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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2009, Article ID 628051, 16 pages doi:10.1155/2009/628051 Research Article Cauchy Means of the Popoviciu Type Matloob Anwar, 1 Naveed Latif, 1 and J. Pe ˇ cari ´ c 1, 2 1 Abdus Salam School of Mathematical Sciences, 68-B New Muslim Town, GC University, Lahore 54000, Pakistan 2 Faculty of Textile Technology, University of Zagreb, Pierottijeva 6, Zagreb 10000, Croatia Correspondence should be addressed to Naveed Latif, sincerehumtum@yahoo.com Received 9 October 2008; Accepted 2 February 2009 Recommended by Wing-Sum Cheung We discuss log-convexity for the differences of the Popoviciu inequalities and introduce some mean value theorems and related results. Also we give the Cauchy means of the Popoviciu type and we show that these means are monotonic. Copyright q 2009 Matloob Anwar et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction and Preliminaries Let fx and px be two positive real valued functions with  b a pxdx  1, then from theory of convex means cf. 1–3, the well-known Jensen inequality gives that for t<0ort>1,  b a pxfx t dx ≥   b a pxfxdx  t , 1.1 and vise versa for 0 <t<1. In 4, Simic has considered the difference D s  D s a, b, f, p  b a pxfx s dx −   b a pxfxdx  s . 1.2 The following result was given in 4see also 5. 2 Journal of Inequalities and Applications Theorem 1.1. Let fx, px be nonnegative and integrable functions for x ∈ a, b,with  b a pxdx  1, then for 0 <r<s<t; r, s, t /  1, one has  D s ss − 1  t−r ≤  D r rr − 1  t−s  D t tt − 1  s−r . 1.3 Remark 1.2. For extension of Theorem 1.1 see cf. 4. Popoviciu 6–8, 9, pages 214-215 has proved the following results. Theorem 1.3. Let φ : a, b → R be convex and f : 0, 1 → R be continuous, increasing, and convex such that a ≤ fx ≤ b for x0, 1.Then  1 0 φfxdx ≤ b  a − 2  f b − a φa 2  f − a b − a 2  b a φxdx, 1.4 where  f   1 0 fxdx. 1.5 If φ is strictly convex, then the equality in 1.4 holds if and only if fxa b − a x − λ  |x − λ| 21 − λ , where λ  b  a − 2  f b − a . 1.6 Theorem 1.4. Let φ : a, b → R be continuous and convex, and let f : 0, 1 → R be convex of order 1, ,n 1 such that a ≤ fx ≤ b for x0, 1. Then  1 0 φfxdx ≤  1 0 φ  U j   f,x  dx for ja  b j  1 ≤  f ≤ j − 1a  b j , 2 ≤ j ≤ n, 1.7  1 0 φfxdx ≤ V   f for a ≤  f ≤ na  b n  1 , 1.8 where U j t, xa  jj  1  t − ja  b j  1    j − 1a  b j − t  x  x j−1 , 1.9 V x b  na − n  1x b − a φa n  1x − a nb − a n1/n  b a φxdx x − a n−1/n . 1.10 Journal of Inequalities and Applications 3 If φ is strictly convex, then equality in 1.7 holds if and only if fxU j   f,x; 1.11 and equality in 1.8 holds if fxa b − a  x − λ  |x − λ| 21 − λ  n , where λ  b  na − n  1  f b − a . 1.12 With the help of the following useful lemmas we prove our results. Lemma 1.5. Define the function ϕ s x ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ x s ss − 1 ,s /  0, 1; − log x, s  0; x log x, s  1. 1.13 Then ϕ  s xx s−2 , that is, ϕ s x is convex for x>0. The following lemma is equivalent to definition of convex function see 9, page 2. Lemma 1.6. If φ is a convex function on I for all s 1 ,s 2 ,s 3 ∈ I for which s 1 <s 2 <s 3 , the following is valid φ  s 1  s 3 − s 2   φ  s 2  s 1 − s 3   φ  s 3  s 2 − s 1  ≥ 0. 