Dedicated to Professor Milosav Marjanović on the occasion of his 80th birthday

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1 THE TEACHING OF MATHEMATICS 2, Vol. XIV, 2, pp SOME CLASSICAL INEQUALITIES AND THEIR APPLICATION TO OLYMPIAD PROBLEMS Zoran Kadelburg Dedicated to Professor Milosav Marjanović on the occasion of his 8th birthday Abstract. The technique introduced by M. Marjanović in [] is used to prove several classical inequalities. Examples of application are given which can be used for preparing students for mathematical competitions. ZDM Subject Classification: H34; AMS Subject Classification: 97H3. Key words and phrases: Chebyshev s inequality; Karamata s inequality; Steffensen s inequality; Jensen s inequality.. Introduction M. Marjanović deduced in [] some refinements of classical inequalities of Karamata [7] and Steffensen [6], making also shorter proofs see also [8]. Similar technique was used in [2] to make a short proof of Chebyshev s inequality [4]. These proofs are reproduced in Section 2 of this paper. It is well known that these classical results can be used in solving some of the most difficult problems in mathematical olympiads in fact, the authors of these problems probably used the mentioned results, and made elementary solutions afterwards. We shall present in Section 3 some examples of this kind, giving solutions based on classical inequalities. 2. Majorization of sequences and functions and its application Recall that, for decreasing finite sequences a = a i n and b = b i n of real numbers, a is said to majorize b, what is denoted by a b, or b a, if the terms of these sequences satisfy the following two conditions: k a i k n b i, for each k {, 2,..., n }; 2 a i = b i. Analogously, for two integrable functions ψ, ψ 2 : [, β] R, ψ is said to majorize ψ 2, what is denoted by ψ ψ 2, or ψ 2 ψ, if the following two conditions are satisfied: x ψ dt x ψ 2 dt for x [, β; 2 ψ dt = ψ 2 dt.

2 98 Z. Kadelburg Example. a If a = a i n is an arbitrary decreasing sequence of nonnegative numbers, having the sum equal to, then,,..., a, a 2,..., a n n, n n,..., for details see Lemma 2 in [2]. b If ψ : [, β] R is a decreasing integrable function, and then ψ ψ 2. ψ 2 x = β x ψ dt for x [, β], c The sequences 4, 4, and 5, 2, 2 are incomparable in the sense of the relation, i.e., none of the two majorizes the other one. Lemma. [5] Let ψ, ψ 2 : [, β] R be two integrable functions, such that ψ ψ 2, and let ϕ: [, β] R be an increasing integrable function. Then ϕψ dx ϕψ 2 dx. Proof. [8] Put ψx = ψ x ψ 2 x and gx = x ψt dt. Then, by the hypothesis, gx for x [, β] and g = gβ =. Using integration by parts in the Stieltjes integral, we get ϕtψt dt = ϕt dgt = ϕtgt β β gt dϕt = gt dϕt. Following [], one can use this lemma to deduce the following classical inequality, which is connected with various names I. Schur [4], G. H. Hardy, J. I. Littlewood, G. Polya [3], H. Weyl [8], and J. Karamata [9]. Following articles [5], [] and [3], we shall call it Karamata s inequality. Theorem. Let a = a i n and b = b i n be two finite decreasing sequences of real numbers from an interval, β. If a b, and if f :, β R is a convex function, then the following inequality holds fa i fb i. Proof. [] The given function f, being convex, is continuous and it can be represented in the form fx = x ϕ dt for an increasing function ϕ. Introduce functions A, B : [, β] R by Ax = min{x, a i }, Bx = min{x, b i }.

3 Some classical inequalities 99 It is easy to see that Ax Bx, for x [, β] and Aa = Ba. Moreover, A x and B x exist everywhere except in a finite set of points. Applying Lemma, we conclude that 2 But, a ϕ dax a ϕ dbx. a an an ϕ dax = n ϕ dx + n ϕ dx + + a n = fa + fa fa n, a a 2 ϕ dx and the similar relation holds for the integral on the right-hand side of 2. This proves Karamata s inequality. If the function f is strictly convex, it can be easily checked that the equality in is obtained if and only if the sequences a i and b i coincide. By the standard technique passing from natural to rational and then to real weights one can deduce the weighted form of Karamata s inequality sometimes called Fuchs inequality, see [2]: λ i fa i λ i fb i k if λ i R + and a i and b i are decreasing sequences, λ i a i k λ i b i for k {, 2,..., n } and n λ i a i = n λ i b i. An immediate consequence is the classical Jensen s inequality: Theorem 2. [6] Let f : [, β] R be a convex function, let x i [, β], i {, 2,..., n} and let λ i [, ] be such that λ i =. Then n 3 f λ i x i λ i fx i. If f is strictly convex, equality in 3 holds if and only if x = x 2 = = x n or all but one λ i s are equal to. Lemma can also be used in proving Steffensen s inequality [6]: Theorem 3. Let f, g : [, a] R, gx, f be decreasing on [, a], and let F x = x f dt. Then a a fg dx F g dx.

