Topics in inequalities

Topics in inequalities

(Parte 2 de 2)

As I know, there is no general criterion to attack the symmetric polynomial inequalities.

However, the author found a result for the homogeneous symmetric polynomial inequalities with degree 3. It’s a direct consequence of Muirhead’s theorem and Schur’s inequality.

Theorem 1. Let P(u,v,w) ∈ R[u,v,w] be a homogeneous symmetric polynomial with degree 3. Then the following two statements are equivalent.

Proof. It will be suffice to prove that (a) implies (b). Let

cyclic

7These inequality and identity are in [JC]. xix cyclic

sym sym sym sym and that the every term on the right hand side is nonnegative. Case 2. q ≤ r : We find that

sym sym cyclic sym and that the every term on the right hand side is nonnegative.

For example, we can apply the theorem 1 to solve the problem 9.

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

Solution. Using x + y + z = 1, we homogenize the given inequality as following :

Exercise 16. (M. S. Klamkin [MEK2]) Determine the maximum and minimum values of

As a corollary, we get the following criterion for homogeneous symmetric polynomial inequalities for the triangles :

Theorem 12. (K. B. Stolarsky) Let P(u,v,w) be a real symmetric form of degree 3.8 If then we have P(a,b,c) ≥ 0, where a,b,c are the lengths of the sides of a triangle.

Proof. Make the Ravi substitution a = y + z, b = z + x, c = x + y and apply the above theorem. We leave the details for the readers. For an alternative proof, see [KBS].

As noted in [KBS], applying Stolarsky’s theorem, we obtain various cubic inequalities in triangle geometry. Recall the 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.

Proof. We only show the right hand side inequality in (j). It’s equivalent to the cubic inequality T(a,b,c) ≥ 0, where

Supplementary Problems Exercise 18. Let x,y,z be positive real numbers. Prove that

xxi

Exercise 20. (APMO 1998) Let a,b,c be positive real numbers. Prove that(

abc

Exercise 2. (IMO Short-listed 1998) Let x,y,z be positive real numbers such that xyz = 1.

Prove that x3

Chapter 3 Normalizations

In the previous chapter, we trasformed non-homogeneous inequalities into homogeneous inequalities. On the other hand, homogeneous inequalities also can be normalized in the various way. We begin with two alternative solutions of the problem 6 :

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

Second Solution. Dividing by a + b + c gives the equivalent inequality

(the weighted) Jensen’s inequality to have

In the above solution, we normalized to x + y + z = 1. We now normalize to xyz = 1.

Third Solution. We make the substitution x = bc and the inequality becomes

which is equivalent to∑cyclic hence, after squaring both sides, equivalent to

∑cyclic

∑cyclic

Using these three inequalities, we get the result.

We now illustrate the normalization technique with fundamental inequalities. Recall the famous Cauchy-Schwarz inequality : 1

Proof. Let A = √ a1 = · = an = 0. Thus, the given inequality clearly holds. So, we now may assume that

1See the exercise 27-29. xxiv

Exercise 25. Prove the Lagrange’s identity :

Using the same idea in the above proof, we find a natural generalization :

Proof. Since the inequality is homogeneous, as in the proof of the theorem 1, we can

To finish the proof, it remains to show the following homogeneous inequality : Theorem 15. (AM-GM Inequality) Let a1,· ,an be positive real numbers. Then, we have

The following simple observation is not tricky :

ma+nb

Hence, for all positive rationals ω1 and ω2 with ω1 + ω2 = 1, we get

xxv

We immediately have

Proof. We can choose two positive rational sequences {ai}i∈N and {bi}i∈N such that lim

From the previous observation, we have an x + bn y ≥ xanybn for all n ∈ N. Now, taking the limits to both sides, we get the result.

Modifying slightly the above arguments, the readers obtain

becomes the AM-GM inequality. Recall that the AM-GM inequality is used to deduce the theorem 12, which is a generalization of the Cauchy-Schwarz inequality. Since we now get the weighted version of the AM-GM inequality, we can establish the weighted version of the Cauchy-Schwarz inequality. It’s called Holder’s Inequality :

xij ωj.

xij ωj or 1 ≥ xij ωj.

The weighted AM-GM inequality provides that n∑ j=1 ωjxij ≥ j=1 ωjxij ≥ xij ωj.

