Topics in inequalities

(Parte 1 de 2)

Hojoo Lee Version 0.3 [2004/03/07]

Introduction

Inequalities are useful in all fields of Mathematics. The purpose of this book is to present standard techniques in the theory of inequalities. In fact, this book is an extended and improved version of the author’s short article Note on Muirhead’s Theorem. Many of the problems are mathematical olympiads problems. I have given sources of the problems at the end of the book. The book is available at http://my.netian.com/∼ideahitme/eng.html

This is an unfinished manuscript. I would greatly appreciate hearing about any errors in the book, even minor ones. You can send all comments to the author at hojoolee@korea.com .

Tentative Plan

Contents

1 Substitutions i 2 Homogenizations xiii 3 Normalizations xi

Chapter 1 Substitutions

The main idea of this chapter is to try to reformulate the problem into an equivalent one. Many problems are simplified by some suitable substitutions. We shall consider various algebraic and trigonometric substitutions. We begin with a classical inequality in triangle geometry.

What is the oldest1 nontrivial geometric inequality ?

In 1765, Euler showed that

Theorem 1. Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Then, we have R ≥ 2r and the equality holds if and only if ABC is equilateral.

abc

Theorem 2. (A. Padoa [AP]) Let a, b, c be the lengths of a triangle. Then, we have and the equality holds if and only if a = b = c.

Proof. We shall use the Ravi Substitution : Since a, b, c are the lengths of a triangle, there are positive reals x, y, z such that a = y + z, b = z + x, c = x + y. (Why?) Then, the inequality proved is (y + z)(z + x)(x + y) ≥ 8xyz for x, y, z > 0. However, we get

Exercise 1. Let ABC be a right traingle. Show that R ≥ (1+√ 2)r. When does the equality holds ?

1The first geometric inequality is the Triangle Inequality : AB + BC ≥ AC 2In this book, [P] stands for the area of the polygon P.

It’s natural to ask that the inequality in the theorem 2 holds for arbitrary positive reals a, b, c? Yes ! It’s possible to prove the inequality without the additional condition that a, b, c are the lengths of a triangle :

Proof. Since the inequality is symmetric in the variables, without loss of generality, we may assume that x ≥ y ≥ z. Then, we have x + y > z and z + x > y. If y + z > x, then x, y, z are the lengths of the sides of a triangle. And by the theorem 2, we get the result. Now, we may assume that y + z ≤ x. However, in this case, we have xyz > 0 ≥ (y + z − x)(z + x − y)(x + y − z).

lim

xnynzn ≥ (yn + zn − xn)(zn + xn − yn)(xn + yn − zn) Now, taking the limits to both sides, we get the result.

Problem 1. (IMO 2000/2) Let a,b,c be positive numbers such that abc = 1. Prove that(

Solution. Since abc = 1, we can make the substitution a = xy , b = yz

We rewrite the given inequality in the terms of x, y, z :(

The Ravi Substitution is so useful when dealing with an inequality for the lengths a, b, c of a triangle. After the Ravi Substitution, we can remove the condition that they are the lengths of the sides of a triangle. Here are some examples :

Problem 2. (IMO 1983/6) Let a, b, c be the lengths of the sides of a triangle. Prove that

However, the Cauchy-schwarz inequality provides that

Problem 3. (IMO 1961/2) Show that, for any triangle with sides a, b, c and area S,

Exercise 2. (Hadwiger-Finsler Inequality) Show that, for any triangle with sides a, b, c and

Here is a two triangles generalization of the Weitzenbock’s Inequality :

If you are faced with an integral that contains square root expressions such as∫ √ 1 − x2 dx,

then trigonometric substitutions such as x = sint, y = tant, z = sect are very useful. When dealing with square root expressions, making a suitable trigonometric substitution simplifies the given inequality :

Problem 4. (Korea 1998) Let x, y, z be the positive reals with x + y + z = xyz. Show that

the function f is not concave up on R+. However, the function f(tanθ) is concave up ! :

) . Using the fact

cosθ

that pi − C = A + B or A + B + C = pi. Hence, it suffices to show the following.

