Putnam Inequalities

7/31/2019 Putnam Inequalities

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PUTNAM TRAINING

INEQUALITIES

(Last updated: October 21, 2011)

Remark. This is a list of exercises on inequalities. Miguel A. Lerma

Exercises

1. If a,b,c > 0, prove that (a2b + b2c + c2a)(ab2 + bc2 + ca2) 9a2b2c2.

2. Prove that n! 0 with xyz = 1. Prove that x + y + z x2 + y2 + z2.6. Show thata21 + b

21 +

a22 + b

22 + +

a2n + b

2n

(a1 + a2 + + an)2 + (b1 + b2 + + bn)2

7. Find the minimum value of the function f(x1, x2, . . . , xn) = x1 + x2 + + xn, wherex1, x2, . . . , xn are positive real numbers such that x1x2 xn = 1.

8. Let x,y,z 0 with xyz = 1. Find the minimum of

S =x2

y + z

+y2

z+ x

+z2

x + y

.

9. If x, y,z > 0, and x + y + z = 1, find the minimum value of

1

x+

1

y+

1

z.

10. Prove that in a triangle with sides a, b, c and opposite angles A, B, C (in radians)the following relation holds:

a A + b B + c C

a + b + c

3.

1


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PUTNAM TRAINING INEQUALITIES 2

11. (Putnam, 2003) Let a1, a2, . . . , an and b1, b2, . . . , bn nonnegative real numbers. Showthat

(a1a2 an)1/n + (b1b2 bn)1/n ((a1 + b1)(a2 + b2) (an + bn))1/n

12. The notation n!(k) means take factorial ofn k times. For example, n!(3) means ((n!)!)!What is bigger, 1999!(2000) or 2000!(1999)?

13. Which is larger, 19991999 or 20001998?

14. Prove that there are no positive integers a, b such that b2 + b + 1 = a2.

15. (Inspired in Putnam 1968, B6) Prove that a polynomial with only real roots and allcoefficients equal to

1 has degree at most 3.

16. (Putnam 1984) Find the minimum value of

(u v)2 +

2 u2 9v

2for 0 < u 0.

17. Show that14n

1

2

3

4

2n 1

2n

0 is decreasing and convex.

25. Prove that f(x) f(a) 0.

26. By the AM-GM inequality we have x1 +1x2

2

x1x2

, . . . Try to prove that those

inequalities is actually equalities.

27. Square both sides of those inequalities.

28. Rearrangement inequality.

29. Rearrangement inequality.


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Solutions

1. Using the Arithmetic Mean-Geometric Mean Inequality on each factor of the LHS wegeta2b + b2c + c2a

3

ab2 + bc2 + ca2

3

3

a3b3c3

3

a3b3c3

= a2b2c2 .

Multiplying by 9 we get the desired inequality.

Another solution consists of using the Cauchy-Schwarz inequality:

(a2b + b2c + c2a)(ab2 + bc2 + ca2) =

(a

b)2 + (b

c)2+ (c

a)2

(

b c)2 + (

c a)2 + (

a b)2

(abc + abc + abc)2= 9a2b2c2 .

2. This result is the Arithmetic Mean-Geometric Mean applied to the set of numbers1, 2, . . . , n:

n

1 2 n < 1 + 2 + + nn

=n(n+1)

2

n=

n + 1

2.

Raising both sides to the nth power we get the desired result.

3. The simplest solution consists of using the weighted power means inequality to the

(weighted) arithmetic and quadratic means of x and y with weightsp

p+q andq

p+q :

p

p + qx +

q

p + qy

p

p + qx2 +

q

p + qy2 ,

hence(px + qy)2 (p + q)(px2 + qy2) .

Or we can use the Cauchy-Schwarz inequality as follows:

(px + qy)2 = (

p

p x +

q

q y)2

{p}2 + {q}2

{p x}2 + {q y}2

(Cauchy-Schwarz)

= (p + q)(px2

+ qy2

) .Finally we use p + q 1 to obtain the desired result.

