inequalities of olympiad caliber · 2012-03-08 · inequalities of olympiad caliber jose luis d´...
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Inequalities of OlympiadCaliber
Jose Luis Dıaz-BarreroRSME Olympiad Committee
BARCELONA TECH
Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA [email protected]
Basic facts to proveinequalities
Hereafter, some useful facts for proving inequalities are presented:
1. If x ≥ y and y ≥ z then holds x ≥ z for any x, y, z ∈ R.
2. If x ≥ y and a ≥ b then x+ a ≥ y + b for any x, y, a, b ∈ R.
3. If x ≥ y then x+ z ≥ y + z for any x, y, z ∈ R.
4. If a ≥ b and c > 0 then ac ≥ bc for any a, b ∈ R.
5. If a ≥ b and c < 0 then bc ≥ ac for any a, b ∈ R.
6. If x ≥ y and a ≥ b then xa ≥ yb for any x, y ∈ R+ or a, b ∈ R+.
7. Let a, b, c ∈ R such that a ≥ b ≥ c then a+ b ≥ a+ c ≥ b+ c.
8. Let x, y, z, t ∈ R such that x ≥ y ≥ z ≥ t then x+y+z ≥ x+y+t ≥ x+z+t ≥y + z + t.
9. If x ∈ R then x2 ≥ 0 with equality if and only if x = 0.
10. If Ai ∈ R+ and xi ∈ R, (1 ≤ i ≤ n) then holds
A1x21 +A2x
22 +A3x
23 + . . .+Anx
2n ≥ 0,
with equality if and only if x1 = x2 = x3 = . . . = xn = 0.
1
Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA [email protected]
Inequalities Warm-up1. Prove the following statements:
1. If ab > 0, then holdsa
b+b
a≥ 2. If ab < 0, then holds
a
b+b
a≤ −2.
2. Let a > 1, b < 1. Then, a+ b > 1 + ab holds.
3. Let a, b be positive numbers, then holds (a+ b)
√a+ b
2≥ a√b+ b
√a.
4. Let a, b be positive numbers, then holds1
2(a+ b) +
1
4≥√a+ b
2.
5. If a > 1, then1
a− 1+
1
a+
1
a+ 1>
3
aholds. (Pietro Mengoli 1625–1686)
6. Let a, b be positive numbers such that a+ b = 1, then holds(a+
1
a
)2
+
(b+
1
b
)2
≥25
2
2. Prove the following statements:
1. Let a, b, c be nonnegative real numbers, then holds6a+ 4b+ 5c ≥ 5
√ab+ 3
√bc+ 7
√ca.
2. Let a, b, c be positive numbers such that a ≤ b ≤ c, then holds
a ≤3
1/a+ 1/b+ 1/c≤ 3√abc ≤
a+ b+ c
3≤
√a2 + b2 + c2
3≤ c
3. Let a, b, c be the length of the sides of a triangle ABC. Then, holds√a(a+ c− b) +
√b(a+ b− c) +
√c(b+ c− a) ≤
√(a2 + b2 + c2)(a+ b+ c).
4. Let a, b, c be positive numbers such that a+ b+ c = 1, then holds(1 +
1
a
)(1 +
1
b
)(1 +
1
c
)≥ 64.
5. Let a, b, c be positive numbers, then holds
a
b+ c+
b
c+ a+
c
a+ b≥
3
2(Nesbitt 1903)
6. Let a, b, c be positive numbers lying in the interval (0, 1]. Then holds
a
1 + b+ ca+
b
1 + c+ ab+
c
1 + a+ bc≤ 1
I hope you enjoy solving the preceding proposals and be sure that I will be verypleased checking and discussing your nice solutions.
2
Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA [email protected]
SET 1
1. Let x, y, z be strictly positive real numbers. Prove that(x
y+
z
3√xyz
)2
+
(y
z+
x
3√xyz
)2
+
(z
x+
y
3√xyz
)2
≥ 12
2. Let x, y, and z be three distinct positive real numbers such that
x+
√y +√z = z +
√y +√x
Prove that 40xz < 1.
3. Let a, b, and c be positive real numbers. Prove that(5a− bb+ c
)2
+
(5b− cc+ a
)2
+
(5c− aa+ b
)2
≥ 12
4. Let a, b, c be positive real numbers such that a+ b+ c = 2. Prove that
bc4√3a2 + 4
+ca
4√3b2 + 4
+ab
4√3c2 + 4
≤2
3
4√3
5. Let a, b, c be three positive numbers such that ab+ bc+ ca = 1. Prove that
(a2 + b2 + c2
)4 (a2 3
√a
b+ c+ b2 3
√b
c+ a+ c2 3
√c
a+ b
)3
≥1
2
6. Let a, b, c be three positive numbers such that a2 + b2 + c2 = 1. Prove that(1
a3(b+ c)5+
1
b3(c+ a)5+
1
c3(a+ b)5
)1/5
≥3
2
I hope you enjoy solving the preceding proposals and be sure that I will be verypleased checking and discussing your nice solutions.
