 3)2
2a2 + (3− a)2+
(b+ 3)2
2b2 + (3− b)2+
(c+ 3)2
2c2 + (3− c)2≤ 8,
adica
f(a) + f(b) + f(c) ≤ 8,
unde
f(x) =(x+ 3)2
2x2 + (3− x)2, x ∈ R+.
Avem
f(x) =x2 + 6x+ 9
3(x2 − 2x+ 3)=
1
3(1 +
8x+ 6
(x− 1)2 + 2)
510
≤ 1
3(1 +
8x+ 6
2) =
4(x+ 1)
3.
Asadar,
f(a) + f(b) + f(c) ≤ 4(a+ 1 + b+ 1 + c+ 1)
3= 8,
ceea ce trebuia demonstrat. Avem egalitate pentru a = b = c.
Ex. 34 Daca a, b, c si x, y, z sunt numere reale nenegative, atunci
[(b+ c)x+ (c+ a)y + (a+ b)z]2 ≥ 4(ab+ bc+ ca)(xy + yz + zx).
Proof. Scriem inegalitatea sub forma
2√
(ab+ bc+ ca)(xy + yz + zx) ≤ (b+ c)x+ (c+ a)y + (a+ b)z.
Deoarece inegalitatea este omogena ın raport cu x, y, z, putem consideraca x+ y+ z = a+ b+ c. In aceasta ipoteza, inegalitatea ceruta se obtinedin inegalitatea mediilor, astfel:
2√
(ab+ bc+ ca)(xy + yz + zx) ≤ (ab+ bc+ ca) + (xy + yz + zx)
=(a+ b+ c)2 − a2 − b2 − c2
2+
(x+ y + z)2 − x2 − y2 − z2
2
= (a+ b+ c)(x+ y + z)− a2 + x2
2− b2 + y2
2− c2 + z2
2
≤ (a+ b+ c)(x+ y + z)− ax− by − cz = (b+ c)x+ (c+ a)y + (a+ b)z.
In cazul abc 6= 0, egalitatea are loc daca si numai dacax
a=y
b=z
c.
Ex. 35 Fie a, b, c numere reale astfel ıncat a+ b+ c > 0 si b2 ≥ 4ac. Aratati ca
4 min{a, b, c} ≤ a+ b+ c ≤ 9
4max{a, b, c}.
Proof. Deoarece conditiile date si inegalitatea ceruta sunt omogene ınraport cu a, b, c, presupunem, fara a pierde din generalitate, ca a+b+c =1.
(A) Sa aratam ca
max{a, b, c} ≥ 4
9. (4.4)
Consideram cazul netrivial b < 49 . Din a + c = 1 − b > 5
9 , rezulta caa ≤ 1
9 implica c > 49 , iar c ≤ 1
9 implica a > 49 , inegalitatea ceruta
fiind adevarata ın ambele cazuri. Ramane de analizat cazul b < 49 si
a, c > 19 . Din b2 ≥ 4ac rezulta ac < 4
81 , deci (59 − c) · c < ac < 481 , iar din
(59−c) ·c <481 , obtinem c > 4
9 . Avem egalitate ın (4.4) pentru a = b = 49
si c = 19 , sau b = c = 4
9 si a = 19 .
(B) Sa aratam ca
min{a, b, c} ≤ 1
4. (4.5)
511
Consideram cazul netrivial a > 14 si b, c > 0. Din b2 ≥ 4ac > c, rezulta
b >√c. Prin urmare,
√c+ c < b+ c = 1− a < 3
4,
iar din√c + c < 3
4 , obtinem c < 14 . Avem egalitate ın (4.5) pentru
a = c = 14 si b = 1
2 .
Ex. 36 Daca a, b, c sunt numere reale pozitive ıncat
7(a2 + b2 + c2) = 11(ab+ bc+ ca),
atunci51
28≤ a
b+ c+
b
c+ a+
c
a+ b≤ 2.
