Abstract

The authors establish some new inequalities for differentiable convex functions, which are similar to the celebrated Hermite-Hadamard's integral inequality for convex functions, and apply these inequalities to construct inequalities for special means of two positive numbers.

1. Introduction

In [1], the following Hermite-Hadamard type inequalities for differentiable convex functions were proved.

Theorem 1.1 (see [1, Theorem 2.2]). Let 𝑓𝐼 be a differentiable mapping on 𝐼,   𝑎,𝑏𝐼 with 𝑎<𝑏. If |𝑓(𝑥)| is convex on [𝑎,𝑏], then ||||𝑓(𝑎)+𝑓(𝑏)21𝑏𝑎𝑏𝑎||||||𝑓𝑓(𝑥)𝑑𝑥(𝑏𝑎)||+||𝑓(𝑎)||(𝑏)8.(1.1)

Theorem 1.2 (see [1, Theorem 2.3]). Let 𝑓𝐼 be a differentiable mapping on 𝐼,𝑎,𝑏𝐼 with 𝑎<𝑏, and let 𝑝>1. If the new mapping |𝑓(𝑥)|𝑝/(𝑝1) is convex on [𝑎,𝑏], then ||||𝑓(𝑎)+𝑓(𝑏)21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥𝑏𝑎2(𝑝+1)1/𝑝||𝑓||(𝑎)𝑝/(𝑝1)+||𝑓||(𝑏)𝑝/(𝑝1)2(𝑝1)/𝑝.(1.2)

In [2], the above inequalities were generalized as follows.

Theorem 1.3 (see [2, Theorems 1 and 2]). Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and let 𝑞1. If |𝑓(𝑥)|𝑞 is convex on [𝑎,𝑏], then ||||𝑓(𝑎)+𝑓(𝑏)21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥𝑏𝑎4||𝑓(||𝑎)𝑞+||𝑓(||𝑏)𝑞21/𝑞,||||𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥𝑏𝑎4||𝑓||(𝑎)𝑞+||𝑓||(𝑏)𝑞21/𝑞.(1.3)

In [3], the above inequalities were further generalized as follows.

Theorem 1.4 (see [3, Theorems 2.3 and 2.4]). Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and let 𝑝>1. If |𝑓(𝑥)|𝑝/(𝑝1) is convex on [𝑎,𝑏], then |||1𝑏𝑎𝑏𝑎𝑓(𝑥)𝑑𝑥𝑓𝑎+𝑏2|||𝑏𝑎416𝑝+11/𝑝×||𝑓||(𝑎)𝑝/(𝑝1)||𝑓+3||(𝑏)𝑝/(𝑝1)(𝑝1)/𝑝+3||𝑓||(𝑎)𝑝/(𝑝1)+||𝑓||(𝑏)𝑝/(𝑝1)(𝑝1)/𝑝,||||1𝑏𝑎𝑏𝑎𝑓(𝑥)𝑑𝑥𝑓𝑎+𝑏2||||𝑏𝑎44𝑝+11/𝑝||𝑓(||+||𝑓𝑎)(||.𝑏)(1.4)

In [4], an inequality similar to the above ones was given as follows.

Theorem 1.5 (see [4, Theorem 3]). Let 𝑓[𝑎,𝑏] be an absolutely continuous mapping on [𝑎,𝑏] whose derivative belongs to 𝐿𝑝[𝑎,𝑏]. Then ||||13𝑓(𝑎)+𝑓(𝑏)2+2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||1𝑓(𝑥)𝑑𝑥62𝑞+1+13(𝑞+1)1/𝑞(𝑏𝑎)1/𝑞𝑓𝑝,(1.5) where (1/𝑝)+(1/𝑞)=1 and 𝑝>1.

Recently, the following inequalities were obtained in [5].

Theorem 1.6. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 for 𝑞1 is convex on [𝑎,𝑏], then ||||16𝑓(𝑎)+𝑓(𝑏)+4𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎212𝑞+1+13(𝑞+1)1/𝑞3||𝑓||(𝑎)𝑞+||𝑓||(𝑏)𝑞41/𝑞+||𝑓||(𝑎)𝑞||𝑓+3||(𝑏)𝑞41/𝑞,||||16𝑓(𝑎)+𝑓(𝑏)+4𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥5(𝑏𝑎)||𝑓7261||(𝑎)𝑞||𝑓+29||(𝑏)𝑞901/𝑞+||𝑓29||(𝑎)𝑞||𝑓+61||(𝑏)𝑞901/𝑞.(1.6)

In this paper, we will establish some new Hermite-Hadamard type integral inequalities for differentiable functions and apply them to derive some inequalities of special means.

