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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID 579567, 17 pages doi:10.1155/2010/579567 Research Article On Refinements of Aczél, Popoviciu, Bellman’s Inequalities and Related Results G. Farid,1 J. Pečarić,1, 2 and Atiq Ur Rehman1 1 Abdus Salam School of Mathematical Sciences, GC University, 68-B, New Muslim Town, Lahore 54600, Pakistan 2 Faculty of Textile Technology, University of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatia Correspondence should be addressed to G. Farid, faridphdsms@hotmail.com Received 11 July 2010; Revised 18 September 2010; Accepted 27 November 2010 Academic Editor: Q. Lan Copyright q 2010 G. Farid 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. We give some refinements of the inequalities of Aczél, Popoviciu, and Bellman. Also, we give some results related to power sums. 1. Introduction The well-known Aczél’s inequality 1 see also 2, page 117 is given in the following result. Theorem 1.1. Let n be a fixed positive integer, and let A, B, ak , bk k 1, . . . , n be real numbers such that A2 − n a2k > 0, B2 − k1 n bk2 > 0, 1.1 k1 then n A − a2k 2 k1 1/2 B − 2 n k1 1/2 bk2 ≤ AB − n ak bk , k1 with equality if and only if the sequences A, a1 , . . . , .an and B, b1 , . . . , bn are proportional. A related result due to Bjelica 3 is stated in the following theorem. 1.2 2 Journal of Inequalities and Applications Theorem 1.2. Let n be a fixed positive integer, and let p, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that Ap − n p ak > 0, Bp − k1 n p bk > 0, 1.3 k1 then, for 0 < p ≤ 2, one has p A − n k1 p ak 1/p p B − n k1 p bk 1/p n ≤ AB − ak bk . 1.4 k1 Note that quotation of the above result in 4, page 58 is mistakenly stated for all p ≥ 1. In 1990, Bjelica 3 proved that the above result is true for 0 < p ≤ 2. Mascioni 5, in 2002, gave the proof for 1 < p ≤ 2 and gave the counter example to show that the above result is not true for p > 2. Dı́az-Barreo et al. 6 mistakenly stated it for positive integer p and gave a refinement of the inequality 1.4 as follows. Theorem 1.3. Let n, p be positive integers, and let A, B, ak , bk , k 1, . . . , n be nonnegative real numbers such that 1.3 is satisfied, then for 1 ≤ j < n, one has p A − n k1 p ak p B − n k1 p bk ≤ RA, B, ak , bk ≤ n AB − ak bk p , 1.5 k1 where ⎞p ⎛ j j n p p p ⎟ ⎜p RA, B, ak , bk ⎝ Ap − ak Bp − bk − ak bk ⎠ . k1 k1 1.6 kj1 Moreover, Dı́az-Barreo et al. 6 stated the above result as Popoviciu’s generalization of Aczél’s inequality given in 7. In fact, generalization of inequality 1.2 attributed to Popoviciu 7 is stated in the following theorem see also 2, page 118. Theorem 1.4. Let n be a fixed positive integer, and let p, q, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that Ap − n p ak > 0, Bq − k1 n q bk > 0. 