Winter Camp 2008 Inequalities Yufei Zhao

InequalitiesYufei Zhao

yufeiz@mit.edu

1 Classical Theorems

Theorem 1. (AM-GM) Let a1, · · · , an be positive real numbers. Then, we have

a1 + · · ·+ ann

≥ n√a1 · · · an.

Theorem 2. (Cauchy-Schwarz) Let a1, · · · , an, b1, · · · , bn be real numbers. Then,

(a12 + · · ·+ an

2)(b12 + · · ·+ bn2) ≥ (a1b1 + · · ·+ anbn)2.

Theorem 3. (Jensen) Let f : [a, b]→ R be a convex function. Then for any x1, x2, . . . , xn ∈ [a, b] and anynonnegative reals ω1, ω2, . . . , ωn with ω1 + ω2 + · · ·+ ωn = 1, we have

ω1f(x1) + ω2f(x2) + · · ·+ ωnf(xn) ≥ f(ω1x1 + ω2x2 + · · ·+ ωnxn).

If f is concave, then the inequality is flipped.

Theorem 4. (Weighted AM-GM) Let ω1, · · · , ωn > 0 with ω1 + · · · + ωn = 1. For all x1, · · · , xn > 0,we have

ω1x1 + ω2x2 + · · ·+ ωnxn ≥ xω11 xω2

2 · · ·xωnn .

Theorem 5. (Schur) Let x, y, z be nonnegative real numbers. For any r > 0, we have∑cyclic

xr(x− y)(x− z) ≥ 0.

Definition 1. (Majorization) Let x = (x1, x2, . . . xn) and y = (y1, y2, . . . , yn) be two sequences of realnumbers. Then x is said to majorize y (denoted x � y) if the following conditions are satisfied

• x1 ≥ x2 ≥ x3 · · · ≥ xn and y1 ≥ y2 ≥ y3 · · · ≥ yn; and

• x1 + x2 + · · ·+ xk ≥ y1 + y2 + · · ·+ yk, for k = 1, 2, . . . , n− 1; and

• x1 + x2 + · · ·+ xn = y1 + y2 + · · ·+ yn.

Theorem 6. (Muirhead)1 Suppose that (a1, . . . , an) � (b1, . . . , bn), and x1, . . . xn are positive real numbers,then ∑

sym

xa11 x

a22 · · ·xan

n ≥∑sym

xb11 xb22 · · ·xbn

n .

where the symmetric sum is taken over all n! permutations of x1, x2, . . . , xn.

Theorem 7. (Karamata’s Majorization inequality) Let f : [a, b] → R be a convex function. Supposethat (x1, · · · , xn) � (y1, · · · , yn), where x1, · · · , xn, y1, · · · , yn ∈ [a, b]. Then, we have

f(x1) + · · ·+ f(xn) ≥ f(y1) + · · ·+ f(yn).

1Practical notes about Muirhead: (1) don’t try to apply Muirhead when there are more than 3 variables, since mostly likelyyou won’t succeed (and never, ever try to use Muirhead when the inequality is only cyclic but not symmetric, since it is incorrectto use Muirhead there) (2) when writing up your solution, it is probably safer to just deduce the inequality using weightedAM-GM by finding the appropriate weights, as this can always be done. The reason is that it is not always clear that Muirheadwill be accepted as a quoted theorem.

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Winter Camp 2008 Inequalities Yufei Zhao

Theorem 8. (Power Mean) Let x1, · · · , xn > 0. The power mean of order r is defined by

M(x1,··· ,xn)(0) = n√x1 · · ·xn , M(x1,··· ,xn)(r) =

(xr1 + · · ·+ xn

r

n

) 1r

(r 6= 0).

Then, M(x1,··· ,xn) : R −→ R is continuous and monotone increasing.

Theorem 9. (Bernoulli) For all r ≥ 1 and x ≥ −1, we have

(1 + x)r ≥ 1 + rx.