2. Lemmas

For establishing new integral inequalities of Hermite-Hadamard type, we need the lemmas below.

Lemma 2.1. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏. If 𝑓𝐿[𝑎,𝑏] and 𝜆,𝜇, then 𝜆𝑓(𝑎)+𝜇𝑓(𝑏)2+2𝜆𝜇2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎=𝑓(𝑥)𝑑𝑥𝑏𝑎410(1𝜆𝑡)𝑓𝑡𝑎+(1𝑡)𝑎+𝑏2+(𝜇𝑡)𝑓𝑡𝑎+𝑏2+(1𝑡)𝑏𝑑𝑡.(2.1)

Proof. Integrating by part and changing variable of definite integral yield 10(1𝜆𝑡)𝑓𝑡𝑎+(1𝑡)𝑎+𝑏22𝑑𝑡=𝑏𝑎(1𝜆𝑡)𝑓𝑡𝑎+(1𝑡)𝑎+𝑏2|||10+10𝑓𝑡𝑎+(1𝑡)𝑎+𝑏2=2𝑑𝑡𝑏𝑎𝜆𝑓(𝑎)+(1𝜆)𝑓𝑎+𝑏24(𝑏𝑎)2𝑎(𝑎+𝑏)/2𝑓(𝑥)𝑑𝑥.(2.2)
Similarly, we have 10(𝜇𝑡)𝑓𝑡𝑎+𝑏22+(1𝑡)𝑏𝑑𝑡=𝑡𝑏𝑎(𝜇𝑡)𝑓𝑎+𝑏2+|||(1𝑡)𝑏10+10𝑓𝑡𝑎+𝑏2+=2(1𝑡)𝑏𝑑𝑡𝑏𝑎(1𝜇)𝑓𝑎+𝑏24+𝜇𝑓(𝑏)(𝑏𝑎)2𝑏(𝑎+𝑏)/2𝑓(𝑥)𝑑𝑥.(2.3) Adding these two equations leads to Lemma 2.1.

Lemma 2.2. For 𝑠>0 and 0𝜉1, one has 10||||𝜉𝑡𝑠𝜉𝑑𝑡=𝑠+1+(1𝜉)𝑠+1,𝑠+110𝑡||||𝜉𝑡𝑠𝜉𝑑𝑡=𝑠+2+(𝑠+1+𝜉)(1𝜉)𝑠+1.(𝑠+1)(𝑠+2)(2.4)

Proof. This follows from a straightforward computation of definite integrals.

3. Some Integral Inequalities of Hermite-Hadamard Type

Now we are in a position to establish some new integral inequalities of Hermite-Hadamard type for differentiable convex functions.

The first main result is Theorem 3.1.

Theorem 3.1. Let 𝑓𝐼 be a differentiable function on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, 0𝜆,𝜇1, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 for 𝑞1 is convex on [𝑎,𝑏], then ||||𝜆f(𝑎)+𝜇𝑓(𝑏)2+2𝜆𝜇2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎8161/𝑞12𝜆+2𝜆211/𝑞×49𝜆+12𝜆22𝜆3||𝑓||(𝑎)𝑞+23𝜆+2𝜆3||𝑓||(𝑏)𝑞1/𝑞+12𝜇+2𝜇21(1/𝑞)×23𝜇+2𝜇3||𝑓||(𝑎)𝑞+49𝜇+12𝜇22𝜇3||𝑓||(𝑏)𝑞1/𝑞.(3.1)