1.7 k1 Also, let 1/p 1/q 1, then, for p > 1, one has p A − n k1 p ak 1/p q B − n k1 q bk 1/q If p < 1 p / 0, then reverse of the inequality 1.8 holds. ≤ AB − n k1 ak bk . 1.8 Journal of Inequalities and Applications 3 The well-known Bellman’s inequality is stated in the following theorem 8 see also 2, pages 118-119. Theorem 1.5. Let n be a fixed positive integer, and let p, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that 1.3 is satisfied. If p ≥ 1, then p A − n k1 p ak 1/p p B − n k1 1/p p bk p ≤ A B − n 1/p p ak bk . 1.9 k1 Dı́az-Barreo et al. 6 gave a refinement of the above inequality for positive integer p. They proved the following result. Theorem 1.6. Let n, p be positive integers, and let A, B, ak , bk , k 1, . . . , n be nonnegative real numbers such that 1.3 is satisfied, then for 1 ≤ j < n, one has n p A − ak 1/p p k1 n p B − bk 1/p p k1 ≤ RA, B, ak , bk ≤ n A Bp − ak bk p 1.10 1/p , k1 where ⎞p ⎤1/p ⎡ ⎛ j j n p p⎟ ⎥ p ⎢⎜ p RA, B, ak , bk ⎣⎝ Ap − ak Bp − bk ⎠ − ak bk p ⎦ . k1 k1 1.11 kj1 In this paper, first we give a simple extension of a Theorem 1.2 with Aczél’s inequality. Further, we give refinements of Theorems 1.2, 1.4, and 1.5. Also, we give some results related to power sums. 2. Main Results To give extension of Theorem 1.2, we will use the result proved by Pečarić and Vasić in 1979 9, page 165. Lemma 2.1. Let p, q, A, ak k 1, . . . , n be nonnegative real numbers such that Ap − then for 0 < p ≤ q, one has n p A − ak p k1 1/p ≤ n q A − ak q k1 n k1 p ak > 0, 1/q . 2.1 4 Journal of Inequalities and Applications Theorem 2.2. Let n be a fixed positive integer, and let p, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that 1.3 is satisfied, then, for 0 < p ≤ 2, one has 1/p n p A − ak n p B − bk p k1 ≤ 1/p p k1 n A2 − a2k 1/2 B2 − k1 n k1 1/2 bk2 2.2 n ≤ AB − ak bk . k1 Proof. By using condition 1.3 in Lemma 2.1 for 0 < p ≤ 2, we have n p A − ak 1/p p n A − a2k ≤ k1 Bp − n k1 p 1/2 , 2 k1 1/p bk n B2 − bk2 ≤ 2.3 1/2 . k1 These imply n p A − ak p k1 1/p n p B − bk 1/p p ≤ A − 2 k1 n k1 1/2 a2k B − 2 n k1 1/2 bk2 . 2.4 Now, applying Azcél’s inequality on right-hand side of the above inequality gives us the required result. Let p and q be positive real numbers such that 1/p 1/q 1, then the well-known Hölder’s inequality states that n ak bk ≤ k1 n k1 p ak 1/p 1/q n q bk , 2.5 k1 where ak , bk k 1, . . . , n are positive real numbers. If 0 < p ≤ q, then the well-known inequality of power sums of order p and q states that n 1/q n 1/p q p bk ≤ bk , k1 2.6 k1 where bk k 1, . . . , n are positive real numbers c.f 9, page 165. Now, if 1 < p ≤ 2, then q ≥ 2 and using inequality 2.6 in 2.5, we get n ak bk ≤ k1 n k1 p ak 1/p 1/p n p bk . k1 2.7 Journal of Inequalities and Applications 5 We use the inequality 2.7 and the Hölder’s inequality to prove the further refinements of the Theorems 1.2 and 1.4. Theorem 2.3. Let j and n be fixed positive integers such that 1 ≤ j < n, and let p, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that 1.3 is satisfied. Let one denote 1/p j p A − ak p M , j p B − bk 1/p p N k1 . 