Definition 2. (Symmetric Means) For given arbitrary real numbers x1, · · · , xn, the coefficient of tn−i inthe polynomial (t+ x1) · · · (t+ xn) is called the i-th elementary symmetric function σi. This means that

(t+ x1) · · · (t+ xn) = σ0tn + σ1t

n−1 + · · ·+ σn−1t+ σn.

For i ∈ {0, 1, · · · , n}, the i-th elementary symmetric mean Si is defined by

Si =σi(ni

) .Theorem 10. Let x1, . . . , xn > 0. For i ∈ {1, · · · , n}, we have

(1) (Newton) Si

Si+1≥ Si−1

Si,

(2) (Maclaurin) Si1i ≥ Si+1

1i+1 .

Theorem 11. (Rearrangement) Let x1 ≥ · · · ≥ xn and y1 ≥ · · · ≥ yn be real numbers. For anypermutation σ of {1, . . . , n}, we have

n∑i=1

xiyi ≥n∑i=1

xiyσ(i) ≥n∑i=1

xiyn+1−i.

Theorem 12. (Chebyshev) Let x1 ≥ · · · ≥ xn and y1 ≥ · · · ≥ yn be real numbers. We have

x1y1 + · · ·+ xnynn

≥(x1 + · · ·+ xn

n

)(y1 + · · ·+ yn

n

).

Theorem 13. (Holder)2 Let x1, · · · , xn, y1, · · · , yn be positive real numbers. Suppose that p > 1 and q > 1satisfy 1

p + 1q = 1. Then, we have

n∑i=1

xiyi ≤

(n∑i=1

xip

) 1p(

n∑i=1

yiq

) 1q

.

More generally, let xij (i = 1, · · · ,m, j = 1, · · ·n) be positive real numbers. Suppose that ω1, · · · , ωn arepositive real numbers satisfying ω1 + · · ·+ ωn = 1. Then, we have

n∏j=1

(m∑i=1

xij

)ωj

≥m∑i=1

n∏j=1

xijωj

.

Theorem 14. (Minkowski)3 If x1, · · · , xn, y1, · · · , yn > 0 and p > 1, then(n∑i=1

xip

) 1p

+

(n∑i=1

yip

) 1p

≥

(n∑i=1

(xi + yi)p

) 1p

2Think of this as generalized Cauchy, as you can use it for more than two sequences.3Think of this as generalized triangle inequality.

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Winter Camp 2008 Inequalities Yufei Zhao

2 A motivating example

In this section we discuss several common techniques in inequalities through following famous problem fromIMO 2001 by Hojoo Lee:

Prove thata√

a2 + 8bc+

b√b2 + 8ca

+c√

c2 + 8ab≥ 1

for all positive real numbers a, b and c.

The official solution is short and mysterious:

Official solution: First we prove that

a√a2 + 8bc

≥ a43

a43 + b

43 + c

43,

or equivalently, that(a

43 + b

43 + c

43 )2 ≥ a 2

3 (a2 + 8bc).

Indeed, this follows from applying AM-GM as follows:

(a43 +b

43 +c

43 )2 = (a

43 )2+(b

43 +c

43 )(a

43 +a

43 +b

43 +c

43 ) ≥ (a

43 )2+2b

23 c

23 ·4a 2

3 b13 c

13 = a

83 +8a

23 bc = a

23 (a2+8bc).

We have similar inequalities for the other two terms, and so

a√a2 + 8bc

+b√

b2 + 8ca+

c√c2 + 8ab

≥ a43

a43 + b

43 + c

43

+b

43

a43 + b

43 + c

43

+c

43

a43 + b

43 + c

43

= 1.

�While the solution looks nice and short, it leaves us wonder how in the world could anyone come up with

it. In particular, where did the exponent 43 come from? Here we provide some motivation.

Isolated fudging. It is not unusual to compare individual terms of an inequality to expressions such as

ar

ar + br + cr

because if the comparison turns out to be successful, we can finish off the problem right away. (Techniques ofthis form are sometimes called isolated fudging, meaning that the effort is focused on manipulating individualterms, as opposed to the inequality as a whole.) So, suppose we guess that it is possible to have an inequalityof the form

a√a2 + 8bc

≥ ar

ar + br + cr.