4 Z. Kadelburg Proof. [] If we denote c = a g dx, then < c a. Let {, x [, c], hx =, x c, a]. Then, it is easy to check that h g, and so, applying Lemma, we obtain Steffensen s inequality in the form c f dx = a fh dx a fg dx. In order to deduce Chebyshev s inequality, we first prove the following lemma [2]. Lemma 2. Let a = a i n, b = b i n and c = c i n be three decreasing sequences of real numbers, such that a b. Then the following inequality holds: a i c i b i c i. Proof. Denote A i = B =. Then we have a i c i b i c i = i j= = a j, B i = i j= a i b i c i = A i B i c i b j, for i {, 2,..., n}, and put A = A i A i B i + B i c i A i B i c i n n = A i B i c i A i B i c i+ i= n = A i B i c i c i+, being A i B i and c i c i+ for each i {, 2,..., n }. In particular, when a and b are decreasing, a b and above inequality holds for convex combinations of points c, c 2,..., c n. n a i = n b i =, the Theorem 4. If x i n and y i n are decreasing sequences of real numbers, then the following inequality holds: n n 4 x i y i n x i y i. Equality in 4 holds if and only if x = x 2 = = x n or y = y 2 = = y n.

5 Some classical inequalities Proof. [2] Without loss of generality, we can assume that the terms x i and y i of the given sequences are nonnegative if, for example, some of x i s or y i s were negative, we would apply the procedure that follows to the terms x i = x i x n and y i = y i y n. Denote X = n x i, a i = x i X and b i = n. Then the sequences a i i b i are decreasing and, by Example, a i b i holds true. Applying Lemma 2 to the sequences a i, b i and taking c i = y i, we obtain i.e., n n n n x i y i x i y i. x i X y i n y i, We state also the following version of Chebyshev s inequality: Theorem 4. Let x i n and y i n be decreasing sequences of real numbers, and let π be an arbitrary permutation of the set {, 2,..., n}. Then the inequality 5 x i y πi x i y i holds. If the sequence x i n is strictly decreasing, then equality in 5 holds if and only if y πi = y i for i {, 2,..., n}. 3. Examples of olympiad problems Problem. [Asian-Pacific Olympiad, 996] Let a, b, c be the length of sides of a triangle. Prove that the inequality a + b c + b + c a + c + a b a + b + c holds. Solution. Suppose, without loss of generality, that a b c and apply Karamata s inequality to the concave function fx = x and sequences a + b c, c + a b, b + c a a, b, c. Problem 2. [3] Prove that the inequality a 3 a 2 + a3 2 a a3 n a a 2 + a a 2 n holds for arbitrary positive numbers a, a 2,..., a n. Solution. Making the substitution x i = log a i, i {, 2,..., n}, we obtain an equivalent inequality e 3x x 2 + e 3x 2 x e 3x n x e 2x + e 2x e 2x n.