3
Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA [email protected]
Solutions Warm-up1. Prove the following statements:
1. If ab > 0, then holdsa
b+b
a≥ 2. If ab < 0, then holds
a
b+b
a≤ −2.
2. Let a > 1, b < 1. Then, a+ b > 1 + ab holds.
3. Let a, b be positive numbers, then holds (a+ b)
√a+ b
2≥ a√b+ b
√a.
4. Let a, b be positive numbers, then holds1
2(a+ b) +
1
4≥√a+ b
2.
5. If a > 1, then1
a− 1+
1
a+
1
a+ 1>
3
aholds. (Pietro Mengoli 1625–1686)
6. Let a, b be positive numbers such that a+ b = 1, then holds(a+
1
a
)2
+
(b+
1
b
)2
≥25
2
Solution 1.1 From the inequality (x − 1)2 ≥ 0 or x2 + 1 ≥ 2x immediately followsx + 1/x ≥ 2 after division by x > 0. Equality holds when x = 1. Likewise, from(x + 1)2 ≥ 0 or x2 + 1 ≥ −2x we get x + 1/x ≤ −2 after division by x < 0 (fact5). Equality holds when x = −1. An alternative proof can be obtained finding themaximum and minimum of the function f(x) = x+ 1/x. Finally, putting x = a/b inthe preceding the inequalities claimed are proven.
Solution 1.2 From a > 1 and b < 1 or a− 1 > 0 and 1− b > 0, we have that
a+ b− 1− ab = a(1− b) + b− 1 = (a− 1)(1− b) > 0
and we are done.
Solution 1.3 We have a + b ≥ 2√ab and
√a+ b
2=
√(√a)2 + (
√b)2
2≥√a+√b
2on account of mean inequalities. Multiplying up the preceding we get the inequalityclaimed. Equality holds when a = b.
Solution 1.4 We observe that1
2(a+ b) +
1
4≥√a+ b
2or equivalently
a+ b
2−√a+ b
2+
1
4=
(1
2−√a+ b
2
)2
≥ 0
4
which trivially holds with equality when a+ b = 1/2.
Solution 1.5 Note that the inequality given is equivalent to1
a− 1+
1
a+ 1>
2
a. Now
applying HM-AM inequality to the positive numbers a− 1 and a+ 1 yields
21
a−1+ 1
a+1
≤(a− 1) + (a+ 1)
2= a
with equality if and only if a−1 = a+1 which is impossible. So, the inequality givenis strict.
Solution 1.6 On account that x2 + y2 ≥ 2(x+y2
)2, we have(
a+1
a
)2
+
(b+
1
b
)2
≥1
2
[(a+
1
a
)+
(b+
1
b
)]2=
1
2
(a+ b+
1
a+
1
b
)2
=1
2
(1 +
1
ab
)2
≥25
2
because from a+ b = 1 immediately follows1
2=a+ b
2≥√ab and
1
ab≥ 4. Equality
holds when a = b = 1/2 and we are done.Notice that this inequality is a generalization of the well-known inequality(
sin2 x+1
sin2 x
)2
+
(cos2 x+
1
cos2 x
)2
≥25
2
2. Prove the following statements:
1. Let a, b, c be nonnegative real numbers, then holds6a+ 4b+ 5c ≥ 5
√ab+ 3
√bc+ 7
√ca.
2. Let a, b, c be positive numbers such that a ≤ b ≤ c, then holds
a ≤3
1/a+ 1/b+ 1/c≤ 3√abc ≤
a+ b+ c
3≤
√a2 + b2 + c2
3≤ c
3. Let a, b, c be the length of the sides of a triangle ABC. Then, holds√a(a+ c− b) +
√b(a+ b− c) +
√c(b+ c− a) ≤
√(a2 + b2 + c2)(a+ b+ c).
4. Let a, b, c be positive numbers such that a+ b+ c = 1, then holds(1 +
1
a
)(1 +
1
b
)(1 +
1
c
)≥ 64.
5. Let a, b, c be positive numbers, then holds
a
b+ c+
b
c+ a+
c
a+ b≥
3
2(Nesbitt 1903)
6. Let a, b, c be positive numbers lying in the interval (0, 1]. Then holds
a
1 + b+ ca+
b
1 + c+ ab+
c
1 + a+ bc≤ 1
5
Solution 2.1 Applying AM-GM inequality, we have
5√ab+ 3
√bc+ 7
√ca ≤ 5
(a+ b
2
)+ 3
(b+ c
2
)+ 7
(c+ a
2
)= 6a+ 4b+ 5c
Equality holds when a = b = c and we are done.
Solution 2.2 We begin proving that a ≤3
1a+ 1
b+ 1
c
. Indeed, this inequality is equi-
valent to ab+ a
c≤ 2 which trivially holds on account of the fact that 0 < a ≤ b ≤ c.