Proof. For 𝑞>1, by Lemma 2.1, the convexity of |𝑓(𝑥)|𝑞 on [𝑎,𝑏], and the noted Hölder's integral inequality, we have ||||𝜆𝑓(𝑎)+𝜇𝑓(𝑏)2+2𝜆𝜇2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎410|||||||𝑓1𝜆𝑡𝑡𝑎+(1𝑡)𝑎+𝑏2|||𝑑𝑡+10|||||||𝑓𝜇𝑡𝑡𝑎+𝑏2|||+(1𝑡)𝑏𝑑𝑡𝑏𝑎410||||1𝜆𝑡𝑑𝑡11/𝑞10||||1𝜆𝑡1+𝑡2||𝑓||(𝑎)𝑞+1𝑡2||𝑓||(𝑏)𝑞𝑑𝑡1/𝑞+10||||𝜇𝑡𝑑𝑡11/𝑞10||||𝑡𝜇𝑡2||𝑓||(𝑎)𝑞+2𝑡2||𝑓||(𝑏)𝑞𝑑𝑡1/𝑞.(3.2) In virtue of Lemma 2.2, a direct calculation yields 10||||1𝜆𝑡1+𝑡2||𝑓||(𝑎)𝑞+1𝑡2||𝑓||(𝑏)𝑞=1𝑑𝑡2||𝑓||(𝑎)𝑞+||𝑓||(𝑏)𝑞10||||11𝜆𝑡𝑑𝑡+2||𝑓||(𝑎)𝑞||𝑓||(𝑏)𝑞10𝑡||||=11𝜆𝑡𝑑𝑡2||𝑓||(𝑎)𝑞+||𝑓||(𝑏)𝑞12𝜆+𝜆2+1||𝑓12||(𝑎)𝑞||𝑓||(𝑏)𝑞(1𝜆)3+(1𝜆)2=1(3𝜆)1249𝜆+12𝜆22𝜆3||𝑓||(𝑎)𝑞+11223𝜆+2𝜆3||𝑓||(𝑏)𝑞,10||||𝑡𝜇𝑡2||𝑓(||𝑎)𝑞+2𝑡2||𝑓(||𝑏)𝑞=1𝑑𝑡2||𝑓||(𝑎)𝑞||𝑓||(𝑏)𝑞10𝑡||||||𝑓𝜇𝑡𝑑𝑡+||(𝑏)𝑞10||||=1𝜇𝑡𝑑𝑡𝜇123+(1𝜇)2(||𝑓2+𝜇)(||𝑎)𝑞||𝑓(||𝑏)𝑞+12𝜇+𝜇2||𝑓(||𝑏)𝑞=11223𝜇+2𝜇3||𝑓||(𝑎)𝑞+11249𝜇+12𝜇22𝜇3||𝑓||(𝑏)𝑞.(3.3)
Substituting the above two equalities into the inequality (3.2) and utilizing Lemma 2.2 result in the inequality (3.1) for 𝑞>1.
For 𝑞=1, from Lemmas 2.1 and 2.2 it follows that ||||𝜆𝑓(𝑎)+𝜇𝑓(𝑏)2+2𝜆𝜇2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎410||||1𝜆𝑡1+𝑡2||𝑓||+(𝑎)1𝑡2||𝑓||+(𝑏)𝑑𝑡10||||𝑡𝜇𝑡2||𝑓||+(𝑎)2𝑡2||𝑓||=(𝑏)𝑑𝑡𝑏𝑎4849𝜆+12𝜆22𝜆3||𝑓||+(𝑎)23𝜆+2𝜆2||𝑓||+(𝑏)23𝜇+2𝜇2||𝑓||+(𝑎)49𝜇+12𝜇22𝜇3||𝑓||,(𝑏)(3.4) which is just equivalent to (3.1) for 𝑞=1. Theorem 3.1 is proved.

If taking 𝜆=𝜇 in Theorem 3.1, we derive the following corollary.

Corollary 3.2. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, 0𝜆1, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 is convex on [𝑎,𝑏] for 𝑞1, then ||||𝜆2[𝑓]+(𝑎)+𝑓(𝑏)(1𝜆)𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎8161/𝑞12𝜆+2𝜆21(1/𝑞)×49𝜆+12𝜆22𝜆3||𝑓||(𝑎)𝑞+23𝜆+2𝜆3||𝑓||(𝑏)𝑞1/𝑞+23𝜆+2𝜆3||𝑓||(𝑎)𝑞+49𝜆+12𝜆22𝜆3||𝑓||(𝑏)𝑞1/𝑞.(3.5)

If letting 𝜆=𝜇=1/2,2/3,1/3, respectively, in Theorem 3.1, we can deduce the inequalities below.