Proof. Since the cosine function is concave up on ( 0, pi

) , Jensen’s inequality gives us

We note that the funtion cosx is not concave up on (0,pi). In fact, it’s concave down on(

. Hence, it seems that the inequality cosA + cosB + cosC ≤ 32 does NOT hold for arbitrary triangles. However, this guess is wrong ! 5

Second Proof. Let BC = a, CA = b, AB = c. Use the Cosine Law to rewrite the given inequality in the terms of a, b, c :

In the theorem 1 and the theorem 2, we see that the geometric inequality R ≥ 2r is equivalent to the algebraic inequality abc ≥ (b + c − a)(c + a − b)(a + b − c). And now we find that, in the proof of the theorem 6, abc ≥ (b + c − a)(c + a − b)(a + b − c) is equivalent to the trigonometric inequality cosA + cosB + cosC ≤ 32 . Then, one may ask that

In any traingle ABC, is there a natural relation between cosA + cosB + cosC and Rr , where R and r are the radii of the circumcircle and incircle of ABC ?

What is the most natural way to find some relations between X = Rr and Y = cosA + cosB + cosC? We already have nice identities :

Hence, we wish to find a nice function F :

Whenever a,b,c are the lengths of the sides of a triangle, we have F(X,Y ) = 0 ⇔F

2ab

We want to find a nice function f such that f(a,b,c) = 0, where f(a,b,c) = F(X,Y ).

which can be simplified as F

. It follows that cosA+cosB +cosC = 1+ rR holds for all right triangles. 6 Hence, it is natural to ask that

Does the identity cosA + cosB + cosC = 1 + rR hold for all triangles ?

Yes, and there is a simple way to verify this :

Theorem 7. Let R and r denote the radii of the circumcircle and incircle of the triangle

Although there is a two-line proof of the identity cosA+cosB +cosC = 1+ rR , I showed some ugly computations. The main reason is to present some motivations how to discover the relation. The identity is NOT TRIVIAL. Students should know that it’s not easy to discover some (new) results. Whenever you encounter some new results, you should think how to discover them. And whenever you meet tricky solutions of some non-trivial problems, you should try to find natural proofs.

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Exercise 4. Let R and r be the radii of the circumcircle and incircle of the triangle ABC with BC = a, CA = b, AB = c. Let s denote the semiperimeter of ABC. Verify the follwing identities 7 :

Using the theorem 7, we can present an alternative proof of the theorem 1 : To show

. Thus, we give another

proof of the theorem 5.

Third Proof. We make the substitution p = cosA, q = cosB, r = cosC. Then, p, q,

, which is

Exercise 5. (a) Let p,q,r be the positive real numbers such that p2 + q2 + r2 + 2pqr = 1. Show that there exist an acute triangle ABC such that p = cosA, q = cosB, r = cosC.

] with

are required to prove

One may assume that A ≥ pi3 or 1−2cosA ≥ 0. Now, notice that cosAcosB+cosB cosC+ cosC cosA−2cosAcosB cosC = cosA(cosB +cosC)+cosB cosC(1−2cosA). We apply

7For more identities, see the exercise 10.

viii cosA(cosB+cosC)+cosB cosC(1−2cosA) ≤ cosA (

In the above solution, we find that cosAcosB + cosB cosC + cosC cosA − 2cosAcosB cosC ≤ 1 2

Exercise 6. Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Let s be the semiperimeter of ABC. Show that

We have seen that certain inequalities can be treated by Ravi substitution and trigonometric substitutions. You can also transform the given inequalities into easier ones through some clever algebraic substitutions.

Problem 6. (IMO 2001/2) Let a, b, c be positive real numbers. Prove that

Solution. To remove the square roots, we make the following substitution :

= 512xyz. This is a contradiction ! 9For a proof, see [WJB].