Next we prove that3√abc ≤
a+ b+ c
3
To do it, we need the inequality a2+b2+c2 ≥ ab+bc+ca which follows immediatelyadding up the trivial inequalities a2 + b2 ≥ 2ab, b2 + c2 ≥ 2bc, c2 + a2 ≥ 2ca, andthe identity a3 + b3 + c3 − 3abc = (a+ b+ c)(a2 + b2 + c2 − ab− bc− ca) (that canbe easily checked). Combining the preceding results, we get a3 + b3 + c3− 3abc ≥ 0.Now putting a = x3, b = y3, c = z3, we have
x3 + y3 + z3 ≥ 3xyz ⇔x3 + y3 + z3
3≥ xyz ⇔
a+ b+ c
3≥ 3√abc
Applying this inequality to the positive numbers 1/a, 1/b and 1/c yields
1a+ 1
b+ 1
c
3≥ 3
√1
abc
Inverting terms immediately follows3
1a+ 1
b+ 1
c
≤ 3√abc. To prove that
a+ b+ c
3≤√
a2+b2+c2
3we square both sides and we get 3(a2 + b2 + c2) ≥ (a + b + c)2 or
equivalently, a2 + b2 + c2 ≥ ab+ bc+ ca. It holds as was proven before. Finally, the
inequality√a2+b2+c2
3≤ c trivially holds after squaring its both sides and rearranging
terms.
Solution 2.3 Since a, b, c are the lengths of the sides of a triangle, then a + b >c, b+ c > a and a+ c > b. Then, we can write
√a(a+ c− b) =
√a(a+ c− b)
√a+ c− b,
√b(a+ b− c) =
√b(a+ b− c)
√a+ b− c,
√c(b+ c− a) =
√c(b+ c− a)
√b+ c− a.
Applying CBS inequality to the vectors ~u =(√
a(a+ c− b),√b(a+ b− c),
√c(b+ c− a)
)and ~v =
(√a+ c− b,
√a+ b− c,
√b+ c− a
)yields(√
a(a+ c− b) +√b(a+ b− c) +
√c(b+ c− a)
)2≤(a(a+ c− b) + b(a+ b− c) + c(b+ c− a)
) (a+ c− b+ a+ b− c+ b+ c− a
)= (a2 + b2 + c2)(a+ b+ c)
from which the statement follows. Equality holds when a = b = c and we are done.
6
Solution 2.4 We have(1 +
1
a
)(1 +
1
b
)(1 +
1
c
)= 1 +
1
a+
1
b+
1
c+
1
ab+
1
bc+
1
ca+
1
abc
≥ 1 + 9 + 27 + 27 = 64
on account that from a + b + c = 1, and applying mean inequalities, immediately
follows1
abc≥ 27,
1
a+
1
b+
1
c≥ 9 and
1
ab+
1
bc+
1
ca=
c
abc+
a
abc+
b
abc=
1
abc≥ 27.
Equality holds when a = b = c = 1/3.
Solution 2.5 We have
a+ b
b+ c+b+ c
a+ b+b+ c
c+ a+c+ a
b+ c+c+ a
a+ b+a+ b
c+ a≥ 6
or
1 +2a
b+ c+ 1 +
2b
c+ a+ 1 +
2c
a+ b≥ 6
from which immediately follows
a
b+ c+
b
c+ a+
c
a+ b≥
3
2
Equality holds when a = b = c.
Solution 2.6 Since 1 + ab = (1− a)(1− b) + a+ b, then
1 + c+ ab = (1− a)(1− b) + a+ b+ c ≥ a+ b+ c
Likewise, we have
1 + a+ bc = (1− b)(1− c) + a+ b+ c ≥ a+ b+ c,
1 + b+ ca = (1− c)(1− a) + a+ b+ c ≥ a+ b+ c.
Therefore,
a
1 + b+ ca+
b
1 + c+ ab+
c
1 + a+ bc≤
a
a+ b+ c+
b
a+ b+ c+
c
a+ b+ c= 1
Equality holds when at least two of the a, b, c are equal to one, and we are done.