1.14 We quote here another useful lemma from log-convexity theory cf. 4. Lemma 1.7. A positive function f is log-convex in the Jensen-sense on an open interval I, that is, for each s, t ∈ I, fsft ≥ f 2  s  t 2  , 1.15 if and only if the relation u 2 fs2uwf  s  t 2   w 2 ft ≥ 0, 1.16 holds for each real u, w and s, t ∈ I. 4 Journal of Inequalities and Applications The following lemma given in 10 gives the relation between Beta function β and Hypergeometric function F. Lemma 1.8. Suppose a, b, c, α, γ ∈ R are such that a  c>b>0 and 0 <α<2γ, β and F are Beta and Hypergeometric functions, respectively. Then  ∞ 0 x b−1 1  αx a 1  γx c dx  γ −b βb, a  c − bF  a, b a  c     γ − α γ  . 1.17 The paper is organized in the following way. After this introduction, in the second section we discuss the log-convexity of differences of the Popoviciu inequalities 1.4, 1.7, and 1.8. In the third section we introduce some mean value theorems and the Cauchy means of the Popoviciu-type and discuss its monotonicity. 2. Main Results Theorem 2.1. Let f : 0, 1 → R be continuous, increasing, and convex such that 0 <a≤ fx ≤ b for x ∈ 0, 1, and let  f be defined in 1.5 and Ω s f ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 ss − 1  b  a − 2  f b − a a s  2  f − a b − a 2 s  1  b s1 − a s1  −  1 0 fx s dx  ,s /  0, 1; 2  f − a b − a   1 0 logfxdx − b  a − 2  f b − a log a − 2  f − a b − a 2 b log b − a log a,s 0; b  a − 2  f b − a a log a   f − a b − a 2  b 2 log b − a 2 log a  −   f − ab  a 2b − a −  1 0 fxlogfxdx, s  1, 2.1 and let Ω s f be positive. One has that Ω s f is log-convex and the following inequality holds for −∞ <r<s<t<∞, Ω t−r s f ≤ Ω t−s r fΩ s−r t f. 2.2 Proof. Consider the function defined by ωxu 2 ϕ s x2uwϕ r xw 2 ϕ t x, 2.3 where r s  t/2, ϕ s is defined by 1.13 and u, w ∈ R. We have ω  xu 2 x s−2  2uwx r−2  w 2 x t−2   ux s/2−1  wx t/2−1  2 > 0,x>0. 2.4 Journal of Inequalities and Applications 5 Therefore, ωx is convex for x>0. Using Theorem 1.3, b  a − 2  f b − a  u 2 ϕ s a2uwϕ r aw 2 ϕ t a   2  f − a b − a 2  b a  u 2 ϕ s x2uwϕ r xw 2 ϕ t x  dx ≥  1 0  u 2 ϕ s fx  2uwϕ r fx  w 2 ϕ t fx  dx, u 2  b  a − 2  f b − a ϕ s a 2  f − a b − a 2  b a ϕ s xdx −  1 0 ϕ s fxdx   2uw  b  a − 2  f b − a ϕ r a 2  f − a b − a 2  b a ϕ r xdx −  1 0 ϕ r fxdx   w 2  b  a − 2  f b − a ϕ t a 2  f − a b − a 2  b a ϕ t xdx −  1 0 ϕ t fxdx  ≥ 0, 2.5 since Ω s f b  a − 2  f b − a ϕ s a 2  f − a b − a 2  b a ϕ s xdx −  1 0 ϕ s fxdx,  f   1 0 fxdx, 2.6 we have u 2 Ω s f2uwΩ r fw 2 Ω t f ≥ 0. 2.7 By Lemma 1.7, we have Ω s fΩ t f ≥ Ω 2 r fΩ 2 st/2 f, 2.8 that is Ω s f is log-convex in the Jensen-sense for s ∈ R. Since lim s → 0 Ω s fΩ 0 f, lim s → 1 Ω s fΩ 1 f. 2.9 This implies Ω s f is continuous, therefore it is log-convex. Since Ω s f is log-convex, that is, log Ω s f is convex, therefore by Lemma 1.6 for −∞ <r<s<t<∞ and taking φ s  log Ω s ,weget log Ω t−r s f ≤ log Ω t−s r flog Ω s−r t f, 2.10 which is equivalent to 2.2. 6 Journal of Inequalities and Applications Theorem 2.2. Let f, Ω s f be defined in Theorem 2.1 and let t, s, u, v be real numbers such that s ≤ u, t ≤ v, s /  t, u /  v, one has  Ω t f Ω s f  1/t−s ≤  Ω v f Ω u f  1/v−u . 