Proof. Din considerente de omogenitate, fara a pierde din generalitate,
presupunem ca b+ c = 2. Fie x = bc, x ≤ (b+ c)2
4≤ 1. Din relatia data,
rezulta
x =7a2 − 22a+ 28
25,
iar x ≤ 1 implica1
7≤ a ≤ 3. Pe de alta parte,
a
b+ c+
b
c+ a+
c
a+ b=
a
b+ c+a(b+ c) + (b+ c)2 − 2bc
a2 + a(b+ c) + bc
=a
2+
2(a+ 2− x)
a2 + 2a+ x=
4a3 + 27a+ 11
8a2 + 7a+ 7,
iar inegalitatile cerute devin astfel
51
28≤ 4a3 + 27a+ 11
8a2 + 7a+ 7≤ 2.
Aceste inegalitati sunt adevarate deoarece
4a3 + 27a+ 11
8a2 + 7a+ 7− 51
28=
(7a− 1)(4a− 7)2
28(8a2 + 7a+ 7)≥ 0
si
2− 4a3 + 27a+ 11
8a2 + 7a+ 7=
(3− a)(2a− 1)2
8a2 + 7a+ 7≥ 0.
Avem egalitate ın inegalitatea stanga pentru 7a = b = c (sau oricepermutare ciclica). In inegalitatea din dreapta, egalitatea are loc pentrua
3= b = c (sau orice permutare ciclica).
Ex. 37 (Cırtoaje) Daca a, b, c sunt numere reale pozitive astfel ıncat a ≥ b ≥ c,atunci
a+ b+ c− 33√abc ≤ 2(
√a−√c)2
512
Proof. Din considerente de omogenitate, fara a pierde din generalitate,presupunem abc = 1. In acest caz, inegalitatea devine astfel
a+ c− b ≥ 4√b− 3.
Din (a− b)(b− c) ≥ 0, rezulta
a+ c− b ≥ ac
b=
1
b2.
Prin urmare, este suficient sa aratam ca
1
b2≥ 4√
b− 3.
Cu notatia x =1√b, rezulta
1
b2− 4√
b+ 3 = x4 − 4x+ 3 = (x− 1)2(x2 + 2x+ 3) ≥ 0.
Cu aceasta, demonstratia este ıncheiata. Egalitatea are loc pentru a =b = c.
Ex. 38 (Cauchy-Schwarz) Daca a1, a2, ..., an si x1, x2, ..., xn sunt numere reale,atunci
(a21 + a22 + ...+ a2n)(x21 + x22 + ...+ x2n) ≥ (a1x1 + a2x2 + ...+ anxn)2.
Proof. Inegalitatea este omogena atat ın raport cu variabilele a1, a2, ..., an,cat si cu variabilele x1, x2, ..., xn. Prin urmare, ın cazul netrivial ın carea21 + a22 + ...+ a2n 6= 0 si x21 + x22 + ...+ x2n 6= 0, putem presupune ca
a21 + a22 + ...+ a2n = n,
x21 + x22 + ...+ x2n = n,
pentru a demonstra ca
n ≥ |a1x1 + a2x2 + ...+ anxn|.
Intr-adevar, avem
n− |a1x1 + a2x2 + ...+ anxn| ≥ n− |a1||x1| − |a2||x2| − ...− |an||xn|
=a21 + a22 + ...+ a2n
2+x21 + x22 + ...+ x2n
2−|a1||x1|−|a2||x2|−...−|an||xn|
=(|a1| − |x1|)2 + (|a2| − |x2|)2 + ...+ (|an| − |xn|)2
2≥ 0.
Avem egalitate atunci cand seturile a1, a2, ..., an si x1, x2, ..., xn suntproportionale, adica aixi+1 = ai+1xi pentru i = 1, 2, ..., n− 1.
513
[Nota 1] Putem demonstra mai simplu inegalitatea Cauchy-Schwarzaratand ca
x21 + x22 + ...+ x2n = a1x1 + a2x2 + ...+ anxn
implicaa21 + a22 + ...+ a2n ≥ a1x1 + a2x2 + ...+ anxn.