Problem 7. (IMO 1995/2) Let a,b,c be positive numbers such that abc = 1. Prove that

, we get xyz = 1. And the inequality takes the form x2

It follows from the Cauchy-schwarz inequality that

so that, by the AM-GM inequality,

Let’s conlculde by giving another solution of the problem 4 :

. Then, we find that a + b + c = abc is equivalent

By the AM-GM inequality, we have

In a like manner, we obtain

Adding these three yields the required result.

Supplementary Problems

Exercise 7. Let f(x,y) be a real polynomial such that, for all θ ∈ R3, f(cosθ,sinθ) = 0.

Show that the polynomial f(x,y) is divisible by x2 + y2 − 1. Exercise 8. Let f(x,y,z) be a real polynomial. Suppose that

Exercise 9. (IMO Unused 1986) Let a,b,c be positive real numbers. Show that

Exercise 10. With the usual notation for a triangle, verify the follwing identities :

Exercise 1. Let a,b,c be the lengths of the sides of a triangle. And let s be the semiperimeter of the triangle. Then, the following inequalities holds.

10For a proof, see [JmhMh]. 11If we assume that there is a triangle ABC with BC = a, CA = b, AB = c, then it’s equivalent to the inequality s2 ≤ 4R2 + 4Rr + 3r2 in the exercise 6.

Exercise 12. (R. Sondat [RS]) Let R,r,s be positive real numbers. Show that a necessary and sufficient condition for the exstence of a triangle with circumradius R, inradius r, and semiperimeter s is s4 − 2(2R2 + 10Rr − r2)s2 + r(4R + r)2 ≤ 0.

Exercise 14. (W. J. Blundon [WJB2],[RAS]) Let R and r denote the radii of the circumcircle and incircle of the triangle ABC. Let s be the semiperimeter of ABC. Show that

Exercise 15. Let G and I be the centroid and incenter of the triangle ABC with inradius r, semiperimeter s, circumradius R. Show that

12It’s equivalent to the Hadwiger-Finsler inequality. 13See the exercise 6. For a solution, see [KWL].

xii

Chapter 2 Homogenizations

Some inequality problems come with constraints such as ab = 1, xyz = 1, x + y + z = 1. Non-homogeneous symmetric inequalities can be transformed into a homogeneous one. Then we apply two powerful theorems : Shur’s Inequality and Muirhead’s Theorem. We begin with the following example.

Problem 8. (Hungary 1996) Let a and b be positive real numbers with a + b = 1. Prove

Solution. Using the condition a+b = 1, we can reduce the given inequality to homogeneous

The above inequality a2b + ab2 ≤ a3 + b3 can be generalized as following :

Proof. Without loss of generality, we can assume that a1 ≥ a2,b1 ≥ b2,a1 ≥ b1. If x or y is zero, then it clearly holds. So, we also assume that both x and y are nonzero. It is easy to check the following :

Remark 1. When does the equality hold in the theorem 8? xiii

Before giving an example for the case of three variables inequality, we introduce two summation notations ∑ cyclic and ∑ sym. Let P(x,y,z) be a three variables function of x, y,

For example, we know that∑ cyclic

sym

Problem 9. (IMO 1984/1) Let x,y,z be nonnegtive real numbers such that x + y + z = 1.

Solution. Using the condition x + y + z = 1, we can reduce the given inequality to homoge- inequality is trivial because it’s equivalent to 0 ≤ xyz + ∑ sym x2y and x,y,z ≥ 0. The right hand side inequality simplifies to 7∑

it will be enough to show that 2∑ case of the following Schur’s inequality :

Theorem 9. (Schur) Let x,y,z be nonnegative real numbers. Then for any r > 0, we have∑

Proof. Since the inequality is symmetric in the three variables, we may assume without loss of generality that x ≥ y ≥ z. Then the given inequality may be rewritten as and every term on the left-hand side is clearly nonnegative. Remark 2. When does the equality hold in Theorem 9?