2.11 Proof. In cf. 9, page 2, we have the following result for convex function f with x 1 ≤ y 1 , x 2 ≤ y 2 , x 1 /  x 2 , y 1 /  y 2 : f  x 2  − f  x 1  x 2 − x 1 ≤ f  y 2  − f  y 1  y 2 − y 1 . 2.12 Since by Theorem 2.1, Ω s f is log-convex, we can set in 2.12: fxlog Ω x and x 1  s, x 2  t, y 1  u, y 2  v.Weget log Ω t f − log Ω s f t − s ≤ log Ω v f − log Ω u f v − u , log  Ω t f Ω s f  1/t−s ≤ log  Ω v f Ω u f  1/v−u , 2.13 and after applying exponential function, we get 2.11. Theorem 2.3. Let f : 0, 1 → R be convex of order 1, ,n 1 such that 0 <a≤ fx ≤ b for x ∈ 0, 1, and let  f be defined in 1.5 and Λ s f  1 0 ϕ s  U j   f,x  dx −  1 0 ϕ s fxdx, 2.14 where U j   f,x   a  jj  1   f − ja  b j  1    j − 1a  b j −  f  x  x j−1 , 2.15 for ja  b j  1 ≤  f ≤ j − 1a  b j , 2 ≤ j ≤ n, 2.16 and let Λ s f be positive. One has that Λ s f is log-convex and the following inequality holds for −∞ <r<s<t<∞, Λ t−r s f ≤ Λ t−s r fΛ s−r t f. 2.17 Proof. As in the proof of Theorem 2.1,weuseTheorem 1.4 instead of Theorem 1.3. Journal of Inequalities and Applications 7 Theorem 2.4. Let f, Λ s f be defined in Theorem 2.3 and t, s, u, v be real numbers such that s ≤ u, t ≤ v, s /  t, u /  v, one has  Λ t f Λ s f  1/t−s ≤  Λ v f Λ u f  1/v−u . 2.18 Proof. Similar to the proof of Theorem 2.2. Lemma 2.5. Let f : 0, 1 → R be convex of order 1, ,n 1 such that 0 <a≤ fx ≤ b for x ∈ 0, 1,  f be defined in 1.5 and V be defined in 1.10, and let β and F are Beta and Hypergeometric functions respectively, and Γ s fV   f  −  1 0 ϕ s fxdx. 2.19 Then Γ s f ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1 ss − 1  b  na − n  1  f b − a a s  n  1  f − a nb − a n1/n a s1/n  b − a b  1/n ×β  1 n , 1  F ⎛ ⎜ ⎜ ⎝ s 1 n 1, 1 n 1 n  1         b−a b ⎞ ⎟ ⎟ ⎠ −  1 0 ft s dt ⎤ ⎥ ⎥ ⎦ ,s /  0, 1; b  na − n  1  f b − a −log a − n  1  f − a nb − a n1/n × ⎡ ⎢ ⎣ n log bb − a 1/n − na 1/n  b − a b  1/n1 β  1 n  1, 1  F ⎛ ⎜ ⎝ 1 n  1, 1 n  1 1 n  2        b − a b ⎞ ⎟ ⎠ ⎤ ⎥ ⎦   1 0 logftdt, s  0; b  na − n  1  f b − a a log a  n  1  f − a nb − a n1/n ×  nab−a 1/n log b n n  1 log bb−a 1/n1 −a 1/n  b−a b  1/n1 β  1 n 1, 1  ×F ⎛ ⎜ ⎝ 1 n  1, 1 n  1 1 n  2        b − a b ⎞ ⎟ ⎠  n n  1  na  ⎤ ⎥ ⎦ −  1 0 ftlogftdt, s  1, 2.20 8 Journal of Inequalities and Applications for a ≤  f ≤ na  b n  1 . 2.21 Proof. First, we solve these three integrals I 1   b a t s t − a n−1/n dt, I 2   b a log t t − a n−1/n dt, I 3   b a t log t t − a n−1/n dt. 2.22 Take I 1   b a t s t − a n−1/n dt. 2.23 Substitute t  a  bx 1  x ,dt b − a 1  x 2 dx, A and limits, when t → a then x → 0, when t → b then x →∞.So, I 1   b a t s t − a n−1/n dt   ∞ 0 a  bx/1  x s a  bx/1  x − a 1−1/n · b − a 1  x 2 dx b − a 1/n  ∞ 0 a  bx s 1  x 1−1/n 1  x s2 x 1−1/n dx, I 1 b − a 1/n a s  ∞ 0 x 1/n−1 1  x s1/n1 1 b/ax −s dx. 2.24 By using Lemma 1.8 with a  s  1/n  1, b  1/n, c  −s, α  1, γ  b/a such that 1/n  1 > 1/n > 0and0< 1 < 2b/a,weget I 1  a s1/n b − a 1/n β  1 n , 1  F ⎛ ⎜ ⎜ ⎝ s  1 n  1, 1 n 1 n  1         b − a b ⎞ ⎟ ⎟ ⎠ . 