Intr-adevar,
a21+a22+...+a
2n−a1x1−a2x2−...−anxn = (a21+a
22+...+a
2n)+(x21+a
22+...+a
2n)
−2(a1x1+a2x2+...+anxn) = (a1−x1)2+(a2−x2)2+...+(an−xn)2 ≥ 0.
[Nota 2] O alta demonstratie a inegalitatii Cauchy-Schwarz poate fidata pornind de la inegalitatea evidenta
(a1t− x1)2 + (a2t− x2)2 + ...+ (ant− xn)2 ≥ 0,
echivalenta cu
(a21 +a22 + ...+a2n)t2−2(a1x1 +a2x2 + ...+anxn)t+x21 +x22 + ...+x2n ≥ 0,
unde t este un numar real arbitrar. In cazul netrivial a21+a22+...+a2n 6= 0,inegalitatea Cauchy-Schwarz rezulta din conditia necesara si suficienta∆ ≤ 0, unde ∆ este discriminantul trinomului de gradul doi ın t dinmembrul stang al inegalitatii.
Ex. 39 (Cırtoaje) Daca a1, a2, ..., an si x1, x2, ..., xn sunt numere reale, atunci
a1x1 + a2x2 + ...+ anxn +√
(a21 + a22 + ...+ a2n)(x21 + x22 + ...+ x2n)
≥ 2
n(a1 + a2 + ...+ an)(x1 + x2 + ...+ xn).
Proof. Deoarece inegalitatea este omogena ın raport cu variabilele x1, x2, ..., xn,fara a pierde din generalitate vom considera ca
x21 + x22 + ...+ x2n = a21 + a22 + ...+ a2n.
Inegalitatea de demonstrat devine astfel
(a1+x1)2+(a2+x2)
2+...+(an+xn)2 ≥ 4
n(a1+a2+...+an)(x1+x2+...+xn).
Deoarece
[(a1+a2+...+an)+(x1+x2+...+xn)]2 ≥ 4(a1+a2+...+an)(x1+x2+...+xn),
este suficient sa aratam ca
(a1+x1)2+(a2+x2)
2+...+(an+xn)2 ≥ 1
n[(a1+a2+...+an)+(x1+x2+...+xn)]2.
Cu notatiile yi = ai + xi, i = 1, 2, ..., n, obtinem inegalitatea cunoscuta
n(y21 + y22 + ...+ y2n) ≥ (y1 + y2 + ...+ yn)2,
care rezulta imediat din inegalitatea Cauchy-Schwarz. Egalitatea are locdaca si numai daca seturile (2x−x1, 2x−x2, ..., 2x−xn) si (a1, a2, ..., an)
sunt proportionale, unde x =x1 + x2 + ...+ xn
n.
514
[Nota] Pentru xi =1
ai, i = 1, 2, ..., n, obtinem urmatoarea inegalitate
n2 + n
√(a21 + a22 + ...+ a2n)
(1
a21+
1
a22+ ...+
1
a2n
)
≥ 2(a1 + a2 + ...+ an)
(1
a1+
1
a2+ ...+
1
an
).
Ex. 40 (Cırtoaje) Fie a, b, c numere reale pozitive astfel ıncat a ≥ b ≥ c. Aratatica
a+ b+ c− 33√abc ≤ 2(a− c)2
a+ 5c.
Proof. Vom arata ca
a+ b+ c− 33√abc ≤ 2a+ c− 3
3√a2c ≤ 2(a− c)2
a+ 5c.
Inegalitatea din dreapta devine egalitate pentru c = 0. Pentru c > 0, dinconsiderente de omegenitate, vom presupune c = 1. In plus, cu notatiaa = x3, x ≥ 1, inegalitatea devine succesiv astfel:
2(x3 − 1)2 ≥ (x3 + 5)(2x3 − 3x2 + 1),
(x− 1)2[2(x2 + x+ 1)2 − (x3 + 5)(2x+ 1)] ≥ 0,
(x− 1)2(x3 + 2x2 − 2x− 1) ≥ 0,
(x− 1)3(x2 + 3x+ 1) ≥ 0,
ultima inegalitate fiind evident adevarata. Scriem acum inegalitatea dinstanga sub forma
a− b− 3 3√ac( 3√a− 3√b) ≥ 0,
sau( 3√a− 3√b)(
3√a2 +
3√ab+
3√b2 − 3 3
√ac) ≥ 0.
