Octogon Mathematical Magazine - Volume 17, No. 1, April 2009 PDF
Preview Octogon Mathematical Magazine - Volume 17, No. 1, April 2009
OCTOGON Mathematical Magazine Brassó Kronstadt Braşov Vol. 17, No. 1, April 2009 Editor-In-Chief Mihály Bencze Str. Hărmanului 6, 505600 Săcele-Négyfalu, Jud. Braşov, Romania Phone: (004)(0268)273632 E-mail: [email protected] [email protected] Editorial Board Šefket Arslanagić, University of Sarajevo, Sarajevo, Bosnia and Preda Mihăilescu, Matematisches Institut, Universitaet Herzegovina Goettingen D.M. Bătineţu-Giurgiu, Editorial Board of Josip Pečaric, University of Zagreb, Gazeta Matematică, Bucharest, Romania Zagreb, Croatia José Luis Díaz-Barrero, Universitat Politechnica de Catalunya, Themistocles M. Rassias, National Technical University of Barcelona, Spain Athens, Athen, Greece Zhao Changjian, China Jiliang University, Hangzhou, China Ovidiu T. Pop, National College Mihai Eminescu, Satu Mare, Romania Constantin Corduneanu, University of Texas József Sándor, Babeş-Bolyai University, at Arlington, Arlington, USA Cluj-Napoca, Romania Sever S. Dragomir, School of Computer Science and Florentin Smarandache, University of Mathematics, Victoria University, Melbourne, Australia New Mexico, New Mexico, USA Péter Körtesi, University of Miskolc, László Zsidó, University of Rome, Miskolc, Hungary Tor Vergata, Roma, Italy Maohua Le, Zhangjiang Normal College, Zhangjiang, China Shanhe Wu, Longyan University, Longyan, China The Octogon Mathematical Magazine is a continuation of Gamma Mathematical Magazine (1978-1989) Manuscripts: Should be sent either to the Editor-in-chief. Instruction for authors are given inside the back cover. ISSN 1222-5657 ISBN 978-973-88255-5-0 © Fulgur Publishers Contents Gao Mingzhe Some new Hilbert type inequalities and applications . . . . . . . 4 Song-Zhimin, Dou-Xiangkai and Yin Li On some new inequalities for the Gamma function . . . . . . . 14 Mih´aly Bencze The integral method in inequalities theory . . . . . . . . . . . . 19 Vlad Ciobotariu-Boer Hermite-Hadamard and Fej´er Inequalities for Wright-Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Mih´aly Bencze, Nicu¸sor Minculete and Ovidiu T. Pop New inequalities for the triangle . . . . . . . . . . . . . . . . . . 70 J.Earnest Lazarus Piriyakumar and R. Helen Helly‘s theorem on fuzzy valued functions . . . . . . . . . . . . 90 Mih´aly Bencze About AM-HM inequality . . . . . . . . . . . . . . . . . . . . . 106 Mih´aly Bencze A Generalization of the logarithmic and the Gauss means . . . 117 J. O. Ad´en´ıran, J. T. Akinmoyewa, A. R. T. S.`ol´ar`ın, T. G. Jaiy´eo.la´ On some algebraic properties of generalized groups . . . . . . . 125 Mih´aly Bencze New identities and inequalities in triangle . . . . . . . . . . . . 135 Mih´aly Bencze and D.M. Ba˘tine¸tu-Giurgiu A cathegory of inequalities . . . . . . . . . . . . . . . . . . . . . 149 O.O. Fabelurin, A. G. Adeagbo-Sheikh On Hardy-type integral inequalities involving many functions . 164 Mih´aly Bencze A method to generate new inequalities in triangle . . . . . . . . 