2.8 k1 i If 0 < p ≤ 2, then n p A − ak 1/p n p B − bk p 1/p p k1 k1 n ≤ MN − 2.9 n ak bk ≤ AB − ak bk . kj1 k1 ii If 1 < p ≤ 2, then n p A − ak 1/p p p B − k1 n k1 ⎛ ≤ MN − ⎝ n p bk 1/p ⎞1/p ⎛ p ak ⎠ ⎝ kj1 n kj1 ⎞1/p p bk ⎠ n ≤ MN − 2.10 ak bk . kj1 Proof. i First of all, we observe that M, N > 0 and also 0 < p ≤ 2, therefore by Theorem 1.2, we have MN ≤ AB − j ak bk . 2.11 k1 We can write p A − n k1 p ak 1/p p B − n k1 p ak 1/p ⎛ ⎝M − p n kj1 ⎞1/p ⎛ p ak ⎠ ⎝N − p n kj1 ⎞1/p p bk ⎠ . 2.12 6 Journal of Inequalities and Applications By applying Theorem 1.2 for 0 < p ≤ 2 on right-hand side of the above equation, we get p A − n k1 p ak 1/p n p B − ak 1/p p ≤ MN − k1 n ak bk . 2.13 kj1 By using inequality 2.11 on right-hand side of the above expression follows the required result. ii Since Ap − n k1 p ak Ap − j n p p ak − ak k1 2.14 kj1 p 1/p p 1/p and denoting a nkj1 ak , b nkj1 bk , then p A − n k1 p ak 1/p p B − n k1 p ak 1/p Mp − ap 1/p N p − bp 1/p . 2.15 It is given that Mp − ap > 0 and N p − bp > 0, therefore by using Theorem 1.2, for n 1, on right-hand side of the above equation, we get n p A − ak 1/p p k1 n p B − ak 1/p p k1 ⎛ ≤ MN − ab MN − ⎝ ⎞1/p ⎛ n p ak ⎠ ⎝ kj1 n 2.16 ⎞1/p p bk ⎠ , kj1 since 1 < p ≤ 2, so by using 2.7 ≤ MN − n ak bk . 2.17 kj1 Theorem 2.4. Let j and n be fixed positive integers such that 1 ≤ j < n, and let p, q, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that 1.7 is satisfied. Also let 1/p1/q 1, M be defined in 2.8 and N j q B − bk q k1 1/q , 2.18 Journal of Inequalities and Applications 7 then, for p > 1, one has n p A − ak 1/p n q B − bk p k1 ⎛ 1/q n − ⎝ ≤ MN q k1 kj1 n − ≤ MN ⎞1/p ⎛ p ak ⎠ ⎞1/q n q ⎝ bk ⎠ kj1 2.19 ak bk kj1 ≤ AB − n ak bk . k1 > 0, therefore by generalized Aczél’s inequality, we have Proof. First of all, note that M, N j ≤ AB − MN ak bk . 2.20 k1 Now, Ap − j n n p p p ak Ap − ak − ak , k1 k1 2.21 kj1 p 1/p q 1/p and denote a nkj1 ak , b nkj1 bk . Then p A − n k1 p ak 1/p p B − n k1 q bk 1/q 1/q q − bq Mp − ap 1/p N . 2.22 q − bq > 0, therefore by using Theorem 1.4, for n 1, on It is given that Mp − ap > 0 and N right-hand side of the above equation, we get n p A − ak 1/p p k1 q B − n k1 q bk ⎛ − ab MN −⎝ ≤ MN 1/q n kj1 ⎞1/p ⎛ p ak ⎠ ⎝ n 2.23 ⎞1/q q bk ⎠ , kj1 by applying Hölder’s inequality − ≤ MN n ak bk , kj1 2.24 8 Journal of Inequalities and Applications by using inequality 2.20 ≤ AB − j k1 AB − n n ak bk − ak bk kj1 2.25 ak bk . k1 In 6, a refinement of Bellman’s inequality is given for positive integer p; here, we give further refinements of Bellman’s inequality for real p ≥ 1. We will use Minkowski’s inequality in the proof and recall that, for real p ≥ 1 and for positive reals ak , bk k 1, . . . , n, the Minkowski’s inequality states that n ak bk p 1/p ≤ n k1 k1 p ak 1/p n p bk 1/p . 