6 2 Z. Kadelburg This is obtained by applying Karamata s inequality to the convex function fx = e x, and the sequences a = 3x x 2, 3x 2 x 3,..., 3x n x and b = 2x, 2x 2,..., 2x n. It is enough to prove that these sequences, when arranged to be decreasing, satisfy a b. Let indices m,..., m n and k,..., k n be chosen so that 6 7 Then {m,..., m n } = {k,..., k n } = {,..., n}, 3x m x m + 3x m2 x m2 + 3x mn x mn +, 2x k 2x k2 2x kn. 3x m x m + 3x k x k + 2x k the first inequality holds because 3x m x m + is, by the choice of numbers m i, the greatest of numbers of the form 3x mi x mi+; the second one follows by the choice of numbers k i. By similar reasons, 3x m x m+ + 3x m2 x m2+ 3x k x k+ + 3x k2 x k2+ 2x k + 2x k2, and, generally, the sum of the first l terms of sequence 6 is not less than the sum of the first l terms of sequence 7, for l {,..., n }. For l = n, obviously, the equality is obtained, and so all the conditions for applying Karamata s inequality are fulfilled. Problem 3. [International Mathematical Olympiad 999] Let n be a fixed integer, n 2. a Determine the minimal constant C such that the inequality 4 x i x j x 2 i + x 2 j C x i i<j n i n is valid for all real numbers x, x 2,..., x n. b For the constant C found in a determine when the equality is obtained. Solution. As far as the given inequality is homogeneous, we can assume that x + x x n =. In this case the inequality can be written as x 3 x + x 3 2 x x 3 n x n C. The function fx = x 3 x is increasing and convex on the segment [, /2]. Let x be the greatest of the given numbers. Then the numbers x 2, x 3,..., x n are not greater than /2. If x [, /2] as well, then from x, x 2,..., x n 2, 2,,... using Theorem, we obtain that fx + fx 2 + fx n f + f + n 2f = If, to the contrary, x > /2, then it is x < /2 and we have that x 2, x 3,..., x n x,,...,. Applying Karamata s inequality once more, we obtain that fx + fx 2 + fx n fx + f x + n 2f = fx + f x.

7 Some classical inequalities 3 It is easy to prove that the function gx = fx + f x has the maximum on the segment [, ] equal to g/2 = /8. Thus, in this case also, fx + fx fx n /8 follows. Equality holds, e.g., for x = x 2 = /2, which proves that C = /8. Problem 4. [G. Szegö, [7]] Let f : [, a ] R be a convex function and a a 2... a 2n+. Then the inequality holds. fa a 2 + a 3 + a 2n+ fa fa 2 + fa 3 + fa 2n+ Solution. [5] Put a = a a 2 + a 3 + a 2n+. Then the given inequality can be rewritten as fa + fa fa 2n+ fa + fa fa 2n. To apply Karamata s inequality it is enough to check that a, a 3,..., a 2n+ a 2,..., a 2n, a. But this follows directly because a 2k a 2k for all k. For another proof of Szegö s inequality see [] or [8]. Problem 5. Let a, b and c be the length of sides of a triangle and let s be its semiperimeter. Prove that for a positive integer n, the inequality a n n 2 b + c + bn c + a + cn 2 a + b s n 3 holds. Solution. [7] Without loss of generality, we can suppose that a b c; then also b + c c + a a + b. Chebyshev s inequality, applied to the sequences a n, b n, c n and b + c, c + a,, implies that a + b a n b + c + bn c + a + cn a + b an + b n + c n 3 a + b + b + c +. c + a By Cauchy-Schwarz inequality, we have 2a + b + c a + b + b + c + 9, c + a and by the mean inequality of order n, a n + b n + c n n a + b + c. 3 3 Now, a n b + c + bn c + a + cn a + b n a + b + c 3 a + b + b + c + c + a 3 n n s 9 = s n. 3

8 4 Z. Kadelburg Problem 6. [University student s competition, Ostrawa 22] Let < x x 2 x n n 2 and =. + x + x 2 + x n Prove that x + x x n n x x2 xn Solution. [7] It is enough to prove that x + x2 + + xn x x2 xn n + + +, x x2 xn or, equivalently, + x 8 x x n xn + x + + x n + x n x x2 xn Take the function fx = x + = x +, x, +. It is easy to check that x x f is increasing on, + and that fx = f for each x >. x By the assumptions, it follows that only x could be less than and + x 2 = x. Hence, x 2. Now it is clear that both in the case x + x + x x and in the case x < fx = f fx 2 fx n. This means that the sequence inequality, we obtain that inequality 8 holds. x n + xk is increasing. Applying Chebyshev s xk k= We leave it to the reader to check that for n > 2 equality holds if and only if x = x 2 = = x n = n. If n = 2, then equality holds for { } x, x 2 t, t t. Problem 7. [Serbian Mathematical Olympiad 27] Let k be a positive integer. Prove that for positive real numbers x, y, z, having the sum equal to, the following inequality holds x k+2 x k+ + y k + z k + y k+2 y k+ + z k + x k + z k+2 z k+ + x k + y k 7.

