International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 809689 | https://doi.org/10.1155/2012/809689

Jaekeun Park, "Hermite-Hadamard and Simpson-Like Type Inequalities for Differentiable (𝛼,𝑚)-Convex Mappings", International Journal of Mathematics and Mathematical Sciences, vol. 2012, Article ID 809689, 12 pages, 2012. https://doi.org/10.1155/2012/809689

Hermite-Hadamard and Simpson-Like Type Inequalities for Differentiable (𝛼,𝑚)-Convex Mappings

Academic Editor: Jewgeni Dshalalow
Received13 Jul 2011
Revised13 Nov 2011
Accepted28 Nov 2011
Published18 Jan 2012

Abstract

The author establish several Hermite-Hadamard and Simpson-like type inequalities for mappings whose first derivative in absolute value aroused to the 𝑞th (𝑞1) power are (𝛼,𝑚)-convex. Some applications to special means of positive real numbers are also given.

1. Introduction

Recall that, for some fixed 𝛼(0,1] and 𝑚[0,1], a mapping 𝑓𝕀[0,) is said to be (𝛼,𝑚)-convex on an interval 𝕀 if the inequality 𝑓(𝑡𝑥+𝑚(1𝑡)𝑦)𝑡𝛼𝑓(𝑥)+𝑚(1𝑡𝛼)𝑓(𝑦)(1.1) holds for 𝑥,𝑦𝕀, and 𝑡[0,1].

Denote by 𝐾𝛼𝑚(𝕀) the set of all (𝛼,𝑚)-convex mappings on 𝕀. For recent results and generalizations concerning 𝑚-convex and (𝛼,𝑚)-convex mappings, see [14].

For the simplicities of notations, for 𝑓𝐾𝛼𝑚(𝕀), let us denote (𝑎𝑚𝑏)𝑆𝑏𝑎1(𝑓)(𝛼,𝑚,𝑟)=𝑟𝑓(𝑎)+(𝑟2)𝑓𝑎+𝑚𝑏21+𝑓(𝑚𝑏)𝑚𝑏𝑎𝑎𝑚𝑏𝑓(𝑥)𝑑𝑥.(1.2)

In [1, 3], Klaričić Bakula and Özdemir et al., proved the following Hadamard’s inequalities for mappings whose second derivative in absolute value aroused to the 𝑞-th (𝑞1) power are (𝛼,𝑚)-convex.

Theorem 1.1. Let 𝑓𝕀[0,𝑏] be a twice differentiable mapping on the interior 𝕀0 of an interval 𝕀 such that 𝑓𝐿1([𝑎,𝑏]), where 𝑎,𝑏𝕀 with 𝑎<𝑏 and 𝑏>0. If |𝑓|𝑞 is (𝛼,𝑚)-convex on [𝑎,𝑏] for (𝛼,𝑚)[0,1]2 and 𝑞1 with 1/𝑝+1/𝑞=1, then the following inequality holds: ||𝑆(𝑎)𝑏𝑎||(𝑓)(𝛼,𝑚,2)𝑚𝑏𝑎2161/𝑝𝜇||𝑓||(𝑎)𝑞||𝑓+𝑚𝜈||(𝑏)𝑞1/𝑞,(1.3) where 1𝜇=𝛼(𝛼+2)(𝛼+3),𝜈=2+5𝛼,6(𝛼+2)(𝛼+3)(1.4)||𝑆(𝑏)𝑏𝑎||(𝑓)(𝛼,𝑚,6)𝑚𝑏𝑎2121/𝑝𝜇0||𝑓||(𝑎)𝑞+𝑚𝜈0||𝑓||(𝑏)𝑞1/𝑞,(1.5) where 𝜇0=𝑞𝑞+𝛼+2Γ(𝛼+2)Γ(𝑞),𝜈Γ(𝑞+𝛼+1)0=1(𝑞𝑞+1)(𝑞+2)𝑞+𝛼+2Γ(𝛼+2)Γ(𝑞),Γ(𝑞+𝛼+1)(1.6) where Γ(𝑥)=0𝑒𝑡𝑡𝑥1𝑑𝑡,𝑥>0.(1.7)