7
Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA [email protected]
Solutions1. Let x, y, z be strictly positive real numbers. Prove that(
x
y+
z
3√xyz
)2
+
(y
z+
x
3√xyz
)2
+
(z
x+
y
3√xyz
)2
≥ 12
Solution 1. Applying AM-QM inequality, yields
1
3
[(x
y+
z
3√xyz
)2
+
(y
z+
x
3√xyz
)2
+
(z
x+
y
3√xyz
)2]
≥1
9
(x
y+y
z+z
x+x+ y + z
3√xyz
)2
from which follows(x
y+
z
3√xyz
)2
+
(y
z+
x
3√xyz
)2
+
(z
x+
y
3√xyz
)2
≥1
3
(x
y+y
z+z
x+x+ y + z
3√xyz
)2
≥1
3
(3 3
√x
y·y
z·z
x+x+ y + z
3√xyz
)2
=1
3
(3 +
x+ y + z
3√xyz
)2
≥ 12
becausex+ y + z
3√xyz
≥ 3 on account of AM-GM inequality. Equality holds when x =
y = z and we are done. �
Solution 2. Setting a =x
3√xyz
, b =y
3√xyz
and c =z
3√xyz
into the statement yields
(a+
b
c
)2
+
(b+
c
a
)2
+
(c+
a
b
)2
≥ 12
To prove the preceding inequality we set ~u =
(a+
b
c, b+
c
a, c+
a
b
), ~v = (1, 1, 1)
into the CBS inequality and we obtain(a+
b
c
)2
+
(b+
c
a
)2
+
(c+
a
b
)2
8
=1
3(12 + 12 + 12)
[(a+
b
c
)2
+
(b+
c
a
)2
+
(c+
a
b
)2]
≥1
3
[(a+
b
c
)+
(b+
c
a
)+
(c+
a
b
)]2Now, taking into account that abc = 1 and applying the AM-GM inequality twice, weget (
a+b
c
)+
(b+
c
a
)+
(c+
a
b
)
≥ 3 3
√(a+
b
c
)(b+
c
a
)(c+
a
b
)≥ 3
3
√23
√ab
c
bc
a
ca
b= 6
Therefore, on account of the preceding, we have
1
3
[(a+
b
c
)+
(b+
c
a
)+
(c+
a
b
)]2≥ 12
and we are done. Note that equality holds when x = y = z. �
2. Let x, y, and z be three distinct positive real numbers such that
x+
√y +√z = z +
√y +√x
Prove that 40xz < 1.
Solution. From x+√y +√z = z +
√y +√x we obtain
x− z =
√y +√x−
√y +√z
and
(x−z)(√
y +√x+
√y +√z
)=
(√y +√x+
√y +√z
)(√y +√x−
√y +√z
)=√x−√z
Since x 6= z, dividing by both sides of the preceding expression by√x−√z yields
(√x+√z) (√
y +√x+
√y +√z
)= 1
On the other hand, since x, y, and z are positive, then 4√x <
√y +√x and 4
√z <√
y +√z. Therefore,
(√x+√z) (
4√x+ 4√z)<(√x+√z) (√
y +√x+
√y +√z
)= 1
Applying AM-GM inequality, we obtain 2 4√xz ≤
√x+√z, and 2 8
√xz ≤ 4
√x+ 4√z.
Multiplying up the last two inequalities, we get
4 8√
(xz)3 ≤(√x+√z) (
4√x+ 4√z)< 1
9
Thus, (xz)3/8 <1
4and
xz <
(1
4
)8/3
=1
3√65536
<1
3√64000
=1
40
from which follows 40xz < 1.�
3. Let a, b, and c be positive real numbers. Prove that(5a− bb+ c
)2
+
(5b− cc+ a
)2
+
(5c− aa+ b
)2
≥ 12
Solution. WLOG we can assume that a ≥ b ≥ c from which immediately follows
that a + b ≥ a + c ≥ b + c and1
b+ c≥
1
c+ a≥
1
a+ b. Since the first and the last
sequences are sorted in the same way, by applying rearrangement inequality, we get
2
(a
b+ c+
b
c+ a+
c
a+ b
)≥ 2
(a
a+ b+
b
b+ c+
c
c+ a
)2
(a
b+ c+
b
c+ a+
c
a+ b
)≥ 2
(b
a+ b+
c
b+ c+
a
c+ a
)a
b+ c+
b
c+ a+
c
a+ b≥
a
a+ b+
b
b+ c+
c
c+ a
Adding up the preceding inequalities, yields
5
(a
b+ c+
b
c+ a+
c
a+ b
)≥ 6 +
a
a+ b+
b
b+ c+
c
c+ a
from which we obtain
1
3
(5a− bb+ c
+5b− cc+ a
+5c− aa+ b
)≥ 2
Taking into account AM-QM inequality, we have√√√√1
3
[(5a− bb+ c
)2
+
(5b− cc+ a
)2
+
(5c− aa+ b
)2]≥
1
3
(5a− bb+ c
+5b− cc+ a
+5c− aa+ b
)≥ 2
from which the statement follows. Equality holds when a = b = c and we are done.�
4. Let a, b, c be positive real numbers such that a+ b+ c = 2. Prove that
bc4√3a2 + 4
+ca
4√3b2 + 4
+ab
4√3c2 + 4
≤2
3
4√3
Solution. Squaring both terms, we get(bc
4√3a2 + 4
+ca
4√3b2 + 4
+ab
4√3c2 + 4
)2
≤4
9
√3
10
Applying CBS to the vectors ~u =(√
bc,√ca,√ab)
and
~v =
(√bc
√3a2 + 4
,
√ca
√3b2 + 4
,
√ab
√3c2 + 4
)we obtain (
bc4√3a2 + 4
+ca
4√3b2 + 4
+ab
4√3c2 + 4
)2
≤ (ab+ bc+ ca)
(bc
√3a2 + 4
+ca
√3b2 + 4
+ab
√3c2 + 4
)From a + b + c = 2 and the well-known inequality (a + b + c)2 ≥ 3(ab + bc + ca)immediately follows ab+ bc+ ca ≤ 4/3. Equality holds when a = b = c = 2/3.