Corollary 3.3. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 is convex on [𝑎,𝑏] for 𝑞1, then ||||12𝑓(𝑎)+𝑓(𝑏)2+𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎116121/𝑞9||𝑓||(𝑎)𝑞||𝑓+3||(𝑏)𝑞1/𝑞+3||𝑓||(𝑎)𝑞||𝑓+9||(𝑏)𝑞1/𝑞,||||13𝑓(𝑎)+𝑓(𝑏)+𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥5(𝑏𝑎)1721801/𝑞||𝑓74||(𝑎)𝑞||𝑓+16||(𝑏)𝑞1/𝑞+||𝑓16||(𝑎)𝑞||𝑓+74||(𝑏)𝑞1/𝑞,||||16𝑓(𝑎)+𝑓(𝑏)+4𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥5(𝑏𝑎)172901/𝑞||𝑓61||(𝑎)𝑞||𝑓+29||(𝑏)𝑞1/𝑞+||𝑓29||(𝑎)𝑞||𝑓+61||(𝑏)𝑞1/𝑞.(3.6)

If setting 𝑞=1 in Corollary 3.3, then one has the following.

Corollary 3.4. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)| is convex on [𝑎,𝑏], then ||||12𝑓(𝑎)+𝑓(𝑏)2+𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎||𝑓16||+||𝑓(𝑎)||,||||1(𝑏)3𝑓(𝑎)+𝑓(𝑏)+𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥5(𝑏𝑎)||𝑓144(||+||𝑓𝑎)(||,||||1𝑏)6𝑓(𝑎)+𝑓(𝑏)+4𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥5(𝑏𝑎)||𝑓72||+||𝑓(𝑎)||.(𝑏)(3.7)

The second main result is Theorem 3.5.

Theorem 3.5. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, 0 ≤ λ, μ  ≤ 1, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 is convex on [𝑎,𝑏] for 𝑞1, then ||||𝜆𝑓(𝑎)+𝜇𝑓(𝑏)2+2𝜆𝜇2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎412(𝑞+1)(𝑞+2)1/𝑞(𝑞+3𝜆)(1𝜆)𝑞+1+(2𝑞+4𝜆)𝜆𝑞+1||𝑓||(𝑎)𝑞+(𝑞+1+𝜆)(1𝜆)𝑞+1+𝜆𝑞+2||𝑓(||𝑏)𝑞1/𝑞+(𝑞+1+𝜇)(1𝜇)𝑞+1+𝜇𝑞+2||𝑓||(𝑎)𝑞+(𝑞+3𝜇)(1𝜇)𝑞+1+(2𝑞+4𝜇)𝜇𝑞+1||𝑓||(𝑏)𝑞1/𝑞.(3.8)

Proof. For 𝑞>1, by the convexity of |𝑓(𝑥)|𝑞 on [𝑎,𝑏], Lemma 2.1, and Hölder's integral inequality, it follows that ||||𝜆𝑓(𝑎)+𝜇𝑓(𝑏)2+2𝜆𝜇2𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎410|||||||𝑓1𝜆𝑡𝑡𝑎+(1𝑡)𝑎+𝑏2|||𝑑𝑡+10|||||||𝑓𝜇𝑡𝑡𝑎+𝑏2|||+(1𝑡)𝑏𝑑𝑡𝑏𝑎410𝑑𝑡11/𝑞10||||1𝜆𝑡𝑞1+𝑡2||𝑓||(𝑎)𝑞+1𝑡2||𝑓||(𝑏)𝑞𝑑𝑡1/𝑞+10𝑑𝑡11/𝑞10||||𝜇𝑡𝑞𝑡2||𝑓||(𝑎)𝑞+2𝑡2||𝑓||(𝑏)𝑞𝑑𝑡1/𝑞=𝑏𝑎410||||1𝜆𝑡𝑞1+𝑡2||𝑓||(𝑎)𝑞+1𝑡2||𝑓||(𝑏)𝑞𝑑𝑡1/𝑞+10||||𝜇𝑡𝑞𝑡2||𝑓||(𝑎)𝑞+2𝑡2||𝑓||(𝑏)𝑞𝑑𝑡1/𝑞.(3.9) By Lemma 2.2, we have 10||||1𝜆𝑡𝑞1+𝑡2||𝑓||(𝑎)𝑞+1𝑡2||𝑓||(𝑏)𝑞=1𝑑𝑡2||𝑓||(𝑎)𝑞+||𝑓||(𝑏)𝑞10||||1𝜆𝑡𝑞1𝑑𝑡+2||𝑓||(𝑎)𝑞||𝑓||(𝑏)𝑞10𝑡||||1𝜆𝑡𝑞=1𝑑𝑡||𝑓2(𝑞+1)||(𝑎)𝑞+||𝑓||(𝑏)𝑞𝜆𝑞+1+(1𝜆)𝑞+1+1||𝑓2(𝑞+1)(𝑞+2)||(𝑎)𝑞||𝑓||(𝑏)𝑞(1𝜆)𝑞+2+(𝑞+2𝜆)𝜆𝑞+1=12(𝑞+1)(𝑞+2)(𝑞+3𝜆)(1𝜆)𝑞+1+(2𝑞+4𝜆)𝜆𝑞+1||𝑓||(𝑎)𝑞+(𝑞+1+𝜆)(1𝜆)𝑞+1+𝜆𝑞+2||𝑓||(𝑏)𝑞,10||||𝜇𝑡𝑞𝑡2||𝑓(||𝑎)𝑞+2𝑡2||𝑓(||𝑏)𝑞=1𝑑𝑡(2(𝑞+1)(𝑞+2)𝑞+1+𝜇)(1𝜇)𝑞+1+𝜇𝑞+2||𝑓(||𝑎)𝑞+(𝑞+3𝜇)(1𝜇)𝑞+1+(2𝑞+4𝜇)𝜇𝑞+1||𝑓||(𝑏)𝑞.(3.10) Substituting these two equalities into the inequality (3.9) yields (3.8) for 𝑞>1.
For 𝑞=1, the proof is the same as the deduction of (3.4). Thus, Theorem 3.5 is proved.