173 Jos´e Luis D´ıaz-Barrero and Eusebi Jarauta-Bragulat Some Related Results to CBS Inequality . . . . . . . . . . . . . 182 Mih´aly Bencze and Wei-Dong Jiang One some new type inequalities in triangle . . . . . . . . . . . . 189 Yu-Lin Wu Two geometric inequalities involved two triangles . . . . . . . . 193 Mih´aly Bencze and Nicu¸sor Minculete Some applications of certain inequalities . . . . . . . . . . . . . 199 Mih´aly Bencze and Zhao Changjian A refinement of Jensen‘s inequality . . . . . . . . . . . . . . . . 209 2 Octogon Mathematical Magazine, Vol. 17, No.1, April 2009 Fuhua Wei and Shanhe Wu Generalizations and analogues of the Nesbitt’s inequality . . . . 215 Hui-Hua Wu and Shanhe Wu Various proofs of the Cauchy-Schwarz inequality . . . . . . . . 221 Mih´aly Bencze About a trigonometrical inequality . . . . . . . . . . . . . . . . 230 Jian Liu On Bergstro¨m’s inequality involving six numbers . . . . . . . . 237 Mih´aly Bencze and Yu-Dong Wu New refinements for some classical inequalities . . . . . . . . . 250 J´ozsef S´andor and Ildik´o Bakcsi On the equation ax2+by2 = z2, where a+b = c2 . . . . . . . . 255 Mih´aly Bencze and Yu-Dong Wu About Dumitru Acu‘s inequality . . . . . . . . . . . . . . . . . . 257 J´ozsef S´andor Euler and music. A forgotten arithmetic function by Euler . . 265 Mih´aly Bencze About a partition inequality . . . . . . . . . . . . . . . . . . . . 272 J´ozsef S´andor A divisibility property of σ (n) . . . . . . . . . . . . . . . . . . 275 k Mih´aly Bencze and D.M. Ba˘tine¸tu-Giurgiu New refinements for AM-HM type inequality . . . . . . . . . . . 277 K.P.Pandeyend Characteristics of triangular numbers . . . . . . . . . . . . . . 282 J´ozsef S´andor A double-inequality for σ (n) . . . . . . . . . . . . . . . . . . . 285 k Nicu¸sor Minculete Improvement of one of Sa´ndor‘s inequalities . . . . . . . . . . . 288 Sˇefket Arslanagi´c About one algebraic inequality . . . . . . . . . . . . . . . . . . . 291 J´ozsef S´andor On certain inequalities for the σ function . . . . . . . . . . . . 294 − J´ozsef S´andor A note on the inequality (x +x +...+x )2 n x2+...+x2 297 1 2 n ≤ 1 n J´ozsef S´andor (cid:0) (cid:1) A note on inequalities for the logarithmic function . . . . . . . 299 J´ozsef S´andor On the inequality (f(x))k < f xk . . . . . . . . . . . . . . . . 302 (cid:0) (cid:1) Contents 3 J´ozsef S´andor A note on Bang‘s and Zsigmond‘s theorems . . . . . . . . . . . 304 Mih´aly Bencze Jo´zsef Wildt International Mathematical Competition . . . . . . 306 Book reviews . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Proposed problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 Open questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 R´obert Sza´sz and Aurel P´al Kup´an Solution of the OQ. 2283 . . . . . . . . . . . . . . . . . . . . . 464 Kramer Alp´ar-Vajk A conjecture on a number theoretical function and the OQ. 1240471 Kramer Alp´ar-Vajk The solution of OQ 1156 . . . . . . . . . . . . . . . . . . . . . 475 Kramer Alp´ar-Vajk The solution of OQ 1141 . . . . . . . . . . . . . . . . . . . . . 