9 Some classical inequalities 5 Solution. The inequality will be proved by several applications of Chebyshev s inequality. Note first, that the expression on left-hand side is symmetric, so we can suppose that x y z. Let us prove now that x k+ + y k + z k y k+ + z k + x k z k+ + x k + y k. It is enough to prove the first inequality, which is equivalent to x k+ + y k y k+ + y k x k x, i.e., to numbers x and y are less than. Since y x, it is x y enough to prove that y x x y, which is equivalent to x x2 y + y 2 = x y x y = x yz, and this inequality obviously holds. Applying Chebyshev s inequality to the triples x k+2, y k+2, z k+2 and x k+ + y k + z k, y k+ + z k + x k, z k+ + x k + y k we obtain that the left-hand side of the given inequality is not less than A = 3 xk+2 +y k+2 +z k+2 x k+ + y k + z k + y k+ + z k + x k + z k+ + x k + y k. A new application of Chebyshev s inequality, this time to the triples x, y, z and x k+, y k+, z k+, gives the inequality x k+2 + y k+2 + z k+2 = x k+ x + y k+ y + z k+ z Thus, 3 xk+ + y k+ + z k+ x + y + z = 3 xk+ + y k+ + z k+. A 3 3 xk+ + y k+ + z k+ x k+ + y k + z k + y k+ + z k + x k + z k+ + x k + y k =: B. But, Cauchy-Schwarz inequality implies that x k+ + y k + z k + y k+ + z k + x k + z k+ + x k + y k and hence B that x k+ + y k + z k + y k+ + z k + x k + z k+ + x k + y k 9, x k+ + y k+ + z k+ x k+ + y k+ + z k+ + 2x k + y k + z k. It remains only to prove 3x k+ + y k+ + z k+ x k + y k + z k. This follows directly by another application of Chebyshev s inequality. Equality holds if and only if x = y = z = /3.,

10 6 Z. Kadelburg REFERENCES [] D. -Dukić, V. Janković, I. Matić, N. Petrović, The IMO Compendium, Springer, 26. [2] L. Fuchs, A new proof of an inequality of Hardy-Littlewood-Pólya, Math. Tidsskr. B. 947, [3] G. H. Hardy, J. E. Littlewood, G. Pólya, Some simple inequalities satisfied by convex function, Messenger Math /29, [4] G. H. Hardy, J. E. Littlewood, G. Pólya, Inequalities, Cambridge University Press, 2nd ed., Cambridge 952. [5] A. I. Hrabrov, Around Mongolian inequality in Russian, Matematicheskoe prosveshchenie, 3, 7 23, [6] J. L. W. V. Jensen, Sur les fonctions convexes et les inégalités entre les valeurs moyennes, Acta Math. 3 96, [7] Z. Kadelburg, D. -Dukić, M. Lukić, I. Matić, Inequalities in Serbian, Društvo matematičara Srbije, Beograd, 23. [8] Z. Kadelburg, D. -Dukić, M. Lukić, I. Matić: Inequalities of Karamata, Schur and Muirhead, and some applications, The Teaching of Mathematics VIII, 25, [9] J. Karamata, Sur une inégalité relative aux fonctions convexes, Publ. Math. Univ. Belgrade 932, [] M. Marjanović, Some inequalities with convex functions, Publ. de l Inst. Math. Belgrade , [] M. Marjanović, Convex and concave functions and corresponding inequalities in Serbian, Nastava Matematike XL, 2 994, pp. and , pp. 4, Društvo matematičara Srbije, Beograd. [2] M. M. Marjanović, Z. Kadelburg, A proof of Chebyshev s inequality, The Teaching of Mathematics X, 2 27, 7 8. [3] D. Nomirovskiĭ, Karamata s Inequality in Russian, Kvant 4 23, [4] I. Schur, Über eine Klasse von Mittelbildungen mit Anwendungen die Determinanten, Theorie Sitzungsler, Berlin, Math. Gesellschaft , 9 2. [5] J. E. Steffensen, On a generalization of certain inequalities by Tchebychef and Jensen, Skand. Aktuarietidskr. 925, [6] J. E. Steffensen, Bounds of certain trigonometric integrals, Tenth Scand. Math. Congress 946, 8 86, Copenhagen, Gjellerups, 947. [7] G. Szegö, Über eine Verallgemeinerung des Dirichletschen Integrals, Math. Zeitschrift 52 95, [8] H. Weyl, Inequalities between the two kinds of eigenvalues of a linear transformation, Proc. Nat. Acad. Sci. U.S.A , University of Belgrade, Faculty of Mathematics, Studentski trg 6/IV, Beograd, Serbia kadelbur@matf.bg.ac.rs