However, we immediately have m∑ j=1 ωjxij = i=1 ωjxij = j=1 ωj xxvi

We now present an IMO problem which can be solved by the technique normalization and homogenization.

Problem 15. (IMO 1999/2) Let n be an integer with n ≥ 2. (a) Determine the least constant C such that the inequality

holds for all real numbers x1,· ,xn ≥ 0. (b) For this constant C, determine when equality holds.

We claim that C = 18 . It suffices to show that

3I slightly modified his solution in [Au99].

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Using x + y = 1, we homogenize the above inequality as following.

Supplementary Problems

Exercise 26. (IMO unused 1991) Let n be a given integer with n ≥ 2. Find the maximum value of ∑

i6=j aiaj i=1 aibi + x i6=j aibj

Exercise 31. Prove the Cauchy-Schwarz inequality for complex numbers 4:n∑

Exercise 32. Prove the complex version of the Lagrange’s identity 5:n∑ xxviii

Appendix

[1] Third solution of the Iran 1996 problem : (Iran 1996) Let x,y,z be positive real numbers. Prove that

Third Solution. (Marcin E. Kuczma [MEK]) We assume that x ≥ y ≥ z ≥ 0 and y > 0(not excluding z = 0). Let F be left hand side of the inequality. And define

It can be verified that

This proves the inequality and shows that it becomes an equality only for x = y = z and for x = y > 0, z = 0.

xxix

References

AI D. S. Mitinovi’c (in cooperation with P. M. Vasic), Analytic Inequalities, Springer

AMN A. M. Nesbitt, Problem 15114, Educational Times (2) 3(1903), 37-38

AP A. Padoa, Period. Mat. (4)5 (1925), 80-85

Au99 A. Storozhev, AMOC Mathematics Contests 1999, Australian Mathematics Trust

EC E. Cesaro, Nouvelle Correspondence Math. 6(1880), 140

ESF Elementare Symmetrische Funktionen, http://hydra.nat.uni-magdeburg.de/math4u/ var/PU4.html

GI O. Bottema, R. Z. Djordjevic, R. R. Janic, D. S. Mitrinovic, P. M. Vasic, Geometric Inequalities, Wolters-Noordhoff Publishing, Groningen 1969

HFS H. F. Sandham, Problem E819, Amer. Math. Monthly 5(1948), 317

JC Ji Chen, Problem 1663, Crux Mathematicorum 18(1992), 188-189

JfdWm J. F. Darling, W. Moser, Problem E1456, Amer. Math. Monthly 68(1961) 294, 230

JmhMh J. M. Habeb, M. Hajja, A Note on Trigonometric Identities, Expositiones Mathematicae 21(2003), 285-290

KBS K. B. Stolarsky, Cubic Triangle Inequalities, Amer. Math. Monthly (1971), 879-881

KWL Kee-Wai Liu, Problem 2186, Crux Mathematicorum with Mathematical Mayhem, 23(1997), 71-72

MC M. Cipu, Problem 2172, Crux Mathematicorum with Mathematical Mayhem, 23(1997), 439-440

MCo M. Colind, Educational Times 13(1870), 30-31

MEK Marcin E. Kuczma, Problem 1940, Crux Mathematicorum with Mathematical Mayhem, 23(1997), 170-171

MP M. Petrovic, Racunanje sa brojnim razmacima, Beograd 1932, 79

MEK2 Marcin E. Kuczma, Problem 1703, Crux Mathematicorum 18(1992), 313-314

NC A note on convexity, Crux Mathematicorum with Mathematical Mayhem, 23(1997), 482-483

PF P. Flor, Uber eine Ungleichung von S. S. Wagner, Elem. Math. 20, 136(1965)

RAS R. A. Satnoianu, A General Method for Establishing Geometric Inequalities in a Triangle, Amer. Math. Monthly 108(2001), 360-364

RS R. Sondat, Nouv. Ann. Math. (3) 10(1891), 43-47 SR S. Reich, Problem E1930, Amer. Math. Monthly 73(1966), 1017-1018

WJB W. J. Blundon, Canad. Math. Bull. 8(1965), 615-626 WJB2 W. J. Blundon, Problem E1935, Amer. Math. Monthly 73(1966), 12 ZsJc Zun Shan, Ji Chen, Problem 1680, Crux Mathematicorum 18(1992), 251 xxxi

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