Ultima inegalitate este adevarata, deoarece
3√a2 +
3√ab+
3√b2 − 3 3
√ac ≥ 3
3√ab− 3 3
√ac = 3 3
√a(
3√b− 3√c) ≥ 0.
Cu aceasta, inegalitatea este demonstrata. Avem egalitate pentru a =b = c, precum si pentru a = b si c = 0.
4.5 Metoda omogenizarii
Metoda omogenizarii poate fi aplicata la demonstrarea inegalitatilor nor-malizate, atunci cand acestea devin mai simple prin omogenizare. Un-eori, pentru demonstrarea unei inegalitati normalizate se poate utilizamai ıntai metoda omogenizarii, apoi metoda normalizarii (a doua nor-malizare fiind, desigur, diferita de prima).
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Ex. 41 Fie a, b, c numere reale nenegative astfel ıncat a + b = a4 + b4. Aratatica
a3 + b3 ≥ a2 + b2.
Proof. Scriem inegalitatea sub forma omogena
(a+ b)(a3 + b3)3 ≥ (a4 + b4)(a2 + b2)3,
care este echivalenta cuA− 3B ≥ 0,
undeA = (a+ b)(a9 + b9)− (a4 + b4)(a6 + b6),
B = a2b2(a4 + b4)(a2 + b2)− a3b3(a+ b)(a3 + b3).
DeoareceA = ab(a3 − b3)(a5 − b5)
siB = a2b2(a− b)(a5 − b5),
rezultaA− 3B = ab(a− b)3(a5 − b5) ≥ 0.
Avem egalitate daca si numai daca a = b = 1.
Ex. 42 Fie a, b, c numere reale nenegative astfel ıncat ab+ bc+ ca = 3. Aratatica
1
a2 + 1+
1
b2 + 1+
1
c2 + 1≥ 3
2.
Proof. In urma dezvoltarii, inegalitatea devine astfel
a2 + b2 + c2 + 3 ≥ a2b2 + b2c2 + c2a2 + 3a2b2c2.
Din cunoscuta inegalitate
(a+ b+ c)(ab+ bc+ ca) ≥ 9abc,
echivalenta cu
a(b− c)2 + b(c− a)2 + c(a− b)2 ≥ 0,
rezultaa+ b+ c ≥ 3abc.
Astfel, este suficient sa aratam ca
a2 + b2 + c2 + 3 ≥ a2b2 + b2c2 + c2a2 + abc(a+ b+ c).
Scriem aceasta inegalitate sub forma omogena
(ab+ bc+ ca)(a2 + b2 + c2) + (ab+ bc+ ca)2 ≥ 3(a2b2 + b2c2 + c2a2)
+3abc(a+ b+ c),
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echivalenta cu
ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2) ≥ 2(a2b2 + b2c2 + c2a2),
sauab(a− b)2 + bc(b− c)2 + ca(c− a)2 ≥ 0.
Avem egalitate atunci cand a = b = c = 1, precum si atunci cand (a, b, c)este o permutare ciclica a tripletului (0,
√3,√
3).
Ex. 43 Fie a, b, c, d numere reale pozitive astfel ıncat a+ b+ c+ d = 4. Aratatica
a
a2 + 1+
b
c2 + 1+
c
d2 + 1+
d
a2 + 1≥ 2.
Proof. Avem
a
b2 + 1= a− ab2
b2 + 1≥ a− ab2
2b= a− ab
2.
In mod similar,
b
c2 + 1≥= b− bc
2,
c
d2 + 1≥= c− cd
2,
d
a2 + 1≥= d− da
2.