Theorem 1.2. Under the same notations in Theorem 2.2, if |𝑓|𝑞 is (𝛼,𝑚)-convex on [𝑎,𝑏] for (𝛼,𝑚)[0,1]2 and 𝑞>1 with 1/𝑝+1/𝑞=1, then the following inequality holds: ||𝑆𝑏𝑎||(𝑓)(𝛼,𝑚,2)𝑚𝑏𝑎8Γ(𝑝+1)Γ(𝑝+3/2)1/𝑝||𝑓||(𝑎)𝑞1||𝑓𝛼+1+𝑚||(𝑏)𝑞𝛼𝛼+11/𝑞.(1.8)

Note that for (𝛼,𝑚){(0,0),(𝛼,0),(1,0),(1,𝑚),(1,1),(𝛼,1)} one obtains the following classes of functions: increasing, 𝛼-starshaped, starshaped, 𝑚-convex, convex, and 𝛼-convex. For the definitions and elementary properties of these classes, see [48].

For recent years, many authors present some new results about Simpson’s inequality for (𝛼,𝑚)-convex mappings and have established error estimations for the Simpson’s inequality: for refinements, counterparts, generalizations, and new Simpson’s type inequalities, see [13, 6].

In [9], Dragomir et al. proved the following theorem.

Theorem 1.3. Let 𝑓𝕀[0,) be an absolutely continuous mapping on [𝑎,𝑏] such that 𝑓𝐿𝑝([𝑎,𝑏]), where 𝑎,𝑏𝕀 with 𝑎<𝑏. Then the following inequality holds: ||𝑆𝑏𝑎||(𝑓)(1,1,6)(𝑏𝑎)1/𝑝62𝑞+1+13(𝑞+1)1/𝑞𝑓𝑝.(1.9)

The readers can estimate the 𝑒𝑟𝑟𝑜𝑟(𝑓) in the generalized Simpson’s formula without going through its higher derivatives which may not exist, not be bounded, or may be hard to find.

In this paper, the author establishes some generalizations of Hermite-Hadamard and Simpson-like type inequalities based on differentiable (𝛼,𝑚)-convex mappings by using the following new identity in Lemma 2.1 and by using these results, obtain some applications to special means of positive real numbers.

2. Generalizations of Simpson-Like Type Inequalities on 𝐾𝛼𝑚(𝕀)

In order to generalize the classical Simpson-like type inequalities and prove them, we need the following lemma [6].

Lemma 2.1. Let 𝑓𝕀[0,𝑏] be a differentiable mapping on the interior 𝕀0 of an interval 𝕀, where 𝑎,𝑏𝕀 with 0𝑎<𝑏 and 𝑏>0. If 𝑓𝐿1([𝑎,𝑏]), then, for 𝑟2 and (0,1) with 1/𝑟(𝑟1)/𝑟, the following equality holds: 𝑆𝑏𝑎(𝑓)(𝛼,𝑚,𝑟)=10𝑝(𝑟,𝑡)𝑓(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡(2.1) for 𝑓𝐾𝛼𝑚([𝑎,𝑏]) and each 𝑡[0,1], where 1𝑝(𝑟,𝑡)=𝑡𝑟1𝑡0,2,𝑡𝑟1𝑟1𝑡2.,1(2.2)

By the similar way as Theorems 1.11.3, we obtain the following theorems.