The following special case of Schur’s inequality is useful :∑

cyclic cyclic

Here is another solution of the problem 1 : xiv

(IMO 2000/2) Let a,b,c be positive numbers such that abc = 1. Prove that(

Second Solution. It is equivalent to the following homogeneous inequality1 :(

cyclic cyclic cyclic which is a special case of Schur’s inequality. Here is another inequality problem with the constraint abc = 1.

Problem 10. (Tournament of Towns 1997) Let a,b,c be positive numbers such that abc = 1.

Prove that 1

Solution. We can rewrite the given inequality as following :

which is equivalent to

and hence to ∑

1For an alterantive homogenization, see the problem 1 in the chpater 1.

And let x,y,z be positive real numbers. Then, we have ∑

Then, applying the theorem 1 twice, we have that ∑

Remark 3. The equality holds if and only if x = y = z. However, if we allow x = 0 or y = 0 or z = 0, 3 then one may easily check that the equality holds if and only if

The following problem can be solved by applying Muirhead’s theorem :

Problem 1. (IMO 1995) Let a,b,c be positive numbers such that abc = 1. Prove that

Solution. It’s equivalent to

Clearing denominators, this becomes∑

sym sym sym sym sym sym sym sym sym and every term on the left hand side is nonnegative by the theorem 3.

2Note the equality in the final equation.

xvi

Problem 12. (Iran 1996) Let x,y,z be positive real numbers. Prove that

Solution. It’s equivalent to4 ∑

sym cyclic sym cyclic

sym sym sym sym cyclic

By Muirhead’s theorem and Schur’s ineuquality, we find that it’s a sum of three terms which are nonnegative.

There is an alternative way4 to attack the above problem without Muirhead’s theorem and Schur’s ineuquality. We need to prove the following lemma.

Proof. They are equivalent to

We leave the details for the readers.

the given inequality in the terms of p, q, r :q

We find that every term on the left hand side is nonnegative by the lemma.

4See also the appendix [1]. 5When does equality hold in each inequality? For more p-q-r inequalities, see the exercise 21 and visit the site [ESF].

xvii

Problem 13. Let x,y,z be nonnegative real numbers with xy + yz + zx = 1. Prove that

First Solution. Using xy + yz + zx = 1, we homogenize the given inequality as following :

sym sym sym sym sym sym sym sym sym sym

By Muirhead’s theorem, we get the result. In the above inequaltiy, witout the condition xy + yz + zx = 1, the equality holds6 if and only if

and the every term on the left hand side is nonnegative by the lemma. Now, we apply Muirhead’s theorem to obtain a geometric inequality [ZsJc] :

Problem 14. If ma,mb,mc are medians and ra,rb,rc the exradii of a traingle, prove that rarb

mamb

+ rbrc mbmc

+ rcra

Solution. Let 2s = a + b + c. Using the well-known identities

we have ∑cyclic rbrc

mbmc =

∑cyclic

Applying the AM-GM inequality, we obtain∑cyclic rbrc mbmc ≥

∑cyclic

∑cyclic

6See the Remark 3. xviii

Thus, it will be suffice to show that∑cyclic

After expanding the above inequality, we see that it becomes2 ∑

cyclic cyclic sym cyclic cyclic

We note that this cannot be proven by just applying Muirhead’s Theorem. Since a, b, c are the sides of a triangle, we can make the Ravi Substitution a = y + z, b = z + x, c = x + y, where x,y,z > 0. After some brute-force algebra, we can rewrite the above inequality as25 ∑

sym sym sym sym sym x4yz

sym

Now, by Muirhead’s theorem, we get the result ! We notice that there are homogeneous symmetric polynomial inequalities which cannot be solved by just applying Muirhead’s theorem. See the following inequality :5 ∑

cyclic sym sym sym cyclic

This holds for all positive real numbers x, y, and z. However, it is not a direct consequence of Muirhead’s theorem because the coefficents of ∑ symx5y and ∑ cyclic x3y3 are too big. In fact, it is equivalent to

(Parte 1 de 2)