Prin urmare, este suficient sa aratam ca
a+ b+ c+ d− ab+ bc+ cd+ da
2≥ 2,
adica4 ≥ ab+ bc+ cd+ da.
Scriem aceasta inegalitate sub forma omogena
(a+ b+ c+ d)2 ≥ 4(ab+ bc+ cd+ da),
echivalenta cu(a− b+ c− d)2 ≥ 0.
Avem egalitate pentru a = b = c = d = 1.
Ex. 44 Fie a si b numere reale nenegative astfel ıncat 2a2 + b2 = 2a+ b. Aratatica
1− ab ≥ a− b3
.
Proof. Pentru a = b = 0, inegalitatea este adevarata. Altfel, scrieminegalitatea ın forma omogena
(2a2 + b2)2
(2a+ b)2− ab ≥ (a− b)(2a2 + b2)
3(2a+ b).
Deoarece(2a2 + b2)2
(2a+ b)2− ab =
(a− b)(4a3 − b3)(2a+ b)2
,
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ramane sa aratam ca
(a− b)[3(4a3 − b3)− (2a+ b)(2a2 + b2)] ≥ 0.
Aceasta este echivalenta cu inegalitatea evidenta
(a− b)2(4a2 + 3ab+ 2b2) ≥ 0.
Egalitatea are loc pentru a = b = 1.
Ex. 45 Fie a, b, c numere reale pozitive astfel ıncat a+ b+ c = 3. Aratati ca
a2 + 3b
a+ b+b2 + 3c
b+ c+c2 + 3a
c+ a≥ 6.
Proof. Prin omogenizare, inegalitatea devine succesiv astfel:∑ a2 + (a+ b+ c)b
a+ b≥ 2(a+ b+ c),
∑(a2 + ab+ b2 + bc
a+ b− a− b
)≥ 0,
∑ b(c− a)
a+ b≥ 0,∑
b(b+ c)(c2 − a2) ≥ 0,
ab3 + bc3 + ca3 − abc(a+ b+ c) ≥ 0.
Ultima inegalitate rezulta din inegalitatea Cauchy-Schwarz, astfel
ab3 + bc3 + ca3 ≥√ab3c+
√abc3 +
√a3bc
c+ a+ b= abc(a+ b+ c).
Cu aceasta, demonstratia este ıncheiata. Avem egalitate pentru a = b =c = 1.
Ex. 46 Fie a si b numere reale astfel ıncat 9a2 + 8ab+ 7b2 ≤ 24. Demonstrati ca
7a+ 5b+ 6ab ≤ 18.
Proof. Observam ca pentru a = b = 1 avem egalitate. Tinand seama deacest fapt, pentru a avea numai expresii omogene de gradul doi ın a si bvom utliza inegalitatile 2a ≤ a2+1 and 2b ≤ b2+1, echivalente respectivcu (a− 1)2 ≥ 0 si (b− 1)2 ≥ 0. Atunci, avem
2(7a+ 5b+ 6ab− 18) ≤ 7(a2 + 1) + 5(b2 + 1) + 12ab− 36
= 7a2 + 5b2 + 12ab− 24 = (9a2 + 8ab+ 7b2 − 24)− 2(a2 + b2 − 2ab)
≤ −2(a− b)2 ≤ 0,
de unde rezulta imediat inegalitatea dorita. Egalitatea are loc daca sinumai daca a = b = 1.
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Ex. 47 Fie a, b, c numere reale pozitive care satisfac abc = 1. Demonstrati ca
81(a2 + 1)(b2 + 1)(c2 + 1) ≤ 8(a+ b+ c)4.
Proof. Deoarece
a2 + 1 = a2 +3√a2b2c2 =
3√a2(
3√a4 +
3√b2c2
),
scriem inegalitatea ın forma omogena
81(
3√a4 +
3√b2c2
)(3√b4 +
3√c2a2
)(3√c4 +
3√a2b2
)≤ 8(a+ b+ c)4.