Theorem 2.2. Let 𝑓𝕀[0,𝑏] be a differentiable mapping on 𝕀0 such that 𝑓𝐿([𝑎,𝑏]), where 𝑎,𝑏𝕀 with 0𝑎<𝑏< and 𝑏>0. If |𝑓|𝐾𝛼𝑚([𝑎,𝑏]), for some (𝛼,𝑚)(0,1]2 and 𝑚𝑏>𝑎, then, for any 𝑟2, the following inequality holds: ||𝑆𝑏𝑎||𝜇(𝑓)(𝛼,𝑚,𝑟)11+𝜇21+𝜇31+𝜇41||𝑓||+𝜈(𝑎)11+𝜈21+𝜈31+𝜈41𝑚||𝑓||,(𝑏)(2.3) where 𝜇11=1(𝛼+1)(𝛼+2)𝑟𝛼+2,𝜇21=1(𝛼+1)(𝛼+2)𝑟𝛼+2+(𝛼+1)𝑟2(𝛼+2)2𝛼+2,𝜇(𝛼+1)(𝛼+2)𝑟31=(𝑟1)𝛼+2(𝛼+1)(𝛼+2)𝑟𝛼+2+2(𝛼+2)3𝑟2𝛼+2,𝜇(𝛼+1)(𝛼+2)𝑟41=(𝑟1)𝛼+2(𝛼+1)(𝛼+2)𝑟𝛼+2+(𝛼+2)𝑟,𝜈(𝛼+1)(𝛼+2)𝑟11=12𝑟2𝜇11,𝜈21=(𝑟2)28𝑟2𝜇21,𝜈31=(𝑟2)28𝑟2𝜇31,𝜈41=12𝑟2𝜇41.(2.4)

Proof. From Lemma 2.1 and using the properties of the modulus, we have the following: ||𝑆𝑏𝑎||(𝑓)(𝛼,𝑚,𝑟)01/𝑟1𝑟||𝑓𝑡||+(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡1/21/𝑟1𝑡𝑟||𝑓||+(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡(𝑟1)/𝑟1/2𝑟1𝑟||𝑓𝑡||+(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡1(𝑟1)/𝑟𝑡𝑟1𝑟||𝑓||(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡.(2.5)
Since |𝑓| is (𝛼,𝑚)-convex on [𝑎,𝑏], we know that for any 𝑡[0,1]||𝑓||(𝑡𝑎+𝑚(1𝑡)𝑏)𝑡𝛼||𝑓||(𝑎)+𝑚(1𝑡𝛼)||𝑓||.(𝑏)(2.6)
By (2.5) and (2.6), we get the following: ||𝑆𝑏𝑎||(𝑓)(𝛼,𝑚,𝑟)01/𝑟1𝑟𝑡𝑡𝛼||𝑓||(𝑎)+𝑚(1𝑡𝛼)||𝑓||+(𝑏)𝑑𝑡1/21/𝑟1𝑡𝑟𝑡𝛼||𝑓||(𝑎)+𝑚(1𝑡𝛼)||𝑓||+(𝑏)𝑑𝑡(𝑟1)/𝑟1/2𝑟1𝑟𝑡𝑡𝛼||𝑓||(𝑎)+𝑚(1𝑡𝛼)||𝑓||+(𝑏)𝑑𝑡1(𝑟1)/𝑟𝑡𝑟1𝑟𝑡𝛼||𝑓||(𝑎)+𝑚(1𝑡𝛼)||𝑓||(𝑏)𝑑𝑡01/2|||1𝑟|||𝑡𝑡𝛼𝑑𝑡+11/2|||𝑟1𝑟|||𝑡𝑡𝛼||𝑓𝑑𝑡||+(𝑎)01/2|||1𝑟|||𝑡(1𝑡𝛼)𝑑𝑡+11/2|||𝑟1𝑟|||𝑡(1𝑡𝛼𝑚||𝑓)𝑑𝑡||=𝜇(𝑏)11+𝜇21+𝜇31+𝜇41||𝑓||+𝜈(𝑎)11+𝜈21+𝜈31+𝜈41𝑚||𝑓||,(𝑏)(2.7) which completes the proof.