2.25 Take second integral I 2   b a log t t − a n−1/n dt, 2.26 Journal of Inequalities and Applications 9 using integration by parts, we have I 2  n log bb − a 1/n − n  b a t −1 t − a 1/n dt. 2.27 Let I 4   b a t −1 t − a 1/n dt. 2.28 By using same substitution A as above, we get I 4   ∞ 0 1  x a  bx ·  a  bx 1  x − a  1/n · b − a 1  x 2 dx  b − a 1/n1 a  ∞ 0 x 1/n1−1 1  x 1/n1 1 b/ax dx. 2.29 By using Lemma 1.8 with a  1/n  1, b  1/n  1, c  1, α  1, γ  b/a such that 1/n  2 > 1/n  1 > 0and0< 1 < 2b/a,weget I 2  n log bb − a 1/n − na 1/n  b − a b  1/n1 β  1 n  1, 1  F ⎛ ⎜ ⎜ ⎝ 1 n  1, 1 n  1 1 n  2         b − a b ⎞ ⎟ ⎟ ⎠ . 2.30 Now, take third integral I 3   b a t log t t − a n−1/n dt. 2.31 Using integration by parts, we get I 3  nab − a 1/n log b  n n  1 log bb − a 1/n1 − n n  1  b a t − a 1/n1 t dt − na  b a t − a 1/n t dt. 2.32 Let I 5   b a t − a 1/n1 t dt. 2.33 By using same substitution A as above, we get I 5   ∞ 0 a  bx/1  x − a 1/n1 a  bx/1  x · b − a 1  x 2 dx  b − a 1/n1 a  ∞ 0 x 1/n1−1 1  x 1/n1 1 b/ax dx. 2.34 10 Journal of Inequalities and Applications By using Lemma 1.8 with a  1/n  1, b  1/n  1, c  1, α  1, γ  b/a such that 1/n  2 > 1/n  1 > 0and0< 1 < 2b/a,weget I 5  a 1/n  b − a b  1/n1 β  1 n  1, 1  F ⎛ ⎜ ⎜ ⎝ 1 n  1, 1 n  1 1 n  2         b − a b ⎞ ⎟ ⎟ ⎠ . 2.35 Let I 6   b a t − a 1/n t dt. 2.36 By using same substitution A as above, we have I 6   ∞ 0 a  bx/1  x − a 1/n a  bx/1  x · b − a 1  x 2 dx  b − a 1/n1 a  ∞ 0 x 1/n1−1 1  x 1/n1 1 b/ax dx. 2.37 By using Lemma 1.8 with a  1/n  1, b  1/n  1, c  1, α  1, γ  b/a such that 1/n  2 > 1/n  1 > 0and0< 1 < 2b/a,weget I 6  a 1/n  b − a a  1/n1 β  1 n  1, 1  F ⎛ ⎜ ⎜ ⎝ 1 n  1, 1 n  1 1 n  2         b − a b ⎞ ⎟ ⎟ ⎠ . 2.38 Then I 3  nab − a 1/n log b  n n  1 log bb − a 1/n1 − a 1/n  b − a a  1/n1 × β  1 n  1, 1  F  1 n  1, 1 n  1, 1 n  2     b − a b  n n  1  na  . 2.39 For s /  0, 1, Γ s f 1 ss − 1  b  na − n  1  f b − a a s  n  1  f − a nb − a n1/n  b a t s t − a n−1/n dt −  1 0 ft s dt  . 2.40 [...]... Proof Since Ωs f is log-convex, therefore by 2.11 we get 3.21 Remark 3.7 Similar results of the Cauchy means and related results can also proved for Theorems 2.3 and 2.6 Acknowledgments This research work is funded by Higher Education Commission Pakistan The research of the third author is supported by the Croatian Ministry of Science, Education and Sports under the Research Grants 117-1170889-0888... Let f, Γs f be defined in Theorem 2.6 and t, s, u, v be real numbers such that s ≤ u, t ≤ v, s / t, u / v, one has 1/ t−s Γt f Γs f Γv f Γu f ≤ 1/ v−u 2.47 Proof Similar to the proof of Theorem 2.2 3 Cauchy Means Let us note that 2.11 has the form of some known inequalities between means e.g., Stolarsky’s means, etc Here we prove that expressions on both sides of 2.11 are also means Lemma 3.1 Let h ∈... Journal of Inequalities and Applications Theorem 2.6 Let f : 0, 1 → R be convex of order 1, , n 1 such that 0 < a ≤ f x ≤ b for x ∈ 0, 1 , f be defined in 1.5 and let Γs f defined in 2.20 be positive One has that Γs f is log-convex and the following inequality holds for −∞ < r < s < t < ∞, Γt−r f ≤ Γt−s f Γs−r f s r t 2.46 Proof As in the proof of Theorem 2.1, we use Theorem 1.4 instead of Theorem 1.3 Theorem... Analytic Methods of Probability Theory, vol 67 of Mathematische Lehrbucher und Monographien, Abteilung 2: Mathematische Monographien, Akademie, Berlin, Germany, ¨ 1985 4 S Simic, “On logarithmic convexity for