Pentru a demonstra noua inegalitate omogena, presupunem ca
a+ b+ c = 3.
Inegalitatea devine astfel
3
√(3√a4 +
3√b2c2
)(3√b4 +
3√c2a2
)(3√c4 +
3√a2b2
)≤ 2.
Utilizand inegalitatea mediilor, este suficient sa aratam ca∑(3√a4 +
3√b2c2
)≤ 6.
Utilizand din nou inegalitatea mediilor, avem
∑(3√a4 +
3√b2c2
)≤∑(
a2 + a+ a
3+bc+ bc+ 1
3
)
=(∑a)2 + 2
∑a+ 3
3= 6.
Cu aceasta, inegalitatea este demonstrata. Avem egalitate daca si numaidaca a = b = c = 1.
Ex. 48 Fie a, b, c numere reale nenegative care satisfac realatia
(a+ b)(b+ c)(c+ a) = 2.
Demonstrati ca
(a2 + bc)(b2 + ca)(c2 + ab) ≤ 1.
Proof. Scriem inegalitatea dorita sub forma omogena
4(a2 + bc)(b2 + ca)(c2 + ab) ≤ (a+ b)2(b+ c)2(c+ a)2.
Deoarece inegalitatea este simetrica ın a, b, c, presupunem, fara a pierdedin generalitate, ca a ≥ b ≥ c. Avem
a2 + bc ≤ a2 + ac ≤ (a+ c)2
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si, din inegalitatea mediilor,
4(b2 + ca)(c2 + ab) ≥ (b2 + ca+ c2 + ab)2.
Atunci, este suficient sa demonstram inegalitatea
b2 + c2 + ab+ ac ≥ (a+ b)(b+ c),
echivalenta cu inegalitatea evidenta c(c−b) ≤ 0. Cu aceasta, demonstratiaeste ıncheiata. Avem egalitate atunci cand unul dintre numerele a, b, ceste 0, iar celelalte doua sunt egale cu 1.
Exercise 1
1 Fie x, y, z ∈ R. Aratati ca
(x2 + y2 + z2)[(x2 + y2 + z2)2 − (xy + yz + zx)2)]
≥ (x+ y + z)2[(x2 + y2 + z2)− (xy + yz + zx)]2.
2 Fie m,n ∈ N+, m > n. Aratati ca
(1 +1
n)n < (1 +
1
m)m
3 Fie a, b, c ∈ R+. Aratati ca
1
a+
1
b+
1
c≤ a8 + b8 + c8
a3b3c3.
4 Fie numerele reale a1, a2, · · · , a100 care satisfac:
(1) a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0;
(2) a1 + a2 ≤ 100;
(3) a3 +a4 + · · ·+a100 ≤ 100. Gasiti maximul pentru a21 +a22 + · · ·+a2100.
5 Fie k, n numere ıntregi pozitive, 1 ≤ k < n. Daca x1, x2, · · · , xk suntnumere reale pozitive astfel ıncat
x1 + x2 + ...+ xk = x1x2...xk,
atunci
xn−11 + xn−12 + · · ·+ xn−1k ≥ kn.
6 Daca a, b, c ∈ R, aratati ca
(a2 + ab+ b2)(b2 + bc+ c2)(c2 + ca+ a2) ≥ (ab+ bc+ ca)3.
7 Fie a, b numere reale pozitive date, iar θ ∈ (0, π2 ). Gasiti maximul pentru
y = a√
sin θ + b√
cos θ.
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8 Fie a1, a2, ..., an numere reale astfel ıncat |ai| ≤ 2 pentru i = 1, 2, ..., n sia31 + a32 + ...+ a3n = 0. Aratati ca
a1 + a2 + ...+ an ≤2n
3.
9 Pentru n ≥ 3, fie x1, x2, · · · , xn ∈ [−1, 1] astfel ıncat
x51 + x52 + ...+ x5n = 0.
Demonstrati ca
x1 + x2 + ...+ xn ≤8n
15.