Corollary 2.3. In Theorem 2.2, (i) if we choose 𝛼=1 and 𝑟=2, then we have the following: ||(𝑚𝑏𝑎)𝑆𝑏𝑎||=||||(𝑓)(1,𝑚,2)𝑓(𝑎)+𝑓(𝑚𝑏)21𝑚𝑏𝑎𝑎𝑚𝑏𝑓||||(𝑥)𝑑𝑥𝑚𝑏𝑎8||𝑓||||𝑓(𝑎)+𝑚||,(𝑏)(2.8) and (ii) if we choose 𝛼=1 and 𝑟=6, then we have the following ||(𝑚𝑏𝑎)𝑆𝑏𝑎||=||||1(𝑓)(1,𝑚,6)6𝑓(𝑎)+4𝑓𝑎+𝑚𝑏21+𝑓(𝑚𝑏)𝑚𝑏𝑎𝑎𝑚𝑏𝑓||||5(𝑥)𝑑𝑥||𝑓72(𝑚𝑏𝑎)||||𝑓(𝑎)+𝑚||.(𝑏)(2.9)

Theorem 2.4. Under the same notations in Theorem 2.2, if |𝑓|𝑞𝐾𝛼𝑚([𝑎,𝑏]), for some (𝛼,𝑚)(0,1]2, 𝑚𝑏>𝑎 and 𝑞>1 with 1/𝑝+1/𝑞=1, then, for any 𝑟2, the following inequality holds: ||𝑆𝑏𝑎||1(𝑓)(𝛼,𝑚,𝑟)2𝑟21/𝑝𝜇11||𝑓||(𝑎)𝑞+𝜈11𝑚||𝑓||(𝑏)𝑞1/𝑞+𝜇41||𝑓||(𝑎)𝑞+𝜈41𝑚||𝑓||(𝑏)𝑞1/𝑞+18𝑟2𝑟21/𝑝𝜇21||𝑓(||𝑎)𝑞+𝜈21𝑚||𝑓(||𝑏)𝑞1/𝑞+𝜇31||𝑓(||𝑎)𝑞+𝜈31𝑚||𝑓(||𝑏)𝑞1/𝑞.(2.10)

Proof. From Lemma 2.1 and using the properties of modulus, we have the following: ||𝑆𝑏𝑎||(𝑓)(𝛼,𝑚,𝑟)01/𝑟1𝑟||𝑓𝑡||+(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡1/21/𝑟1𝑡𝑟||𝑓||+(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡(𝑟1)/𝑟1/2𝑟1𝑟||𝑓𝑡||+(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡1(𝑟1)/𝑟𝑡𝑟1𝑟||𝑓||(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡.(2.11)
Using the power-mean integral inequality and (𝛼,𝑚)-convexity of |𝑓|𝑞 for any 𝑡[0,1], we have the following(a)01/𝑟1𝑟||𝑓𝑡||(𝑡𝑎+𝑚(1𝑡)𝑏)𝑞𝑑𝑡𝜇11||𝑓||(𝑎)𝑞+𝜈11𝑚||𝑓||(𝑏)𝑞,(2.12)(b)1/21/𝑟1𝑡𝑟||||𝑓(𝑡𝑎+𝑚(1𝑡)𝑏)𝑞𝑑𝑡𝜇21||𝑓||(𝑎)𝑞+𝜈21𝑚||𝑓||(𝑏)𝑞,(2.13)(c)(𝑟1)/𝑟1/2𝑟1𝑟||||𝑡𝑓(𝑡𝑎+𝑚(1𝑡)𝑏)𝑞𝑑𝑡𝜇31||𝑓||(𝑎)𝑞+𝜈31𝑚||𝑓||(𝑏)𝑞,(2.14)(d)1(𝑟1)/𝑟𝑡𝑟1𝑟||||𝑓(𝑡𝑎+𝑚(1𝑡)𝑏)𝑞𝑑𝑡𝜇41||𝑓||(𝑎)𝑞+𝜈41𝑚||𝑓||(𝑏)𝑞.(2.15)
By the similar way as the above inequalities (a)–(d), we have the following:(a′)01/𝑟1𝑟||𝑓𝑡||1(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡2𝑟21/𝑝𝜇11||𝑓||(𝑎)𝑞+𝜈11𝑚||𝑓||(𝑏)𝑞1/𝑞,(2.16)(b′)1/21/𝑟1𝑡𝑟||𝑓||1(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡8𝑟2𝑟21/𝑝𝜇21||𝑓||(𝑎)𝑞+𝜈21𝑚||𝑓||(𝑏)1/𝑞,(2.17)(c′)(𝑟1)/𝑟1/2𝑟1𝑟||𝑓𝑡||1(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡8𝑟2𝑟21/𝑝𝜇31||𝑓||(𝑎)𝑞+𝜈31𝑚||𝑓||(𝑏)𝑞1/𝑞,(2.18)(d′)1(𝑟1)/𝑟𝑡𝑟1𝑟||𝑓||1(𝑡𝑎+𝑚(1𝑡)𝑏)𝑑𝑡2𝑟21/𝑝𝜇41||𝑓||(𝑎)𝑞+𝜈41𝑚||𝑓||(𝑏)𝑞1/𝑞.(2.19)
By (2.11) and (2.16)–(2.19) the assertion (2.10) holds.