differences of power means, ” Journal of Inequalities and Applications, vol 2007, Article ID 37359, 8 pages, 2007 5 M Anwar and J Peˇ ari´ , “On logarithmic convexity for differences of power means and... combination of k and l should also satisfy 3.2 14 Journal of Inequalities and Applications Let η be defined as follows: k ξ 2 η k ξ b a − 2f 2 a b−a 2 f − a b2 ba 3 b−a a2 1 − f x 2 dx 3.10 0 Obviously, we have Ψ m 0 3.11 On the other hand, there is an ξ ∈ I1 such that Ψ m η m ξ 0 3.12 By using the linearity property of the operator η Ψ l η k ξ −Ψ k η l ξ 0 3.13 Now Ψ l / 0 and η l ξ / 0, we have from the. .. s dx dx 1/ t−s ≤ b 3.16 Journal of Inequalities and Applications 15 In fact, similar result can also be given for 3.7 Namely, suppose that k /l has inverse function Then from 3.7 we have −1 Dk a L Dl a k l ξ L b k a b l a x dx − x dx − 1 k f x 0 1 l f x 0 dx dx 3.17 The expression on the right-hand side of 3.17 is also a mean From the inequality 3.16 , we can define means Mt,s f as follows: Dat L bt... Journal of Inequalities and Applications 1 0 f x μa log a 2 , H denotes 1 0 s log f x dx, H denotes μ log a 2 , G denotes 2μ log a, K denotes f x log f x μ b 2 1 f 0 dx, and O denotes a − 2f , b−a ν 2 f −a b−a 2 x log f x dx, where 3.20 In our next result we prove that this new mean is monotonic Theorem 3.6 Let t ≤ u, r ≤ s, then the following inequality is valid Mt,r f; w ≤ Mu,s f; w 3.21 Proof Since... f x ≤ b for x ∈ 0, 1 , then for −∞ < s / t / 0, 1 / s < ∞ there exists ξ ∈ I1 such that ξt−s where D denotes b Proof Set k x s s−1 t t−1 Dat L bt 1 − at 1 / t 1 − Das L bs 1 − as 1 / s 1 − 1 0 1 0 f x t dx f x s dx , 3.15 a − 2f / b − a and L denotes 2 f − a / b − a 2 ϕt x and l x ϕs x , t / s / 0, 1 in 3.7 we get 3.15 Remark 3.5 Since the function ξ → ξt−s is invertible, therefore from 3.15 we have... 2 a 1 − f x 2 dx 0 By combining 3.4 and 3.5 and using the fact that for m ≤ ρ ≤ M there exists ξ ∈ I1 such ρ we get 3.2 that h ξ Theorem 3.3 Let k, l ∈ C2 I1 and satisfy 3.2 , f be a continuous, increasing and convex such that a ≤ f x ≤ b for x ∈ 0, 1 , f be defined in 1.5 , and b−a f x /a x − λ |x − λ| , 2 1−λ where λ a − 2f , b−a b 3.6 then there exists ξ ∈ I1 such that k ξ l ξ 2 f −a / b−a b a... Lemma 3.1 Let h ∈ C2 I be such that h is bounded, that is, m ≤ h ≤ M Then the functions φ1 , φ2 defined by φ1 t M 2 t −h t , 2 φ2 t h t − m 2 t , 2 3.1 are convex functions Theorem 3.2 Let h ∈ C2 I1 , I1 is a compact interval in R and f be a continuous, increasing and convex such that a ≤ f x ≤ b for x ∈ 0, 1 , f be defined in 1.5 then ∃ ξ ∈ a, b , ξ / 0 such that b 2 f −a a − 2f h a b−a h ξ 2 b−a b . log-convexity for the differences of the Popoviciu inequalities and introduce some mean value theorems and related results. Also we give the Cauchy means of the Popoviciu type and we show that these means. discuss the log-convexity of differences of the Popoviciu inequalities 1.4, 1.7, and 1.8. In the third section we introduce some mean value theorems and the Cauchy means of the Popoviciu- type. and the following inequality holds for −∞ <r<s<t<∞, Γ t−r s f ≤ Γ t−s r fΓ s−r t f. 2.46 Proof. As in the proof of Theorem 2.1,weuseTheorem 1.4 instead of Theorem 1.3. Theorem