2.26 k1 Theorem 2.5. Let j and n be fixed positive integers such that 1 ≤ j < n, and let p, A, B, ak , bk k 1, . . . , n be nonnegative real numbers such that 1.3 is satisfied. Also let M and N be defined in 2.8. If p ≥ 1, then n p A − ak 1/p p p B − k1 n k1 p bk 1/p ⎧⎛ ⎞1/p ⎛ ⎞1/p ⎫p ⎤1/p ⎪ ⎪ n n ⎬ ⎨ p⎠ p⎠ ⎢ ⎥ p ⎝ ⎝ ≤ ⎣M N − ak bk ⎦ ⎪ ⎪ ⎭ ⎩ kj1 kj1 ⎡ ⎛ n ≤ ⎝M Np − ⎞1/p 2.27 ak bk p ⎠ kj1 ≤ n A B − ak bk p 1/p p . k1 Proof. First of all, note that M, N > 0 and p ≥ 1, therefore by using Bellman’s inequality, we have MN ≤ p A B − j k1 1/p p ak bk . 2.28 Journal of Inequalities and Applications 9 Now, Ap − j n n p p p ak Ap − ak − ak , k1 k1 2.29 kj1 p 1/p p 1/p and denote a nkj1 ak , b nkj1 bk . Then n p A − ak 1/p p n p B − bk 1/p p k1 Mp − ap 1/p N p − bp 1/p . 2.30 k1 It is given that Mp − ap > 0 and N p − bp > 0, therefore by using Bellman’s inequality, for n 1, on right-hand side of the above equation, we get p A − n k1 p ak 1/p q B − n k1 q bk 1/q ⎧⎛ ⎞1/p ⎛ ⎞1/p ⎫p ⎤1/p ⎪ ⎪ n n ⎨ ⎬ p p ⎢ ⎥ ⎣M Np − ⎝ ak ⎠ ⎝ bk ⎠ ⎦ , ⎪ ⎪ ⎩ kj1 ⎭ kj1 ⎡ ≤ M Np − a bp !1/p 2.31 by applying Minkowski’s inequality ⎡ ≤ ⎣M N − p n ⎤1/p p⎦ ak bk , 2.32 kj1 and by using inequality 2.28 ⎤1/p j n p p ≤ ⎣A B − ak bk − ak bk ⎦ ⎡ p k1 " A Bp − n kj1 #1/p ak bk p 2.33 . k1 Remark 2.6. In 10, Hu and Xu gave the generalized results related to Theorems 2.4 and 2.5. 3. Some Further Remarks on Power Sums The following theorem 9, page 152 is very useful to give results related to power sums in connection with results given in 11, 12. 10 Journal of Inequalities and Applications Theorem 3.1. Let x1 , . . . , xn ∈ I n , where I 0, a is interval in Ê and x1 − x2 − · · · − xn ∈ I. Also let f : I → Ê be a function such that fx/x is increasing on I, then f n x1 − xi ≤ fx1 − n fxi . i2 3.1 i2 Remark 3.2. If fx/x is strictly increasing on I, then strict inequality holds in 3.1. Here, it is important to note that if we consider p, q ∈ Ê, p / 0, fx xq/p , 3.2 then fx/x is increasing on 0, ∞ for 0 < p ≤ q. By using it in Theorem 3.1, we get x1 − n q/p xi q/p ≤ x1 i2 − n i2 q/p xi . 