On the other hand,
√3
(bc
√3a2 + 4
)=
bc√a2 + 4/3
≤bc
√a2 + ab+ bc+ ca
=bc√
(a+ b)(a+ c)≤
1
2
(bc
a+ b+
bc
a+ c
)The last inequality holds in account of HM-GM inequality. Likewise,
√3
(ca
√3b2 + 4
)≤
1
2
(ca
b+ c+
ca
b+ a
)and
√3
(ab
√3c2 + 4
)≤
1
2
(ab
c+ a+
ab
c+ b
)Adding the preceding inequalities, we obtain
√3
(bc
√3a2 + 4
+ca
√3b2 + 4
+ab
√3c2 + 4
)
≤1
2
(ca
b+ c+
ca
b+ a+
bc
a+ b+
bc
a+ c+
ab
c+ a+
ab
c+ b
)=
1
2(a+ b+ c) = 1
Therefore,bc
√3a2 + 4
+ca
√3b2 + 4
+ab
√3c2 + 4
≤√3
3
Equality holds when a = b = c = 2/3. From the preceding, we have(bc
4√3a2 + 4
+ca
4√3b2 + 4
+ab
4√3c2 + 4
)2
≤4
3·√3
3
and the statement follows. Equality holds when a = b = c = 2/3, and we are done.�
5. Let a, b, c be three positive numbers such that ab+ bc+ ca = 1. Prove that
(a2 + b2 + c2
)4 (a2 3
√a
b+ c+ b2 3
√b
c+ a+ c2 3
√c
a+ b
)3
≥1
2
11
Solution 1. First we observe that from a2+ b2+ c2 ≥ ab+ bc+ ca and the constrain,immediately follows that a2 + b2 + c2 ≥ 1. Multiplying and dividing the LHS of theinequality claimed by (a2 + b2 + c2)3 we have
(a2 + b2 + c2)7
a2 3
√ab+c
+ b2 3
√bc+a
+ c2 3
√ca+b
a2 + b2 + c2
3
≥ (a2 + b2 + c2)7
(a2 + b2 + c2
a2 b+ca
+ b2 c+ab
+ c2 a+bc
)
=(a2 + b2 + c2)8
2(ab+ bc+ ca)=
(a2 + b2 + c2)8
2≥
1
2
In the preceding we have used the inequality f(1/3) ≥ f(−1), where
f(α) =
(w1a
α + w2bα + w3c
α
w1 + w2 + w3
)1/α
is the mean powered inequality with weights w1 =a2
a2 + b2 + c2, w2 =
b2
a2 + b2 + c2,
and w3 =c2
a2 + b2 + c2respectively. Equality holds when a = b = c = 1/
√3, and we
are done. �
Solution 2. First, using the constrain, we write the inequality claimed in the mostconvenient form
a2 3
√a
b+ c+ b2 3
√b
c+ a+ c2 3
√c
a+ b≥
13√2
(ab+ bc+ ca
a2 + b2 + c2
)4/3
=13√2
(ab+ bc+ ca
a2 + b2 + c2
)4/3
(ab+ bc+ ca)
To prove the preceding inequality, we consider the function f : [0,+∞)→ R definedby f(t) = t7/3 which is convex, as can be easily checked. Applying Jensen’s inequali-
ty, with q1 =(b+ c)2
A, q2 =
(c+ a)2
A, q3 =
(a+ b)2
A, where A = (a+ b)2+(b+ c)2+
(c+ a)2; and x1 =a
b+ c, x2 =
b
c+ a, x3 =
c
a+ b, we have
q1f
(a
b+ c
)+ q2f
(b
c+ a
)+ q3f
(c
a+ b
)
≥ f(q1
a
b+ c+ q2
b
c+ a+ q3
c
a+ b
),
or equivalently,
(b+ c)2
A
(a
b+ c
)7/3
+(c+ a)2
A
(b
c+ a
)7/3
+(a+ b)2
A
(c
a+ b
)7/3
≥((a(b+ c) + b(c+ a) + c(a+ b)
A
)7/3
12
Rearranging terms, and after simplification, we obtain
a2 3
√a
b+ c+ b2 3
√b
c+ a+ c2 3
√c
a+ b≥
[2(ab+ bc+ ca)
]7/3[2(a2 + b2 + c2 + ab+ bc+ ca)
]4/3
≥
[2(ab+ bc+ ca)
]7/3[4(a2 + b2 + c2)
]4/3 =13√2
(ab+ bc+ ca
a2 + b2 + c2
)4/3
(ab+ bc+ ca)
on account of the well-known inequality a2 + b2 + c2 ≥ ab+ bc+ ca. Equality holdswhen a = b = c = 1/
√3, and we are done.