Now how can we pick the candidates for r? Blind guess and check will probably get us nowhere. Luckily,there is a method that will give you the unique candidate for r (though with no promise that this r willwork). Be prepared, this method will require some calculus.

Suppose that some r works. Let us consider the function

f(a, b, c) =a√

a2 + 8bc− ar

ar + br + cr.

What do we know about f(a, b, c)? Because of the inequality that we want, we need f(a, b, c) ≥ 0 for alla, b, c > 0. Also, by considering the point of equality, we see that (1, 1, 1) must be a local minimum of f . Soconsider the partial derivative of f with respect to a (denoted ∂f/∂a), meaning that we differentiate f withrespect to a while treating the other variables as if they were constants. Since (1, 1, 1) is a local minimum,this partial derivative when evaluated at (1, 1, 1) must give zero. So let’s do this computation. (If you knowsome multivariable calculus, it may be instructive to think in terms of ∇f = 0 at (1, 1, 1).)

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Winter Camp 2008 Inequalities Yufei Zhao

We have∂f

∂a=

√a2 + 8bc− a2

√a2+8bc

a2 + 8bc− rar−1(ar + br + cr)− ar · rar−1

(ar + br + cr)2.

Evaluating at (a, b, c) = (1, 1, 1) and setting the value to zero, we get 3− 13

9 − 3r−r9 = 0, which gives r = 4

3 .Aha! Now we found the candidate for r, we can now go through the steps of the official solution (which

were easy, given we know what r is). �Note that while we used calculus to motivate our solution, we do not need to include any

calculus in the solution! In fact, calculus is best avoided in olympiad solutions as it is generally viewedunfavorably. However, since we are not required to provide the motivation to our proof, we do not need toworry about this issue. Amazing, isn’t it?

Turning the table around. Seeing that the inequality is homogeneous (meaning that the transformation(a, b, c) 7→ (ka, kb, kc) does not change anything), it is natural to impose a constraint on it. So let us assumewithout the loss of generality that abc = 1

8 , so that we need to prove

a√a2 + 1

a

+b√

b2 + 1b

+c√

c2 + 1c

≥ 1.

This does not seem any easier. Now, let us turn the table around and switch the roles of the constraint andthe inequality.

Letx =

a√a2 + 1

a

, y =b√

b2 + 1b

, z =c√

c2 + 1c

.

Note that a√a2+ 1

a

= 1√1+ 1

a3

, so we can write a, b, c in terms of x, y, z:

a3 =x2

1− x2, b3 =

y2

1− y2, c3 =

z2

1− z2.

So, the inequality that we wish to prove is x+y+z ≥ 1 assuming abc = 18 . By considering the contrapositive,

it suffices to prove that abc < 18 assuming x+ y + z < 1. That is, we need to prove that

x2y2z2

(1− x2)(1− y2)(1− z2)<

183

given x+ y + z < 1. The square roots are now gone! The new inequality turns out to be extremely easy, asit is merely a straightforward application of AM-GM. Indeed, we have

1− x2 > (x+ y + z)2 − x2 = y2 + z2 + xy + xy + yz + yz + xz + xz ≥ 8x12 y

34 z

34

and similarly for 1− y2 and 1− z2. Setting this back into about inequality gives the desired result. �

How to use Jensen. Since the inequality is homogeneous, we can assume that a+ b+ c = 1. Note thatthe function x 7→ 1√

xis convex. So we can use Jensen’s inequality as follows:

a · 1√a2 + 8bc

+ b · 1√b2 + 8ca

+ c · 1√c2 + 8ab

≥ 1√a(a2 + 8bc) + b(b2 + 8ca) + c(c2 + 8ab)

=1√

a3 + b3 + c3 + 24abc.

So it remains to prove that1√

a3 + b3 + c3 + 24abc≥ 1

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Winter Camp 2008 Inequalities Yufei Zhao

or equivalently,a3 + b3 + c3 + 24abc ≤ (a+ b+ c)3.