3.3 p This implies Lemma 2.1 by substitution, xi → xi . In this section, we use Theorem 3.1 to give some results related to power sums as given in 11–13, but here we will discuss only the nonweighted case. In 11, we introduced Cauchy means related to power sums; here, we restate the means without weights. Let x x1 , . . . , xn be a positive n-tuple, then for r, s, t ∈ 0, ∞ we defined ⎫1/t−r ⎧ ⎨ r − s $n xs %t/s − n xt ⎬ i1 i1 i i Ast,r x , ⎩ t − s $n xs %r/s − n xr ⎭ i1 i i1 i t/ r, r / s, t / s, ⎫1/s−r ⎧ ⎨ r − s $n xs % log n xs − s n xs log xi ⎬ i1 i i1 i i1 i Ass,r x Asr,s x , $n s %r/s n r ⎭ ⎩ s x − x i1 i1 i s/ r, i 3.4 ⎛ ⎜ 1 Asr,r x exp⎝ s − r Ass,s x $n s i1 xi s $n exp ⎞ n r s x − s x log x i ⎟ i1 i i1 i ⎠, % n r ' s r/s − i1 xi i1 xi log &$ n n s/ r, %$ %2 $ %2 log ni1 xis − s2 ni1 xis log xi % $n s % ) . n s s i1 xi log i1 xi − s i1 xi log xi xs ($ni i1 2s %r/s We proved that Ast,r x is monotonically increasing with respect to t and r. In this section, we give exponential convexity of a positive difference of the inequality 3.1 by using parameterized class of functions. We define new means and discuss their relation to the means defined in 11. Also, we prove mean value theorem of Cauchy type. It is worthwhile to recall the following. Journal of Inequalities and Applications Definition 3.3. A function h : a, b → 11 Ê is exponentially convex if it is continuous and n % $ ui uj h xi xj ≥ 0, 3.5 i,j1 for all n ∈ Æ and all choices ui ∈ Ê, i 1, 2, . . . , n, and xi ∈ a, b, such that xi xj ∈ a, b, 1 ≤ i, j ≤ n. Proposition 3.4. Let f : a, b → Ê. The following propositions are equivalent: i f is exponentially convex, ii f is continuous and n * xi xj vi vj f 2 i,j1 + ≥ 0, 3.6 for every vi ∈ Ê and for every xi ∈ a, b, 1 ≤ i ≤ n. Corollary 3.5. If h : a, b → 0, ∞ is exponentially convex function, then h is a log-convex function. 3.1. Exponential Convexity Lemma 3.6. Let t ∈ Ê and ϕt : 0, ∞ → Ê be the function defined as ⎧ ⎪ ⎨ xt , ϕt x t − 1 ⎪ ⎩x log x, t/ 1, 3.7 t 1, then ϕt x/x is strictly increasing function on 0, ∞ for each t ∈ Ê. Proof. Since * ϕt x x + xt−2 > 0, ∀x ∈ 0, ∞, 3.8 therefore ϕt x/x is strictly increasing function on 0, ∞ for each t ∈ Ê. Theorem 3.7. Let x x1 , . . . , xn be a positive n-tuple n ≥ 2 such that x1 − x2 − · · · − xn > 0, and let n n Λt x ϕt x1 − ϕt xi − ϕt x1 − xi . i2 i2 3.9 12 Journal of Inequalities and Applications a For m ∈ Æ , let p1 , . . . , pm be arbitrary real numbers, then the matrix , Λpi pj /2 where 1 ≤ i, j ≤ m 3.10 is a positive semidefinite matrix. b The function t → Λt , t ∈ Ê is exponentially convex. c The function t → Λt , t ∈ Ê is log convex. Proof. a Define a function Fx n % pi pj , where pij 2 $ ui uj ϕpij x, i,j1 3.11 then * Fx x + n ui xpi −2/2 2 ≥0 ∀x ∈ 0, ∞. 3.12 i1 This implies that Fx/x is increasing function on 0, ∞. So using F in the place of f in 3.1, we have n ui uj Λϕpij ≥ 0. 