�
6. Let a, b, c be three positive numbers such that a2 + b2 + c2 = 1. Prove that(1
a3(b+ c)5+
1
b3(c+ a)5+
1
c3(a+ b)5
)1/5
≥3
2
Solution 1. To prove the inequality claimed we consider the function f : R → Rdefined by f(t) = t5. This function is convex in [0,+∞) as can be easily checked.Applying Jensen’s inequality, namely
3∑k=1
qkf(xk) ≥ f(
3∑k=1
qkxk
)
with
q1 =a2
a2 + b2 + c2, q2 =
b2
a2 + b2 + c2, q3 =
c2
a2 + b2 + c2
and
x1 =1
a(b+ c), x2 =
1
b(c+ a), x3 =
1
c(a+ b),
yields
1
a2 + b2 + c2
[a2
(1
a(b+ c)
)5
+ b2(
1
b(c+ a)
)5
+ c2(
1
c(a+ b)
)5]
≥
(ab+c
+ bc+a
+ ca+b
)5(a2 + b2 + c2)5
or equivalently,
1
a3(b+ c)5+
1
b3(c+ a)5+
1
c3(a+ b)5≥(
a
b+ c+
b
c+ a+
c
a+ b
)5
where we have used the constrain a2 + b2 + c2 = 1. Now, it is easy to see that
a
b+ c+
b
c+ a+
c
a+ b≥
3
2
13
Indeed, WLOG we can assume that a ≥ b ≥ c from which follows a+b ≥ a+c ≥ b+cand 1
b+c≥ 1
c+a≥ 1
a+b. Now, applying rearrangement inequality, we get
a
b+ c+
b
c+ a+
c
a+ b≥
a
a+ b+
b
b+ c+
c
c+ aa
b+ c+
b
c+ a+
c
a+ b≥
b
a+ b+
c
b+ c+
a
c+ a
Adding the preceding expressions, we obtain
a
b+ c+
b
c+ a+
c
a+ b≥
3
2(Nesbitt’s Inequality)
Equality holds when a = b = c.
Finally, substituting this result in the preceding the statement follows. Equalityholds when a = b = c =
√3/3, and we are done. �
Solution 2. We recall that Holder’s inequality can be stated as follows: Let a1, a2, ..., anand b1, b2, ..., bn, (n ≥ 2), be sequences of positive real numbers and let p, q ∈ R∗+ suchthat 1
p+ 1
q= 1, then the following inequality holds(
n∑i=1
api
) 1p(
n∑i=1
bqi
) 1q
≥n∑i=1
aibi
Particularizing the preceding result for n = 3 with p = 5/4, q = 5 for which (1/p +1/q = 1), we get (
3∑i=1
a5/4i
) 45(
3∑i=1
b5i
) 15
≥3∑i=1
aibi
Putting in the above result a1 = a8/5, a2 = b8/5, a3 = c8/5, b1 =1
a3/5(b+ c), b2 =
1
b3/5(c+ a), b3 =
1
c3/5(a+ b)yields
[(a8/5
)5/4+(b8/5
)5/4+(c8/5
)5/4]4/5 [( 1
a3/5(b + c)
)5
+
(1
b3/5(c + a)
)5
+
(1
c3/5(a + b)
)5]1/5
≥a
b+ c+
b
c+ a+
c
a+ b
or equivalently,
(a2+b2+c2)4/5(
1
a3(b+ c)5+
1
b3(c+ a)5+
1
c3(a+ b)5
)1/5
≥a
b+ c+
b
c+ a+
c
a+ b
from which the statement follows on account of the constrain and Nesbitt’s inequal-ity. Equality holds when when a = b = c =
√3/3, and we are done.
�
Solution 3 by Emilio Fernandez Moral. On account of mean inequalities we have(1
a3(b+ c)5+
1
b3(c+ a)5+
1
c3(a+ b)5
)1/5
≥(
33√a3b3c3(a+ b)5(b+ c)5(c+ a)5
)1/5
14
=5
√3
abc
13√(a+ b)(b+ c)(c+ a)
≥31/5+1/2
2 (abc)1/5≥
3
2
because 3√(a+ b)(b+ c)(c+ a) ≤
(a+ b) + (b+ c) + (c+ a)
3=
2(a+ b+ c)
3≤
2
√1
3and (abc)1/5 = (a2b2c2)1/10 =
(3√a2b2c2
)3/10≤(a2 + b2 + c2
3
)3/10
=
(1
3
)3/10
and 31/5+1/2+3/10 = 3. Equality holds when a = b = c =√3/3. �
15
Jose Luis Dıaz-BarreroRSME Olympic CommitteeUPC–BARCELONA [email protected]
Some ResultsHereafter, some classical discrete inequalities are stated and proven. We begin with
Theorem 1 (General Mean Inequalities) Let a1, a2, . . . , an be positive real numbers.Then, the function f : R→ R defined by
f(α) =
(aα1 + aα2 + . . .+ aαn
n
)1/α
is nondecreasing and the following limits hold
limα→−∞
f(α) = min1≤i≤n
{ai
}, lim
α→0f(α) = n
√a1a2 . . . an, lim
α→+∞f(α) = max
1≤i≤n
{ai
}
Proof. Let 0 < a < b. Consider the function g : [0,+∞)→ R defined by g(α) = αb/a.Since
g′′(α) =b
a
(b
a− 1
)αb/a−2 ≥ 0, (α ≥ 0),
then g is convex in [0,+∞). Applying Jensen’s inequality, we get
g
1
n
n∑j=1
αaj
≤ 1
n
n∑j=1
g(αaj )
or 1
n
n∑j=1
αaj
b/a ≤ 1
n
n∑j=1
αbj
from which follows f(a) ≤ f(b). For negative values of α consider the functionh(α) = (−α)b/a which is convex in (−∞, 0).