4. By the power means inequality:

a + b + c

3

a +

b +

c

3

2

M1(a,b,c)

M1/2(a,b,c)

From here the desired result follows.


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5. We have:

x + y + z = (x + y + z) 3

xyz (xyz = 1)

(x + y + z)2

3(AM-GM inequality)

x2 + y2 + z2 . (power means inequality)6. The result can be obtained by using Minkowskis inequality repeatedly:

a21 + b21 +

a22 + b

22 + +

a2n + b

2n

(a1 + a2)2 + (b1 + b2)2 + +

a2n + b

2n

(a1 + a2 + a3)2 + (b1 + b2 + b3)2 + + a2n + b2n. . .

(a1 + a2 + + an)2 + (b1 + b2 + + bn)2Another way to think about it is geometrically. Consider a sequence of points in theplane Pk = (xk, yk), k = 0, . . . , n, such that

(xk, yk) = (xk1 + ak, yk1 + bk) for k = 1, . . . , n .

Then the left hand side of the inequality is the sum of the distances between twoconsecutive points, while the right hand side is the distance between the first one andthe last one:

d(P0, P1) + d(P1, P2) + + d(Pn1, Pn) d(P0, Pn) .7. By the Arithmetic Mean-Geometric Mean Inequality

1 = n

x1x2 . . . xn x1 + x2 + + xnn

,

Hence f(x1, x2, . . . , xn) n. On the other hand f(1, 1, . . . , 1) = n, so the minimumvalue is n.

8. For x = y = z = 1 we see that S = 3/2. We will prove that in fact 3/2 is the minimumvalue of S by showing that S 3/2.Note that

S =

x

y + z

2+

y

z+ x

2+

z

x + y

2.

Hence by the Cauchy-Schwarz inequality:

S (u2 + v2 + w2)

xuy + z

+yv

z+ x+

zwx + y

2.

Writing u =

y + z, v =

z+ x, w =

x + y we get

S 2(x + y + z) (x + y + z)2 ,


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hence, dividing by 2(x + y + z) and using the Arithmetic Mean-Geometric Meaninequality:

S 12

(x + y + z) 12

3 3xyz = 32

.

9. By the Arithmetic Mean-Harmonic Mean inequality:

31x

+ 1y

+ 1z

x + y + z3

=1

3,

hence

9 1x

+1

y+

1

z.

On the other hand for x = y = z = 1/3 the sum is 9, so the minimum value is 9.10. Assume a b c, A B C. Then

0 (a b)(A B) + (a c)(A C) + (b c)(B C)= 3(aA + bB + cC) (a + b + c)(A + B + C) .

Using A + B + C = and dividing by 3(a + b + c) we get the desired result.- Remark: We could have used also Chebyshevs Inequality:

a A + b B + c C

3

a + b + c

3

A + B + C

3

.

11. Assume ai + bi > 0 for each i (otherwise both sides are zero). Then by the ArithmeticMean-Geometric Mean inequality

a1 an(a1 + b1) (an + bn)

1/n 1

n

a1

a1 + b1+ + an

an + bn,

and similarly with the roles ofa and b reversed. Adding both inequalities and clearingdenominators we get the desired result.(Remark: The result is known as superadditivity of the geometric mean.)

12. We have that n! is increasing for n 1, i.e., 1 n < m = n! < m! So 1999! > 2000= (1999!)! > 2000! = ((1999!)!)! > (2000!)! = . . . = 1999!

(2000)

>2000!(1999).

13. Consider the function f(x) = (1999 x) ln (1999 + x). Its derivative is f(x) = ln (1999 + x) + 1999 x

1999 + x, which is negative for 0 x 1, because in that interval

1999 x1999 + x

1 = ln e < ln (1999 + x) .

Hence f is decreasing in [0, 1] and f(0) > f(1), i.e., 1999 ln 1999 > 1998 ln 2000.Consequently 19991999 > 20001998.