Corollary 2.5. In Theorem 2.4, (i) if we choose 𝛼=1 and 𝑟=2, then we have that ||||𝑓(𝑎)+𝑓(𝑚𝑏)21𝑚𝑏𝑎𝑎𝑚𝑏𝑓||||6(𝑥)𝑑𝑥1/𝑞8||𝑓(𝑚𝑏𝑎)||(𝑎)𝑞||𝑓+5𝑚||(𝑏)𝑞1/𝑞+5||𝑓||(𝑎)𝑞||𝑓+𝑚||(𝑏)𝑞1/𝑞,(2.20) and (ii) if we choose 𝛼=1 and 𝑟=6, then we have that ||||16𝑓(𝑎)+4𝑓𝑎+𝑚𝑏21+𝑓(𝑚𝑏)𝑚𝑏𝑎𝑎𝑚𝑏𝑓||||1(𝑥)𝑑𝑥(𝑚𝑏𝑎)172181/𝑞||𝑓||(𝑎)𝑞||𝑓+17𝑚||(𝑏)𝑞1/𝑞+||𝑓17||(𝑎)𝑞||𝑓+𝑚||(𝑏)𝑞1/𝑞+25228852/𝑞×7||𝑓(||𝑎)𝑞||𝑓+11𝑚(||𝑏)𝑞1/𝑞+||𝑓11(||𝑎)𝑞||𝑓+7𝑚(||𝑏)𝑞1/𝑞.(2.21)

Theorem 2.6. Under the same notations in Theorem 2.2, if |𝑓|𝑞𝐾𝛼𝑚([𝑎,𝑏]), for some (𝛼,𝑚)(0,1]2, 𝑚𝑏>𝑎 and 𝑞>1 with 1/𝑝+1/𝑞=1, then, for any 𝑟2, the following inequality holds: ||𝑆𝑏𝑎||1(𝑓)(𝛼,𝑚,𝑟)𝑟𝑝+11/𝑝𝜇12||𝑓||(𝑎)𝑞+𝜈12𝑚||𝑓||(𝑏)𝑞1/𝑞+𝜇42||𝑓||(𝑎)𝑞+𝜈42𝑚||𝑓||(𝑏)𝑞1/𝑞+(𝑟2)𝑝+12𝑝+1𝑟𝑝+11/𝑝𝜇22||𝑓||(𝑎)𝑞+𝜈22𝑚||𝑓||(𝑏)𝑞1/𝑞+𝜇32||𝑓||(𝑎)𝑞+𝜈32𝑚||𝑓||(𝑏)𝑞1/𝑞,(2.22) where 𝜇12=1𝑟𝛼+1(,𝜇𝛼+1)22=𝑟𝛼+12𝛼+12𝛼+1𝑟𝛼+1,𝜇(𝛼+1)32=2𝛼+1(𝑟1)𝛼+1𝑟𝛼+12𝛼+1𝑟𝛼+1,𝜇(𝛼+1)42=𝑟𝛼+1(𝑟1)𝛼+1𝑟𝛼+1(,𝜈𝛼+1)12=1𝑟𝜇12,𝜈22=𝑟22𝑟+𝜇22,𝜈32=𝑟22𝑟+𝜇32,𝜈42=1𝑟𝜇42.(2.23)