Let L1 = limα→−∞
f(α). Then, lnL1 = limα→−∞
ln
(aα1 + aα2 + . . .+ aαn
n
)1/α
. That is,
lnL1 = limα→−∞
[1
αln
(aα1 + aα2 + . . .+ aαn
n
)]= lim
α→−∞
aα1 ln a1 + aα2 ln a2 + . . .+ aαn ln an
aα1 + aα2 + . . .+ aαn
= ln
(min
1≤i≤n
{ai
})
16
from which follows L1 = min1≤i≤n
{ai
}. Likewise, we obtain that L2 = lim
α→+∞f(α) =
max1≤i≤n
{ai
}. Finally, denoting by L = lim
α→0f(α), we have
lnL = limα→0
[1
αln
(aα1 + aα2 + . . .+ aαn
n
)]= lim
α→0
aα1 ln a1 + aα2 ln a2 + . . .+ aαn ln an
aα1 + aα2 + . . .+ aαn
=ln a1 + ln a2 + . . .+ ln an
n= ln n
√a1a2 . . . an
from which immediately follows limα→0
f(α) = n√a1a2 . . . an and the proof is complete.
�In 1821 Cauchy published his famous inequality as the second of the two noteson the theory of inequalities that formed the final part of his book Cours d’AnalyseAlgebrique. Namely,
Theorem 2 Let a1, a2, ..., an, b1, b2, ..., bn be any real numbers. Then, the followinginequality holds: (
n∑k=1
akbk
)2
≤(
n∑k=1
a2k
)(n∑k=1
b2k
)
Proof. Consider the quadratic polynomial A(x) =
n∑k=1
(akx− bk)2. Then,
A(x) = x2n∑k=1
a2k − 2x
n∑k=1
akbk +
n∑k=1
b2k ≥ 0
Since the preceding equation is nonnegative its discriminant must be less or equalthan zero. That is, (
n∑k=1
akbk
)2
−(
n∑k=1
a2k
)(n∑k=1
b2k
)≤ 0
from which the statement follows. Equality holds when the n-tuples (a1, a2, . . . , an)and (b1, b2, . . . , bn) are proportional. This completes the proof.
�A generalization of CBS inequality is the well known inequality of Holder. It is statedand proved in the following theorem.
Theorem 3 Let a1, a2, ..., an and b1, b2, ..., bn, (n ≥ 2) be sequences of positive realnumbers and let p, q ∈ R∗+ such that 1
p+ 1
q= 1, then the following inequality holds(
n∑i=1
api
) 1p(
n∑i=1
bqi
) 1q
≥n∑i=1
aibi.
Proof 1. We will argue by induction. Assume that the inequality holds for n and wehave to prove it for n+ 1. In fact,
n+1∑i=1
aibi =
n∑i=1
aibi + an+1bn+1 ≤(
n∑i=1
api
) 1p(
n∑i=1
bqi
) 1q
+ an+1bn+1
17
≤(
n∑i=1
api + apn+1
) 1p(
n∑i=1
bqi + bqn+1
) 1q
=
(n+1∑i=1
api
) 1p(n+1∑i=1
bqi
) 1q
.
To complete the inductive process, we must verify that the case when n = 2 alsoholds. Indeed, the inequality
a1b1 + a2b2 ≤ (ap1 + ap2)1p (bq1 + bq2)
1q ,
immediate follows from the fact that the function
f(x) = (ap1 + ap2)1p (bq1 + xq)
1q − a1b1 − a2x,
is strictly positive for all x > 0. This completes the proof.
Proof 2. First, we write the inequality in the most convenient form∑ni=1 aibi(∑n
i=1 api
) 1p(∑n
i=1 bqi
) 1q
≤ 1
or equivalently,n∑i=1
(api∑ni=1 a
pi
) 1p(
bqi∑ni=1 b
qi
) 1q
≤ 1.
Now, using the power mean inequality,
x1αy
1β ≤
1
αx+
1
βy,
we have
n∑i=1
(api∑ni=1 a
pi
) 1p(
bqi∑ni=1 b
qi
) 1q
≤n∑i=1
[1
p
api∑ni=1 a
pi
+1
q
bqi∑ni=1 b
qi
]
=1
p
∑ni=1 a
pi∑n
i=1 api
+1
q
∑ni=1 b
qi∑n
i=1 bqi
=1
p+
1
q= 1,
as desired. Equality holds if and only if the n-tuples (ap1, ap2, ..., a
pn) and (bq1, b
q2, ..., b
qn)
are proportional. �Notice that for p = q = 2, we get the inequality
(a1b1 + a2b2 + ...+ anbn)2 ≤ (a2
1 + a22 + ...+ a2
n)(b21 + b22 + ...+ b2n)
This is the Cauchy-Bunyakowski-Schwarz inequality.