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14. We have b2 < b2 + b + 1

a2< b2 + 2b + 1 = (b + 1)2. But b2 and (b + 1)2 are consecutive

squares, so there cannot be a square strictly between them.

15. We may assume that the leading coefficient is +1. The sum of the squares of the rootsof xn + a1x

n1 + + an is a21 2a2. The product of the squares of the roots is a2n.Using the Arithmetic Mean-Geometric Mean inequality we have

a21 2a2n

n

a2n .

Since the coefficients are 1 that inequality is (1 2)/n 1, hence n 3.Remark: x3 x2 x + 1 = (x + 1)(x 1)2 is an example of 3th degree polynomialwith all coefficients equal to

1 and only real roots.

16. The given function is the square of the distance between a point of the quarter ofcircle x2 + y2 = 2 in the open first quadrant and a point of the half hyperbola xy = 9in that quadrant. The tangents to the curves at (1, 1) and (3, 3) separate the curves,and both are perpendicular to x = y, so those points are at the minimum distance,and the answer is (3 1)2 + (3 1)2 = 8.

17. Let

P =

1

2

3

4

2n 1

2n

, Q =

2

3

4

5

2n 22n 1

.

We have P Q = 12n

. Also 12

< 23

< 34

< 45

< < 2n12n

, hence 2P Q, so2P2 P Q = 1

2n, and from here we get P 1

4n.

On the other hand we have P < Q2n

2n + 1< Q, hence P2 < P Q =

1

2n, and from here

P 0. We have f(x) = 1/x

e1/x< 0, f(x) =

e1/x( 2x3

+ 1x4

) > 0, hence f is decreasing and convex.By convexity, we have

1

2(f(e) + f()) f e+

2

.

On the other hand we have (e + )/2 < 3, and since f is decreasing, f( e+2

) > f(3),and from here the result follows.

25. We have:f(x) f(a) =

ni=1

(x ai)2 n

i=1

(a ai)2

=n

i=1

(x ai)2 (a ai)2

=

ni=1

(x2 2aix a2 + 2aia)

= nx2 2nax + na2

= n(x a)2

0 ,hence f(x) f(a) for every x.

26. By the Geometric Mean-Arithmetic Mean inequality

x1 +1

x2 2

x1x2

, . . . , x100 +1

x1 2

x100x1

.

Multiplying we getx1 +

1

x2

x2 +

1

x3

x100 +

1

x1

2100 .

From the system of equations we getx1 +

1

x2

x2 +

1

x3

x100 +

1

x1

= 2100 ,

so all those inequalities are equalities, i.e.,

x1 +1

x2= 2

x1x2

=

x1 1x2

2= 0 = x1 = 1

x2,

and analogously: x2 = 1/x3, . . . , x100 = 1/x1. Hence x1 = 1/x2, x2 = 1/x3, . . . ,x100 = 1/x1, and from here we get x1 = 2, x2 = 1/2, . . . , x99 = 2, x100 = 1/2.


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27. Squaring the inequalities and moving their left hand sides to the right we get

0

c2

(a

b)2 = (c + a

b)(c

a + b)

0 a2 (b c)2 = (a + b c)(a b + c)0 b2 (c a)2 = (b + c a)(b c + a) .

Multiplying them together we get:

0 (a + b c)2(a b + c)2(a + b + c)2 ,hence, one of the factors must be zero.

28. The answer is 1. In fact, the sequences (sin3 x, cos3 x) and (1/ sin x, 1/ cos x) areoppositely sorted, hence by the rearrangement inequality:

sin

3

x/ cos x + cos

3

x/ sin x sin3

x/ sin x + cos

3

x/ cos x= sin2 x + cos2 x = 1 .

Equality is attained at x = /4.

29. By the rearrangement inequality we have for k = 2, 3, , n:a1

s a1 +a2

s a2 + +an

s an a1

s ak +a2

s ak+1 + +an

s ak1 ,were the denominators on the right hand side are a cyclic permutation of sa1, , san. Adding those n 1 inequalities we get the desired result.