Proof. Suppose that 𝑞>1. From Lemma 2.1, using the Hölder integral inequality, we get the following: ||𝑆𝑏𝑎||(𝑓)(𝛼,𝑚,𝑟)01/𝑟1𝑟𝑡𝑝𝑑𝑡1/𝑝01/𝑟||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞+1/21/𝑟1𝑡𝑟𝑝𝑑𝑡1/𝑝1/21/𝑟||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞+(𝑟1)/𝑟1/2𝑟1𝑟𝑡𝑝𝑑𝑡1/𝑝(𝑟1)/𝑟1/2||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞+1(𝑟1)/𝑟𝑡𝑟1𝑟𝑝𝑑𝑡1/𝑝1(𝑟1)/𝑟||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞1𝑟𝑝+11/𝑝01/𝑟||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞+12𝑝+1𝑟2𝑟𝑝+11/𝑝1/21/𝑟||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞+12𝑝+1𝑟2𝑟𝑝+11/𝑝(𝑟1)/𝑟1/2||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞+1𝑟𝑝+11/𝑝1(𝑟1)/𝑟||𝑓||(𝑡𝑏+𝑚(1𝑡)𝑎)𝑞𝑑𝑡1/𝑞,(2.24) where we have used the fact that 1/2<(1/(𝑝+1))1/𝑝<1.
Since |𝑓|𝑞𝐾𝑚𝛼([𝑎,𝑏]) for some fixed 𝛼(0,1] and 𝑚[0,1], we have the followings: 01/𝑟||𝑓||(𝑡𝑏+(1𝑡)𝑎)𝑞𝑑𝑡𝜇12||𝑓||(𝑎)𝑞+𝜈12𝑚||𝑓||(𝑏)𝑞,1/21/𝑟||𝑓||(𝑡𝑏+(1𝑡)𝑎)𝑞𝑑𝑡𝜇22||𝑓||(𝑎)𝑞+𝜈22𝑚||𝑓||(𝑏)𝑞,(𝑟1)/𝑟1/2||𝑓||(𝑡𝑏+(1𝑡)𝑎)𝑞𝑑𝑡𝜇32||𝑓||(𝑎)𝑞+𝜈32𝑚||𝑓||(𝑏)𝑞,1(𝑟1)/𝑟||𝑓||(𝑡𝑏+(1𝑡)𝑎)𝑞𝑑𝑡𝜇42||𝑓||(𝑎)𝑞+𝜈42𝑚||𝑓||(𝑏)𝑞.(2.25)
Hence, if we combine the inequalities in (2.24)-(2.25), we get the desired result.

Corollary 2.7. In Theorem 2.6, (i) if we choose 𝛼=1 and 𝑟=2, then we have that ||||𝑓(𝑎)+𝑓(𝑚𝑏)21𝑚𝑏𝑎𝑎𝑚𝑏𝑓||||1(𝑥)𝑑𝑥41+1/𝑞||𝑓(𝑚𝑏𝑎)||(𝑎)𝑞||𝑓+3𝑚||(𝑏)𝑞1/𝑞+3||𝑓||(𝑎)𝑞||𝑓+𝑚||(𝑏)𝑞1/𝑞,(2.26) and (ii) if we choose 𝛼=1 and 𝑟=6, then we have ||||16𝑓(𝑎)+4𝑓𝑎+𝑚𝑏21+𝑓(𝑚𝑏)𝑚𝑏𝑎𝑎𝑚𝑏||||1𝑓(𝑥)𝑑𝑥(𝑚𝑏𝑎)61+2/𝑞||𝑓||(𝑎)𝑞||𝑓+11𝑚||(𝑏)𝑞1/𝑞+||𝑓11||(𝑎)𝑞||𝑓+𝑚||(𝑏)𝑞1/𝑞+132+1/𝑞||𝑓||(𝑎)𝑞||𝑓+2𝑚||(𝑏)𝑞1/𝑞+2||𝑓||(𝑎)𝑞||𝑓+𝑚||(𝑏)𝑞1/𝑞,(2.27) where we have used the fact that (1/2)1/𝑞<1.

