IMFM Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia Preprint series Vol. 48 (2010), 1129 ISSN 2232-2094 VARIABLE ZAGREB INDICES AND KARAMATA'S INEQUALITY Vesna Andova Mirko Petrusevski Ljubljana, September 15, 2010 Variable Zagreb indices and Karamata's inequality Vesna Andova" and Mirko Petrusevski6 io CO CD Jh CD CO Institute of Mathematics and Physics, Faculty of Electrical Engineering and Information Technologies, Ss. Cyril and Methodius Univ., CD Ruger Boskovik, P. O. Box 574, 1000 Skopje, Macedonia, e-mail: vesna.andova@gmail.com 6Institute of Mathematics and Informatics, Faculty of Mechanical Engineering, Ss. Cyril and Methodius Univ. Ruger Boskovik, PO Box 574, 1000 Skopje, Macedonia e-mail: mirko.petrusevski@gmail.com Abstract For a simple graph G with n vertices and m edges, the inequality M1(G)/n < M2(G)/m, where M1(G) and M2(G) are the first and the second Zagreb indices of G, is known as Zagreb indices inequality. Generalization of these indices gives first AM1(G) and second AM2(G) variable Zagreb indices. Vukicevic in [13] has given an incomplete proof for the claim: for all simple graphs and all A £ [0, 1 ], holds aMi(G)/n 0 for A € [0,1] and f (i,j) < 0 for A € R\[0,1]. Lemma 1.2. Let i,j,k and j be different natural numbers and let A € [0, 1 ]. Then the function g(i,j,k,l) defined by (4) in non-negative. Without loss of generality, we may assume that i = max{j, k, l} and that k > l. Now, there are three possible orderings: (a) i > j > k > l, (b) i > k > j > l, (c) i > k > l > j. m The cases (a) and (b) are proven in [13], even more for these orderings holds g(i,j, k,l) > 0 for all A € [0,1]. dg(% j k l) The incompleteness in the proof of Lemma 1.2 is for the third ordering. Namely, -—-— is di not non-negative [13] in the case (c). 2 Proof of Theorem 1.1 By the above discussion, one can easily see that the main problem here is determining the sign of g for A £ [0, 2]. In order to do that, we will use some already known results [8]. Lemma 2.1. [Karamata's inequality] Let U C R be an open interval and f : U ^ U be a convex CD function. Let ai > a2 > ... > an and bi > 62 > • • • > bn belong to U are such that ai + a2 + ... + ai > bi + b2 + ... + bi for every i £ {1, 2,... ,n} with equality for i = n. Then f (a1) + f (a2) + ... + f (an) > f (bl) + f (b2) + ... + f (bn). Since monotonicity of the a's only strengthens the majorizing conditions a1 + a2 + ... + ai > b1 + b2 + ... + bi for every i £ {1, 2,... ,n} with equality for i = n, we have that the same inequality holds without any restrictions on order on the a's. If in addition U = R and the function f is non-decreasing on U, then the majorizing conditions can be further relaxed from "with equality for i = n". Namely if a1 + a2 + ... + an >b1 + b2 + ... + bn, then we take a'n = b1 + b2 + ... + bn — a1 — a2 — ... — an-1. With a'n instead of an we have that all the needed for Karamata's is satisfied and f (an) > f (a'n), which goes on our hand. These comments explain how the following is derived from Lemma 2.1. CSI Lemma 2.2. [Majorizing inequality] Let f : R ^ R be a non-decreasing convex function. Let ai,a2,... ,an and b1 > b2 > ... > bn be reals such that a1 + a2 + ... + ai > b1 + b2 + ... + bi for every i £ {1, 2, .. ., n}. Then f (ai) + f (02) + ... + f (an) > f (bi) + f (ki) + ... + f (bn). CO CD CD CO A Lemma 2.2 will be are use to prove the following result. Theorem 2.1. Let a, b,c,d £ R+ and x £ [0, i]. Then axbx(^ + + cxdx(1 + i) > a2x-i + b2x-i + c2x-i + d2x-i. c d a b Proof. Put A = — logt a, B = — logt b, C = — logt c, D = — logt d, for a fixed real t > 1. Then this inequality takes on the form tA-(C+D)x + tB-(c + D)x + t°-(A+B)x + tD-(A+B)x > t(i-2x)A + t(i_2x)B + t(i_2x)C + t(i_2x)D _ Put ai = A — (C + D)x, a2 = B — (C + D)x, a3 = C — (A + B)x, a4 = D — (A + B)x and bi = (1 — 2x)A, b2 = (1 — 2x)B, b3 = (1 — 2x)C, b4 = (1 — 2x)D. CO Without loss of generality we can take that A > B,C, D and C > D. There are three cases to be considered regarding how B is positioned to C, D: (1) If B > C, then A > B > C > D. Since x £ [0,1/2], i.e., x, 1 — 2x > 0, it is obvious that bi > b2 > b3 > b4 and ai > ^^ bi, for j = 1, 2, 3, 4. So now the sequences ai: a2, a3, a4 and bi, b2, b3, b4 satisfy the (conditions for the majorizing inequality. Similarly to case (1) the orderings of a's and b's for the other two cases are: (2) if C > B > D, then A > C > B > D, and ai, a3, a2, a4 and bi, b3, b2, b4; (3) if D > B, then A > C > D > B, and ai, a3, a4, a2 and bi, b3, b4, b2; and they satisfy the conditions for the majorizing inequality. □ Proof. (of the Theorem 1.1) By Theorem 2.1 we have that the function g(i,j, k, l) is non-negative for any positive integers i,j, k, l and any A £ [0, i]. By Lemma 1.1, the function f (i,j) is also non-negative for A £ [0, 2] C [0,1] . Since f and g are non-negative for A £ [0, 2] we have xM2/m —x Mi/n > 0 i.e., xMi/n