This is again extremely easy, as AM-GM gives

(a+ b+ c)3 − a3 − b3 − c3 = 3(a2b+ b2a+ b2c+ c2b+ c2a+ a2c) + 6abc ≥ 24abc.

�

3 Problems

The following problems are selected from a packet written by Thomas Mildorf, which can be found athttp://web.mit.edu/tmildorf/www/Inequalities.pdf. Solutions can also be found there.

1. Show that for positive reals a, b, c(a2b+ b2c+ c2a

) (ab2 + bc2 + ca2

)≥ 9a2b2c2.

2. Let a, b, c be positive reals such that abc = 1. Prove that

a+ b+ c ≤ a2 + b2 + c2.

3. Let P (x) be a polynomial with positive coefficients. Prove that if

P

(1x

)≥ 1P (x)

holds for x = 1, then it holds for all x > 0.

4. Show that for all positive reals a, b, c, d,

1a

+1b

+4c

+16d≥ 64a+ b+ c+ d

.

5. (USAMO 1980/5) Show that for all non-negative reals a, b, c ≤ 1,

a

b+ c+ 1+

b

c+ a+ 1+

c

a+ b+ 1+ (1− a)(1− b)(1− c) ≤ 1.

6. (USAMO 1977/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q, prove that

(a+ b+ c+ d+ e)(

1a

+1b

+1c

+1d

+1e

)≤ 25 + 6

(√p

q−√q

p

)2

and determine when equality holds.

7. Let a, b, c, be non-negative reals such that a+ b+ c = 1. Prove that

a3 + b3 + c3 + 6abc ≥ 14.

8. (IMO 1995/2) a, b, c are positive reals with abc = 1. Prove that

1a3(b+ c)

+1

b3(c+ a)+

1c3(a+ b)

≥ 32.

9. Let a, b, c be positive reals such that abc = 1. Show that

2(a+ 1)2 + b2 + 1

+2

(b+ 1)2 + c2 + 1+

2(c+ 1)2 + a2 + 1

≤ 1.

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Winter Camp 2008 Inequalities Yufei Zhao

10. (USAMO 1998/3) Let a0, . . . , an be real numbers in the interval (0, π2 ) 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.

11. (Romanian TST) Let a, b, x, y, z be positive reals. Show that

x

ay + bz+

y

az + bx+

z

ax+ by≥ 3a+ b

.

12. The numbers x1, x2, . . . , xn obey −1 ≤ x1, x2, . . . , xn ≤ 1 and x 31 + x 3

2 + · · ·+ x 3n = 0. Prove that

x1 + x2 + · · ·+ xn ≤n

3.

13. (Turkey) Let n ≥ 2 be an integer, and x1, x2, . . . , xn positive reals such that x21 + x2

2 + · · · + x2n = 1.

Determine the smallest possible value of

x51

x2 + x3 + · · ·+ xn+

x52

x3 + · · ·+ xn + x1+ · · ·+ x5

n

x1 + · · ·+ xn−1.

14. (Poland 1995) Let n be a positive integer. Compute the minimum value of the sum

x1 +x2

2

2+x3

3

3+ · · ·+ xnn

n,

where x1, x2, . . . , xn are positive reals such that

1x1

+1x2

+ · · ·+ 1xn

= n.

15. Prove that for all positive reals a, b, c, d,

a4b+ b4c+ c4d+ d4a ≥ abcd(a+ b+ c+ d).

16. (Vietnam 1998) Let x1, . . . , xn be positive reals such that

1x1 + 1998

+1

x2 + 1998+ · · ·+ 1

xn + 1998=

11998

Prove thatn√x1x2 · · ·xnn− 1

≥ 1998.

17. (MOP 2002) Let a, b, c be positive reals. Prove that(2ab+ c

) 23

+(

2bc+ a

) 23

+(

2ca+ b

) 23

≥ 3

18. (Iran 1996) Show that for all positive real numbers a, b, c,

(ab+ bc+ ca)(

1(a+ b)2

+1

(b+ c)2+

1(c+ a)2

)≥ 9

4

6