3. Applications to Special Means

Now using the results of Section 2, we give some applications to the following special means of positive real numbers 𝑎,𝑏+ with 𝑏𝑎.(1) The arithmetic mean: 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2.(2)The geometric mean: 𝐺(𝑎,𝑏)=𝑎𝑏.(3)The logarithmic mean: 𝐿(𝑎,𝑏)=(𝑏𝑎)/(ln𝑏ln𝑎) for 𝑎𝑏.(4)The harmonic mean: 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏).(5)The power mean: 𝑀𝑟(𝑎,𝑏)=((𝑎𝑟+𝑏𝑟)/2)1/𝑟, 𝑟1,𝑎,𝑏.(6)The generalized logarithmic mean: 𝐿𝑛𝑏(𝑎,𝑏)=𝑛+1𝑎𝑛+1(𝑏𝑎)(𝑛+1)1/𝑛,𝑎𝑏.(3.1)(7)The identric mean:1𝐼(𝑎,𝑏)=𝑎𝑎=𝑏𝑒𝑏𝑏𝑎𝑎1/(𝑏𝑎)𝑎𝑏.(3.2)

Proposition 3.1. For 𝑛(,0)[1,){1} and [𝑎,𝑏][0,𝑏] with 𝑏>0, we have the following inequalities: ||(𝑎)𝐴(𝑎𝑛,𝑚𝑛𝑏𝑛)𝐿𝑛𝑛||(𝑎,𝑚𝑏)𝑚𝑏𝑎8|𝑛|𝑀𝑛1𝑛1𝑎,𝑚1/(𝑛1)𝑏.|||1(𝑏)3𝐴(𝑎𝑛,𝑚𝑛𝑏𝑛2)+3𝐴𝑛(𝑎,𝑚𝑏)𝐿𝑛𝑛|||5(𝑎,𝑚𝑏)72(𝑚𝑏𝑎)|𝑛|𝑀𝑛1𝑛1𝑎,𝑚1/(𝑛1)𝑏.(3.3)

Proof. The assertions follow from Corollary 2.3 for 𝑓(𝑥)=𝑥𝑛.

Proposition 3.2. For [𝑎,𝑏][0,𝑏], we have the following inequalities: ||𝐻(𝑎)1(𝑎,𝑚𝑏)𝐿1||(𝑎,𝑚𝑏)𝑚𝑏𝑎8𝑎2𝑏2𝑀22𝑚1/2,|||1𝑎,𝑏(𝑏)3𝐻12(𝑎,𝑚𝑏)+3𝐴1(𝑎,𝑚𝑏)𝐿1|||(𝑎,𝑚𝑏)5(𝑚𝑏𝑎)72𝑎2𝑏2𝑀22𝑚1/2.𝑎,𝑏(3.4)

Proof. The assertions follow from Corollary 2.3 for 𝑓(𝑥)=1/𝑥.