3.13 i,j1 Hence, the given matrix is positive semidefinite. b Since after some computation we have that limt → 1 Λt Λ1 so t → Λt is continuous on Ê, then by Proposition 3.4, we have that t → Λt is exponentially convex. c Since ϕt x/x is strictly increasing function on 0, ∞, so by Remark 3.2, we have ϕt x1 − n i2 xi < ϕt x1 − n ϕt xi , i2 it follows that Λt x > 0. Now, by Corollary 3.5, we have that t → Λt is log convex. Let us introduce the following. 3.14 Journal of Inequalities and Applications 13 Definition 3.8. Let x x1 , . . . , xn be a positive n-tuple n ≥ 2 such that x1s − x2s − · · · − xns > 0 for s ∈ 0, ∞, then for t, r, s ∈ 0, ∞, we define ⎫1/t−r ⎧ ⎨ r − s xt − n xt − $xs − n xs %t/s ⎬ i2 i2 i i 1 1 s , Ct,r x ⎩ t − s xr − n xr − $xs − n xs %r/s ⎭ i2 1 i 1 i2 t/ r, r / s, t / s, i s Cs,r x s Cr,s x % $ % ⎞1/s−r $ sx1s log x1 − s ni2 xis log xi − x1s − ni2 xis log x1s − ni2 xis − s r ⎠ ⎝ , %r/s $ s x1r − ni2 xir − x1s − ni2 xis ⎛ s/ r, s Cr,r x ⎛ ⎜ exp⎝ 1 s − r sx1r ⎞ $ s n s %r/s $ s n s % log x1 − s log xi − x1 − i2 xi log x1 − i1 xi ⎟ & ⎠, n r $ s n s %r/s ' r s x1 − i2 xi − x1 − i2 xi n r i2 xi s / r, s Cs,s x exp $ %2 $ %$ %2 s2 x1s log x1 2 − s2 ni2 xis log xi − x1s − ni2 xis logx1s − ni2 xis % $ %) . ( $ 2s sx1s log x1 − s ni2 xis log xi − x1s − ni2 xis log x1s − ni2 xis 3.15 s s s s s x Cr,s x limt → s Ct,r x limt → s Cr,t x, Cr,r x Remark 3.9. Let us note that Cs,r s s s limt → r Ct,r x, and Cs,s x limr → s Cr,r x. 1/s s Remark 3.10. If in Ct,r x we substitute x1 by ni1 xis , then we get Ast,r x, and if we 1/s s substitute x1 by xis − ni2 xis in Ast,r x, we get Ct,r x. In 11, we have the following lemma. Lemma 3.11. Let f be a log-convex function and assume that if x1 ≤ y1 , x2 ≤ y2 , x1 / x2 , y1 / y2 , then the following inequality is valid: * fx2 fx1 +1/x2 −x1 $ % 1/y2 −y1 f y2 ≤ . $ % f y1 3.16 Theorem 3.12. Let x x1 , . . . , xn be positive n-tuple n ≥ 2 such that x1s − x2s − · · · − xns > 0 for s ∈ 0, ∞, then for r, t, u, v ∈ 0, ∞ such that r ≤ u, t ≤ v, one has s s Ct,r x ≤ Cv,u x. 3.17 14 Journal of Inequalities and Applications Proof. Let Λs be defined by 3.9. Now taking x1 r, x2 t, y1 u, y2 v, where r/ t, u / v, r, t, u, v / 1, and fs Λs in Lemma 3.11, we have %t 1/t−r n t $ n t r − 1 x1 − i2 xi − x1 − i2 xi $ % t − 1 x1r − ni2 xir − x1 − ni2 xi r ≤ %v 1/v−u n v $ n v u − 1 x1 − i2 xi − x1 − i2 xi . % $ v − 1 x1u − ni2 xiu − x1 − ni2 xi u 3.18 Since s > 0, by substituting xi xis , t t/s, r r/s, u u/s, and v v/s, where r, t, u, v / s, in above inequality, we get ⎛ $ %t/s ⎞s/t−r x1t − ni2 xit − x1s − ni2 xis − s r ⎝ ⎠ t − s xr − n xr − $xs − n xs %r/s 1 i2 i ⎛ 1 i2 i %v/s ⎞s/v−u x1v − i2 xiv − x1s − i2 xis − s u ⎠ ≤⎝ . v − s xu − n xu − $xs − n xs %u/s i2 i i2 i 1 1 n $ n 3.19 By raising power 1/s, we get 3.17 for r, t, u, v / s, r / t and u / v. From Remark 3.9, we get that 3.17 is also valid for r t or u v or r, t, u, v s. 1/s s Remark 3.13. If we substitute x1 by ni1 xis , then monotonicity of Ct,r x implies