Using Holder’s inequality can be easily proven the following inequality of Minkowski.
Theorem 4 Let a1, a2, ..., an and b1, b2, ..., bn, (n ≥ 2) be sequences of positive realnumbers and let p > 1, be a real number, then(
n∑i=1
(ai + bi)p
) 1p
≤(
n∑i=1
api
) 1p
+
(n∑i=1
bpi
) 1p
When p < 1 inequality reverses.
18
Proof. Let be q = pp−1
, then 1p+ 1
q= 1, and applying Holder’s inequality, we obtain
n∑i=1
(ai + bi)p =
n∑i=1
ai(ai + bi)p−1 +
n∑i=1
bi(ai + bi)p−1
≤(
n∑i=1
api
) 1p(
n∑i=1
(ai + bi)(p−1)q
) 1q
+
(n∑i=1
bpi
) 1p(
n∑i=1
(ai + bi)(p−1)q
) 1q
=
( n∑i=1
api
) 1p
+
(n∑i=1
bpi
) 1p
( n∑i=1
(ai + bi)p
)p−1p
Multiplying the preceding inequality by(∑n
i=1(ai + bi)p)−p−1
p , we get(n∑i=1
(ai + bi)p
) 1p
≤(
n∑i=1
api
) 1p
+
(n∑i=1
bpi
) 1p
and the proof is complete. Equality holds if and only if (a1, a2, ..., an) and (b1, b2, ..., bn)are proportional.
�
A most useful result is presented in the following
Theorem 5 (Rearrangement) Let a1, a2, · · · , an and b1, b2, · · · , bn be sequences ofpositive real numbers and let c1, c2, · · · , cn be a permutation of b1, b2, · · · , bn. Thesum S = a1b1 + a2b2 + · · ·+ anbn is maximal if the two sequences a1, a2, · · · , an andb1, b2, · · · , bn are sorted in the same way and minimal if the two sequences are sortedoppositely, one increasing and the other decreasing.
Proof. Let ai > aj. Consider the sums
S1 = a1c1 + · · ·+ aici + · · ·+ ajcj + · · ·+ ancn
S2 = a1c1 + · · ·+ aicj + · · ·+ ajci + · · ·+ ancn
We have obtained S2 from S1 by switching the positions of ci and cj. Then
S1 − S2 = aici + ajcj − aicj − ajci = (ai − aj)(ci − cj)
Therefore,ci > cj ⇒ S1 > S2 and ci < cj ⇒ S1 < S2.
This completes the proof. �
In the sequel, after defining the concepts of convex and concave function, the ine-quality of Jensen is presented.
Definition. Let f : I ⊆ R → R be a real function. We say that f is convex (concave)on I if for all a1, a2 ∈ I and for all t ∈ [0, 1], we have
f(ta1 + (1− t)a2) ≤ f(a1) + (1− t)f(a2)
When f is concave the inequality reverses.
19
Theorem 6 Let f : I ⊆ R→ R be a convex function. Let q1, q2, . . . , qn be nonnegativereal numbers such that q1 + q2 + . . .+ qn = 1. Then for all ak ∈ I (1 ≤ k ≤ n) holds
f
(n∑k=1
qkak
)≤
n∑k=1
qkf(ak)
Equality holds when a1 = a2 = . . . = an. If f is concave the preceding inequalityreverses.
Proof. We will argue by mathematical induction. For n = 2, we have
f(q1a1 + q2a2) ≤ q1f(a1) + q2f(a2)
The preceding inequality holds because f is convex and q1 + q2 = 1. Assume that
f
(n∑k=1
qkak
)≤
n∑k=1
qkf(ak)
and we have to see that
f
(n+1∑k=1
qkak
)≤
n+1∑k=1
qkf(ak)
when q1 + q2 + . . . + qn + qn+1 = 1. Let q =∑nk=1 qk. Then, we have
n∑k=1
qk
q= 1,
n∑k=1
(qk
q
)ak ∈ I and q + qn+1 = 1. Now, taking into account the case when n = 2
and the inductive hypothesis, yields
f
(n+1∑k=1
qkak
)= f
[q
(n∑k=1
(qk
q
)ak
)+ qn+1an+1
]
≤ qf
(n∑k=1
(qk
q
)ak
)+ qn+1f(an+1)
≤ q
n∑k=1
(qk
q
)f(ak) + qn+1f(an+1)
=
n+1∑k=1
qkf(ak)
and by the principle of mathematical induction (PMI) the statement is proven. Equa-lity holds when a1 = a2 = . . . = an.
�
20
Think about
“Pulchra dicuntur quae visa placent– beauty is that which, be-ing seen, pleases”. This definition, applies well to Mathematicalbeauty in which lack of understanding is so often responsible oflack of pleasure.
Thomas Aquina’s definition of Beauty