476 Gabriel T. Pr˘ajitura and Tsvetomira Radeva A logarithmic equation (OQ 19) . . . . . . . . . . . . . . . . . . 477 OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No.1, April 2009, pp 4-13 4 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon Some new Hilbert type inequalities and applications Gao Mingzhe1 ABSTRACT. In this paper it is shown that some new Hilbert type integral inequalities can be established by introducing a proper logarithm function. And the constant factor is proved to be the best possible. In particular, for case , the classical Hilbert inequality and its equivalent form are obtained. As applications, some new inequalities which they are equivalent each other are built. 1. INTRODUCTION Let f(x),g(x) L2(0,+ ). Then ∈ ∞ 1 1 2 2 ∞ ∞f(x)g(x) ∞ ∞ dxdy π f2(x)dx g2(x)dx (1.1) x+y ≤ Z0 Z0 Z0 Z0 This is the famous Hilbert integral inequality, where the coefficient π is the best possible. In the papers [1-2], the following inequality of the form 1 1 ∞ ∞ ln x f(x)g(y) ∞ 2 ∞ 2 y dxdy π2 f2(x)dx g2(x)dx (1.2) (cid:16) (cid:17)x y ≤ Z0 Z0 − Z0 Z0 was established, and the coefficient π2 is also thebestpossible. Owing to the importance of the Hilbert inequality and the Hilbert type inequality in analysis and applications, some mathematicians have been studying them. Recently, various improvements and extensions of (1.1) and (1.2) appear in a great deal of papers (see [3]-[8]etc.). The aim of the present paper is to build some new Hilbert type integral inequalities by introducing a proper integral kernel function and by using the 1Received: 20.02.2009 2000 Mathematics Subject Classification. 26D15 Keywordsandphrases. Hilberttypeintegralinequality,logarithmfunction,euler number, the best constant, equivalent inequalities. Some new Hilbert type inequalities and applications 5 technique of analysis, and to discuss the constant factor of which is related to the Euler number, and then to study some equivalent forms of them. In the sake of convenience, we introduce some notations and define some functions. Let 0 < α < 1 and n be a positive integer. Define a function ζ by ∗ ∞ ( 1)k ζ∗(n,α) = (α−+k)n. (1.3) k=0 X And further define the function ζ by 2 1 ζ = (2n)! 2ζ 2n+1, ,(n N ) (1.4) 2 ∗ 0 2 ∈ (cid:26) (cid:18) (cid:19)(cid:27) In order to prove our main results, we need the following lemmas. Lemma 1.1. Let 0 < α < 1 and n be a nonnegative integer. Then 1 n 1 1 tα 1 ln dt = n!ζ (n+1,α). (1.5) − ∗ t 1+t Z (cid:18) (cid:19) 0 where ζ is defined by (1.3). ∗ This result has been given in the paper [9]. Hence its proof is omitted here. Lemma 1.2. With the assumptions as Lemma 1.1, then ∞ 1 2n 1 uα 1 ln du = (2n)! ζ (2n+1,α)+ζ (2n+1,1 α) (1.6) − ∗ ∗ u 1+u { − } Z (cid:18) (cid:19) 0 where ζ is defined by (1.3). ∗ Proof. It is easy to deduce that 1 ∞ 1 2n 1 1 2n 1 uα 1 ln du = uα 1 ln du+ − − u 1+u u 1+u Z (cid:18) (cid:19) Z (cid:18) (cid:19) 0 0 1 ∞ 1 2n 1 1 2n 1 + uα 1 ln du = uα 1 ln du+ − − u 1+u u 1+u Z (cid:18) (cid:19) Z (cid:18) (cid:19) 1 0 6 Gao Mingzhe 1 1 2n 1 1 1 + v α(lnv)2n dv = uα 1 ln du+ − − 1+v u 1+u Z Z (cid:18) (cid:19) 0 0 1 2n 1 1 + v(1 α) 1 ln dv. − − v 1+v Z (cid:18) (cid:19) 0 By using Lemma 1.1, the equality (1.6) is obtained at once. 0 Throughout the paper, we define ln x = 1, when x = y. y (cid:16) (cid:17) 2. MAIN RESULTS We are ready now to formulate our main results. Theorem 2.1. Let f and g be two real functions, and n be a nonnegative integer, If ∞ ∞ f2(x)dx < + and g2(x)dx < + , then ∞ ∞ Z Z 0 0 2n ∞ ∞ ln x f(x)g(y) y dxdy (cid:16) (cid:17)x+y ≤ Z Z 0 0 1 1 2 2 ∞ ∞ π2n+1E f2(x)dx g2(x)dx , (2.1) n ≤ (cid:0) (cid:1)Z0 Z0 where the constant factor π2n+1E is thebest possible, and that E = 1 and n 0 E is the Euler number, viz. E = 1, E = 5, E = 61, E = 1385, n 1 2 3 4 E = 50521, etc. 5 Proof. We may apply the Cauchy inequality to estimate the left-hand side of (2.1) as follows: 2n ∞ ∞ ln x f(x)g(y) y dxdy = (cid:16) (cid:17)x+y Z Z 0 0 Some new Hilbert type inequalities and applications 7 1 1 = ∞ ∞ ln xy 2n 2 x 41 f(x) ln xy 2n 2 y 41 gydxdy (cid:16)x+(cid:17)y y (cid:16)x+(cid:17)y x ≤ Z0 Z0 (cid:18) (cid:19) (cid:16) (cid:17) 1 1 ∞ ∞ ln xy 2n x 21 2 ∞ ∞ ln xy 2n x 21 2 f2(x)dxdy g2(x)dxdy = ≤ (cid:16)x+(cid:17)y y (cid:16)x+(cid:17)y y Z0 Z0 (cid:18) (cid:19) Z0 Z0 (cid:18) (cid:19) 1 1 2 2 ∞ ∞ = ω(x)f2(x)dx ω(x)g2(x)dx (2.2) Z Z 0 0 lnx 2n 1 where ω(x) = ∞ y x 2 dy, “ x+”y y 0 By using LemmRa 1.2, it i(cid:16)s e(cid:17)asy to deduce that 2n ∞ ln xy x 21 ∞ 1 1 2n 1 ω(x) = x(cid:16) 1+(cid:17)y y dy = u−2 ln u 1+udu = ζ2. (2.3) Z x (cid:18) (cid:19) Z (cid:18) (cid:19) 0 0 (cid:0) (cid:1) where ζ is defined by (1.4). Based on (1.3) and (1.4), we have 2 1 ∞ ( 1)k ζ2 = (2n)! 2ζ∗ 2n+1, 2 = (2n)!2 1 +−k 2n+1 = (cid:26) (cid:18) (cid:19)(cid:27) k=0 2 X ∞ ( 1)k (cid:0) (cid:1) = (2n)!22n+2 − . (2k+1)2n+1 k=0 X It is known from the paper [10] that ∞ ( 1)k π2n+1 − = E . (2.4) 1 +k 2n+1 22n+2(2n)! n k=0 2 X where En is the Euler num(cid:0)ber,vi(cid:1)z. E1 = 1, E2 = 5, E3 = 61, E4 = 1385, E = 50521, etc. 5 Since ∞ (2−k1+)1k = π4, we can define E0 = 1. As thus, the relation (2.4) is also k=0 valid wPhen n = 0. So, we get from (2.3) and (2.4) that 8 Gao Mingzhe ω(x) = π2n+1E , (2.5) n It follows from (2.2) and (2.5) that the inequality (2.1) is valid. It remains to need only to show that π2n+1E in (2.1) is the best possible. n 0 if x (0,1) ε > 0. Define two functions by f(x) = ∈ and 1+ε ∀ ( x− 2 if x [1, ) ∈ ∞ 0 if y (0,1) e g(y) = ∈ . It is easy to deduce that 1+ε ( y− 2 if y [1, ) ∈ ∞ + + e ∞ ∞ 1 f2(x)dx = g2(y)dy = . ε Z Z 0 0 f e If π2n+1E is not the best possible, then there exists C > 0, such that n C < π2n+1E and n 2n ∞ ∞ ln x f(x)g(y) y S f,g = dxdy (cid:16) (cid:17)x+y ≤ (cid:16) (cid:17) Z0 Z0 e e e e 1 1 2 2 ∞ ∞ C C f2(x)dx g2(y)dy = . (2.6) ≤ ε Z Z 0 0 f e On the other hand, we have ∞ ∞ x−1+2ε ln xy 2ny−1+2ε S f,g = n o(cid:26)(cid:16) (cid:17) (cid:27)dxdy = x+y (cid:16) (cid:17) Z0 Z0 e e ∞ ∞ ln xy 2ny−1+2ε 1+ε = (cid:16) x(cid:17)+y dy x− 2 dx = Z0 Z0 n o = ∞ ∞ ln u1 2nu−1+2εdu x 1 ε dx = − − 1+u Z0 Z0 (cid:0) (cid:1) (cid:8) (cid:9) 1 ∞ 1+ε 1 2n 1 = u− 2 ln du. (2.7) ε u 1+u Z (cid:18) (cid:19) 0
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