As the derivation of corollaries of Theorem 3.1, we can obtain the following corollaries of Theorem 3.5.

Corollary 3.6. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, 0𝜆1, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 is convex for 𝑞1 on [𝑎,𝑏], then ||||𝜆𝑓(𝑎)+𝑓(𝑏)2+(1𝜆)𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎412(𝑞+1)(𝑞+2)1/𝑞×(𝑞+3𝜆)(1𝜆)𝑞+1+(2𝑞+4𝜆)𝜆𝑞+1||𝑓||(𝑎)𝑞+(𝑞+1+𝜆)(1𝜆)𝑞+1+𝜆𝑞+2||𝑓||(𝑏)𝑞1/𝑞+(𝑞+1+𝜆)(1𝜆)𝑞+1+𝜆𝑞+2||𝑓||(𝑎)𝑞+(𝑞+3𝜆)(1𝜆)𝑞+1+(2𝑞+4𝜆)𝜆𝑞+1||𝑓||(𝑏)𝑞1/𝑞.(3.11)

Corollary 3.7. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 is convex for 𝑞1 on [𝑎,𝑏], then ||||12𝑓(𝑎)+𝑓(𝑏)2+𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎814(𝑞+1)(𝑞+2)1/𝑞||𝑓(3𝑞+6)||(𝑎)𝑞||𝑓+(𝑞+2)||(𝑏)𝑞1/𝑞+||𝑓(𝑞+2)||(𝑎)𝑞||𝑓+(3𝑞+6)||(𝑏)𝑞1/𝑞,||||16𝑓(𝑎)+𝑓(𝑏)+4𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥𝑏𝑎11218(𝑞+1)(𝑞+2)1/𝑞×11+6𝑞+(3𝑞+8)2𝑞+1||𝑓||(𝑎)𝑞+1+(3𝑞+4)2𝑞+1||𝑓||(𝑏)𝑞1/𝑞+1+(3𝑞+4)2𝑞+1||𝑓||(𝑎)𝑞+11+6𝑞+(3𝑞+8)2𝑞+1||𝑓||(𝑏)𝑞1/𝑞.(3.12)

Corollary 3.8. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)| is convex on [𝑎,𝑏], then ||||12𝑓(𝑎)+𝑓(𝑏)2+𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎||𝑓16||+||𝑓(𝑎)||,||||1(𝑏)6𝑓(𝑎)+𝑓(𝑏)+4𝑓𝑎+𝑏21𝑏𝑎𝑏𝑎||||𝑓(𝑥)𝑑𝑥2(𝑏𝑎)||𝑓27(||||𝑓𝑎)+32(||.𝑏)(3.13)