the n s s s 1/s monotonicity of At,r x, and if we substitute x1 by xi − i2 xi , then monotonicity of s x. Ast,r x implies monotonicity of Ct,r 3.2. Mean Value Theorems We will use the following lemma 11 to prove the related mean value theorems of Cauchy type. Lemma 3.14. Let f ∈ C1 I, where I 0, a such that m≤ xf x − fx ≤ M. x2 3.20 Consider the functions φ1 , φ2 defined as φ1 x Mx2 − fx, φ2 x fx − mx2 , then φi x/x for i 1, 2 are monotonically increasing functions. 3.21 Journal of Inequalities and Applications 15 Theorem 3.15. Let x1 , . . . , xn ∈ I n , where I is a compact interval such that I ⊆ 0, ∞ and x1 − x2 − · · · − xn ∈ I. If f ∈ C1 I, then there exists ξ ∈ I such that fx1 − n fxi − f x1 − n i2 fxi i2 ⎧ 2 ⎫ n n ⎬ ξf ξ − fξ ⎨ 2 2 x . − x − x − x 1 i 1 i ⎩ ⎭ ξ2 i2 i2 3.22 Proof. Since I is compact and f ∈ CI, therefore let . M max / xf x − fx : x ∈ I , x2 . m min / xf x − fx : x ∈ I . x2 3.23 In Theorem 3.1, setting f φ1 and f φ2 , respectively, as defined in Lemma 3.14, we get the following inequalities: fx1 − n fxi − f x1 − n i2 xi i2 ⎧ ⎨ n ≤ M x12 − xi2 − ⎩ i2 n x1 − xi i2 2 ⎫ ⎬ ⎭ ⎧ 2 ⎫ n n n n ⎨ ⎬ . fx1 − fxi − f x1 − xi ≥ m x12 − xi2 − x1 − xi ⎩ ⎭ i2 i2 i2 i2 , 3.24 If fx x2 , then fx/x is strictly increasing function on I, therefore by Theorem 3.1, we have x12 n − xi2 − n x1 − xi i2 2 3.25 > 0. i2 Now, by combining inequalities 3.24, we get m≤ fx1 − x12 − n $ n i2 fxi − f x1 − n 2 $ n i2 xi − x1 − i2 i2 xi %2 xi % ≤ M. 3.26 Finally, by condition 3.20, there exists ξ ∈ I, such that fx1 − x12 − as required. % $ n i2 fxi − f x1 − i2 xi %2 n 2 $ n i2 xi − x1 − i2 xi n ξf ξ − fξ ξ2 3.27 16 Journal of Inequalities and Applications Theorem 3.16. Let x1 , . . . , xn ∈ I n , where I is a compact interval such that I ⊆ 0, ∞ and x1 − x2 − · · · − xn ∈ I. If f, g ∈ C1 I, then there exists ξ ∈ I such that the following equality is true: fx1 − gx1 − % $ n i2 fxi − f x1 − i2 xi % $ n n i2 gxi − g x1 − i2 xi n ξf ξ − fξ ξg ξ − gξ 3.28 provided that the denominators are nonzero. Proof. Let a function k ∈ C1 I be defined as k c1 f − c2 g, 3.29 where c1 and c2 are defined as n n c1 gx1 − gxi − g x1 − xi , i2 c2 fx1 − n i2 fxi − f x1 − i2 n 3.30 xi . i2 Then, using Theorem 3.15, with f k, we have 2 ⎫ $ % % ⎧ $ n n ⎬ ⎨ c2 ξg ξ − gξ c1 ξf ξ − fξ 2 2 . x − − x − x − x 0 1 i ⎭ ⎩ 1 i2 i ξ2 ξ2 i2 Since x12 − n i2 xi2 − x1 − n i2 3.31 2 xi > 0, therefore 3.31 gives c2 ξf ξ − fξ . c1 ξg ξ − gξ 3.32 Putting in 3.30, we get 3.28. Acknowledgments This research was partially funded by Higher Education Commission, Pakistan. The research of the second author was supported by the Croatian Ministry of Science, Education and Sports under the Research Grant no. 117-1170889-0888. References 1 J.. 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