10 Fie x1, x2, · · · , xn numere reale astfel ıncat x1+x2+...+xn = na. Aratatica
n∑k=1
(xk − a)2 ≤ 1
2(n∑k=1
|xk − a|)2.
11 Daca x, y, z ≥ 0, atunci
x(y + z − x)2 + y(z + x− y)2 + z(x+ y − z)2 ≥ 3xyz.
12 Daca x, y, z sunt numere reale pozitive, atunci
1
(x+ y)2+
1
(y + z)2+
1
(z + x)2≥ 9
4(xy + yz + zx).
13 (Cırtoaje, Ji Chen) Daca a, b, c si x, y, z sunt numere reale nenegative,atunci
2
(a+ b)(x+ y)+
2
(b+ c)(y + z)+
2
(c+ a)(z + x)≥ 9
(b+ c)x+ (c+ a)y + (a+ b)z.
14 Fie x, y, z numere pozitive astfel ıncat xyz = 1. Demonstrati ca
x
x3 + y + z+
y
y3 + z + x+
z
z3 + x+ y≤ 1.
15 Daca a, b, c sunt numere reale pozitive astfel ıncat a+ b+ c = 3, atunci
6(a
b+b
c+c
a+ 3 ≥ 7(a2 + b2 + c2).
16 Daca a, b, c,d sunt numere reale nenegative astfel ıncat a+ b+ c+d = 4,atunci
a2bc+ b2cd+ c2da+ d2ab ≤ 4.
17 Daca a, b, c sunt numere reale nenegative, cel mult unul egal cu 0, atunci
1
b2 − bc+ c2+
1
c2 − ca+ a2+
1
a2 − ab+ b2≥ 12
(a+ b+ c)2.
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18 Daca a, b, c sunt numere reale nenegative, cel mult unul egal cu 0, atunci
ab− bc+ ca
b2 + c2+bc− ca+ ab
c2 + a2+ca− ab+ bc
a2 + b2≥ 3
2.
19 (Cırtoaje) Daca a ≥ b ≥ c > 0, atunci
3(a− c)2
4(a+ b+ c)≤ a+ b+ c− 3
3√abc ≤ 4(a− c)2
a+ b+ c.
20 Fie a, b, c numere reale pozitive astfel ıncat abc = 1. Aratati ca
3 + a
(1 + a)2+
3 + b
(1 + b)2+
3 + c
(1 + c)2≥ 3.
21 Daca a, b, c, d sunt numere reale pozitive astfel ıncat a + b + c + d = 4,atunci
1
ab+
1
bc+
1
cd+
1
da≥ a2 + b2 + c2 + d2.
22 Daca 0 < a ≤ b ≤ c ≤ 0, atunci
a
b+b
c+c
a≥ 2a
b+ c+
2b
c+ a+
2c
a+ b.
23 Daca a, b, c, d 6= 13 sunt numere reale pozitive care satisfac relatia
abcd = 1,
atunci
1
(3a− 1)2+
1
(3b− 1)2+
1
(3c− 1)2+
1
(3d− 1)2≥ 1.
24 Daca a, b, c sunt numere reale pozitive care satisfac relatia ab+bc+ca = 3,atunci
bc+ 2
a2 + 2+ca+ 2
b2 + 2+ab+ 2
c2 + 2≥ 3.
25 Fie a, b, c numere reale nenegative, dintre care cel mult unul egal cu 0.Demonstrati ca
a(b+ c)
b2 + bc+ c2+
b(c+ a)
c2 + ca+ a2+
c(a+ b)
a2 + ab+ b2≥ 2.
26 Fie a, b, c, d, e numere reale astfel ıncat a+b+c+d+e = 0. Demonstratica
a2 + b2 + c2 + d2 + e2 ≥ 3(ab+ bc+ cd+ de+ ea).
27 Daca a1, a2, ...an ∈ [1, 2] , atunci
n∑i=1
3
ai + 2ai+1≥
n∑i=1
2
ai + ai+1,
unde an+1 = a1.
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