Proposition 3.3. For 𝑛(,0)[1,){1} and [𝑎,𝑏][0,𝑏], we have the following inequalities: ||(𝑎)𝐴(𝑎𝑛,𝑚𝑛𝑏𝑛)𝐿𝑛𝑛||3(𝑎,𝑚𝑏)1/𝑞8𝐴(𝑚𝑏𝑎)|𝑛|×1/𝑞𝑎(𝑛1)𝑞,5𝑚𝑏(𝑛1)𝑞1/𝑞+𝐴1/𝑞5𝑎(𝑛1)𝑞,𝑚𝑏(𝑛1)𝑞1/𝑞,|||1(𝑏)3𝐴(𝑎𝑛,𝑚𝑛𝑏𝑛2)+3𝐴𝑛(𝑎,𝑚𝑏)𝐿𝑛𝑛|||1(𝑎,𝑚𝑏)17291/𝑞𝐴(𝑚𝑏𝑎)|𝑛|×1/𝑞𝑎(𝑛1)𝑞,17𝑚𝑏(𝑛1)𝑞+𝐴1/𝑞17𝑎(𝑛1)𝑞,𝑚𝑏(𝑛1)𝑞+252288251/𝑞(×𝐴𝑚𝑏𝑎)|𝑛|1/𝑞7𝑎(𝑛1)𝑞,11𝑚𝑏(𝑛1)𝑞+𝐴1/𝑞11𝑎(𝑛1)𝑞,7𝑚𝑏(𝑛1)𝑞.(3.5)

Proof. The assertions follow from Corollary 2.5 for 𝑓(𝑥)=𝑥𝑛.

Proposition 3.4. For 𝑛(,0)[1,){1} and [𝑎,𝑏][0,𝑏], we have the following inequalities: ||(𝑎)𝐴(𝑎𝑛,𝑚𝑛𝑏𝑛)𝐿𝑛𝑛||1(𝑎,𝑏)22+1/𝑞×𝐴|𝑛|(𝑚𝑏𝑎)1/𝑞𝑎(𝑛1)𝑞,3𝑚𝑏(𝑛1)𝑞+𝐴1/𝑞3𝑎(𝑛1)𝑞,𝑚𝑏(𝑛1)𝑞,|||1(𝑏)3𝐴(𝑎𝑛,𝑚𝑛𝑏𝑛2)+3𝐴𝑛(𝑎,𝑚𝑏)𝐿𝑛𝑛|||1(𝑎,𝑚𝑏)61+1/𝑞131/𝑞|𝐴𝑛|(𝑚𝑏𝑎)×1/𝑞𝑎(𝑛1)𝑞,11𝑚𝑏(𝑛1)𝑞+𝐴1/𝑞11𝑎(𝑛1)𝑞,𝑚𝑏(𝑛1)𝑞+132+1/𝑞21/𝑞𝐴|𝑛|(𝑚𝑏𝑎)×1/𝑞𝑎(𝑛1)𝑞,2𝑚𝑏(𝑛1)𝑞+𝐴1/𝑞2𝑎(𝑛1)𝑞,𝑚𝑏(𝑛1)𝑞.(3.6)

Proof. The assertions follow from Corollary 2.7 for 𝑓(𝑥)=𝑥𝑛.

Proposition 3.5. For [𝑎,𝑏][0,𝑏], we have the following inequalities: ||||(𝑎)ln𝐼(𝑎,𝑏)||||𝐺(𝑎,𝑏)𝑏𝑎1𝑎𝑏41+1/𝑞(𝑎𝑞+3𝑏𝑞)1/𝑞+(3𝑎𝑞+𝑏𝑞)1/𝑞,(||||𝑏)ln𝐼(𝑎,𝑏)𝐺1/3(𝑎,𝑏)𝐴2/3||||(𝑎,𝑏)𝑏𝑎1𝑎𝑏61+2/𝑞(𝑎𝑞+11𝑏𝑞)1/𝑞+(11𝑎𝑞+𝑏𝑞)1/𝑞+132+1/𝑞(𝑎𝑞+2𝑏𝑞)1/𝑞+(2𝑎𝑞+𝑏𝑞)1/𝑞.(3.7)

Proof. The assertions follow from Corollary 2.7 for 𝑓(𝑥)=ln𝑥 and 𝑚=1.

Acknowledgment

The author is so indebted to the referee who read carefully through the paper very well and mentioned many scientifically and expressional mistakes.

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Copyright © 2012 Jaekeun Park. 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.

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