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Main Inequalities

1. Arithmetic Mean, Geometric Mean, Harmonic Mean Inequalities. Let a1, . . . , anbe positive numbers. Then the following inequalities hold:

n1

a1+ + 1

an

na1 a2 an a1 + + ann

Harmonic Mean

Geometric Mean

Arithmetic Mean

In all cases equality holds if and only if a1 = = an.

2. Power Means Inequality. The AM-GM, GM-HM and AM-HM inequalities are partic-

ular cases of a more general kind of inequality called Power Means Inequality.Let r be a non-zero real number. We define the r-mean or rth power mean of positivenumbers a1, . . . , an as follows:

Mr(a1, . . . , an) =

1

n

ni=1

ari

1/r.

The ordinary arithmetic mean is M1, M2 is the quadratic mean, M1 is the harmonic mean.Furthermore we define the 0-mean to be equal to the geometric mean:

M0(a1, . . . , an) = n

i=1

ai1/n .Then, for any real numbers r, s such that r < s, the following inequality holds:

Mr(a1, . . . , an) Ms(a1, . . . , an) .Equality holds if and only if a1 = = an.

2.1. Power Means Sub/Superadditivity. Let a1, . . . , an, b1, . . . , bn be positive real numbers.

(a) Ifr > 1, then the r-mean is subadditive, i.e.:

Mr(a1 + b1, . . . , an + bn)

Mr(a1, . . . , an) + Mr(b1, . . . , bn) .

(b) Ifr < 1, then the r-mean is superadditive, i.e.:

Mr(a1 + b1, . . . , an + bn) Mr(a1, . . . , an) + Mr(b1, . . . , bn) .Equality holds if and only if (a1, . . . , an) and (b1, . . . , bn) are proportional.

3. Cauchy-Schwarz. ni=1

aibi

2 n

i=1

a2i

ni=1

b2i

.


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4. Holder. If p > 1 and 1/p + 1/q = 1 then

ni=1

aibi ni=1

|ai|p1/p ni=1

|bi|p1/q .(For p = q = 2 we get Cauchy-Schwarz.)

5. Minkowski. If p > 1 then ni=1

|ai + bi|p1/p

n

i=1

|ai|p1/p

+ n

i=1

|bi|p1/p

.

Equality holds iff (a1, . . . , an) and (b1, . . . , bn) are proportional.

6. Norm Monotonicity. If ai > 0 (i = 1, . . . , n), s > t > 0, then ni=1

asi

1/s n

i=1

ati

1/t.

7. Chebyshev. Let a1, a2, . . . , an and b1, b2, . . . , bn be sequences of real numbers which aremonotonic in the same direction (we have a1 a2 an and b1 b2 bn, or wecould reverse all inequalities.) Then

1

n

ni=1

aibi 1

n

ni=1

ai

1n

ni=1

bi

.

8. Schur. If x,y,x are positive real numbers and k is a real number such that k 1, thenxk(x y)(x z) + yk(y x)(y z) + zk(z x)(z y) 0 .

For k = 1 the inequality becomes

x3 + y3 + z3 + 3xyz xy(x + y) + yz(y + z) + zx(z+ x) .9. Weighted Power Means Inequality. Let w1, . . . , wn be positive real numbers suchthat w1 + + wn = 1. Let r be a non-zero real number. We define the rth weighted powermean of non-negative numbers a1, . . . , an as follows:

Mrw(a1, . . . , an) = ni=1

wiari1/r

.

As r 0 the rth weighted power mean tends to:

M0w(a1, . . . , an) = ni=1

awii

.

which we call 0th weighted power mean. If wi = 1/n we get the ordinary rth power means.Then for any real numbers r, s such that r < s, the following inequality holds:

Mrw(a1, . . . , an) Msw(a1, . . . , an) .


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