Corollary 3.9. Let 𝑓𝐼 be differentiable on 𝐼, 𝑎,𝑏𝐼 with 𝑎<𝑏, and 𝑓𝐿[𝑎,𝑏]. If |𝑓(𝑥)|𝑞 is convex for 𝑞1 on [𝑎,𝑏] and 𝑓(𝑎)+𝑓(𝑏)2=𝑓𝑎+𝑏2(3.14) then ||||𝑓(𝑎)+𝑓(𝑏)21𝑏𝑎𝑏𝑎𝑓||||=||||𝑓(𝑥)𝑑𝑥𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎1164(𝑞+1)(𝑞+2)1/𝑞×||𝑓(2𝑞+5)||(𝑎)𝑞||𝑓+(2𝑞+3)||(𝑏)𝑞1/𝑞+||𝑓||(𝑎)𝑞||𝑓+(4𝑞+7)||(𝑏)𝑞1/𝑞+||𝑓(4𝑞+7)||(𝑎)𝑞+||𝑓||(𝑏)𝑞1/𝑞+||𝑓(2𝑞+3)||(𝑎)𝑞||𝑓+(2𝑞+5)||(𝑏)𝑞1/𝑞.(3.15) In particular, when 𝑞=1, one has ||||𝑓(𝑎)+𝑓(𝑏)21𝑏𝑎𝑏𝑎𝑓||||=||||𝑓(𝑥)𝑑𝑥𝑎+𝑏21𝑏𝑎𝑏𝑎𝑓||||(𝑥)𝑑𝑥𝑏𝑎||𝑓16||+||𝑓(𝑎)||.(𝑏)(3.16)

4. Applications to Means

For two positive numbers 𝑎>0 and 𝑏>0, define 𝐴(𝑎,𝑏)=𝑎+𝑏2,𝐺(𝑎,𝑏)=𝑎𝑏,𝐻(𝑎,𝑏)=2𝑎𝑏,𝑏𝑎+𝑏𝐿(𝑎,𝑏)=𝑠+1𝑎𝑠+1(𝑠+1)(𝑏𝑎)1/𝑠1,a𝑏𝑎,𝑎=𝑏,𝐼(𝑎,𝑏)=𝑒𝑏𝑏𝑎𝑎1/(𝑏𝑎)𝐻,𝑎𝑏𝑎,𝑎=𝑏,𝜔,𝑠𝑎(𝑎,𝑏)=𝑠+(𝑎𝑏)𝑠/2𝜔+𝑏𝑠𝜔+21/𝑠,𝑠0𝑎𝑏,𝑠=0(4.1) for 0𝜔<. These means are, respectively, called the arithmetic, geometric, harmonic, generalized logarithmic, identric, and Heronian means of two positive number 𝑎 and 𝑏.

Applying Theorem 3.1 to 𝑓(𝑥)=𝑥𝑠 for 𝑠0 and 𝑥>0 leads to the following inequalities for means.

Theorem 4.1. Let 𝑎,𝑏>0, 𝑞1, either 𝑠>1 and (𝑠1)𝑞1 or 𝑠<0, then ||||𝐴(𝑎𝑠,𝑏𝑠)+𝐴𝑠(𝑎,𝑏)2𝐿𝑠||||(𝑎,𝑏)𝑏𝑎16|𝑠|𝑞14(𝑞+1)(𝑞+2)1/𝑞×(2𝑞+5)𝑎(𝑠1)𝑞+(2𝑞+3)𝑏(𝑠1)𝑞1/𝑞+𝑎(𝑠1)𝑞+(4𝑞+7)𝑏(𝑠1)𝑞1/𝑞+(4𝑞+7)𝑎(𝑠1)𝑞+𝑏(𝑠1)𝑞1/𝑞+(2𝑞+3)𝑎(𝑠1)𝑞+(2𝑞+5)𝑏(𝑠1)𝑞1/𝑞.(4.2) In particular, if 𝑠2 or 𝑠<0, then ||||𝐴(𝑎𝑠,𝑏𝑠)+𝐴𝑠(𝑎,𝑏)2𝐿𝑠||||(𝑎,𝑏)𝑏𝑎8𝑎|𝑠|𝐴𝑠1,𝑏𝑠1.(4.3)

Taking 𝑓(𝑥)=ln𝑥 for 𝑥>0 in Theorem 3.1 results in the following inequalities for means.

Theorem 4.2. For 𝑎,𝑏>0 and 𝑞1, one has |||ln𝐺(𝑎,𝑏)+ln𝐴(𝑎,𝑏)2|||ln𝐼(𝑎,𝑏)𝑏𝑎1164(𝑞+1)(𝑞+2)1/𝑞2𝑞+5𝑎𝑞+2𝑞+3𝑏𝑞1/𝑞+1𝑎𝑞+4𝑞+7𝑏𝑞1/𝑞+4𝑞+7𝑎𝑞+1𝑏𝑞1/𝑞+2𝑞+3𝑎𝑞+2𝑞+5𝑏𝑞1/𝑞.(4.4) In particular, |||ln𝐴(𝑎,𝑏)+ln𝐺(𝑎,𝑏)2|||ln𝐼(𝑎,𝑏)𝑏𝑎81.𝐻(𝑎,𝑏)(4.5)

Finally, we can establish an inequality for Heronian mean as follows.

Theorem 4.3. For 𝑏>𝑎>0, 𝜔0, and 𝑠4 or 0𝑠<1, one has |||||𝐻𝑠𝜔,𝑠(𝑎,𝑏)𝐻(𝑎𝑠,𝑏𝑠)+𝐻(𝑠/2)+1𝜔,(𝑠/2)+1𝑏𝑎+𝑎𝑏,1𝐻𝑠𝜔,𝑠𝐿𝑎2,𝑏2𝐺2|||||(𝑎,𝑏),1𝑏𝑎4𝑎|𝑠|𝐴(𝑎,𝑏)𝐴2(𝑠1),𝑏2(𝑠1)(𝜔+2)𝐺2𝑠+𝑎(𝑎,𝑏)𝜔𝐴(𝑎,𝑏)𝐴𝑠2,𝑏𝑠22(𝜔+2)𝐺𝑠.(𝑎,𝑏)(4.6)

Proof. Let 𝑓(𝑥)=(𝑥𝑠+𝜔𝑥𝑠/2+1)/(𝜔+2) for 𝑥>0 and 𝑠(1,4). Then 𝑓𝑠(𝑥)=𝑥𝜔+2𝑠1+𝜔2𝑥𝑠/21,||𝑓||(𝑥)=𝑥(𝑠/2)3|𝑠|𝜔𝜔+28(𝑠2)(𝑠4)+(𝑠1)(𝑠2)𝑥𝑠/20.(4.7) In virtue of Corollary 3.3, it follows that ||||12𝑓(𝑏/𝑎)+𝑓(𝑎/𝑏)2+𝑓𝑏/𝑎+𝑎/𝑏21𝑏/𝑎𝑎/𝑏𝑏/𝑎𝑎/𝑏𝑓||||=||||1(𝑥)𝑑𝑥212𝑏𝑠+𝜔(𝑎𝑏)𝑠/2+𝑎𝑠𝑎𝑠+𝑎(𝜔+2)𝑠+𝜔(𝑎𝑏)𝑠/2+𝑏𝑠𝑏s+(𝜔+2)(𝑏/𝑎+𝑎/𝑏)𝑠+𝜔(𝑏/𝑎+𝑎/𝑏)𝑠/2+11𝜔+2𝜔+2(𝑏/𝑎)𝑠+1(𝑎/𝑏)𝑠+1(𝑠+1)(𝑏/𝑎𝑎/𝑏)+𝜔(𝑏/𝑎)𝑠/2+1(𝑎/𝑏)𝑠/2+1(||||=|||||𝐻𝑠/2+1)(𝑏/𝑎𝑎/𝑏)+1𝑠𝜔,𝑠(𝑎,𝑏)𝐻(𝑎𝑠,𝑏𝑠)+𝐻(𝑠/2)+1𝜔,(𝑠/2)+1𝑏𝑎+𝑎𝑏,1𝐻𝑠𝜔,𝑠𝐿𝑎2,𝑏2𝐺2|||||.(𝑎,𝑏),1(4.8) On the other hand, we have 𝑏/𝑎𝑎/𝑏|||𝑓16𝑎𝑏|||+|||𝑓𝑏𝑎|||=𝑏2𝑎2𝑏16(𝜔+2)𝑎𝑏|𝑠|𝑎𝑠1+𝑎𝑏𝑠1+𝜔2𝑏|𝑠|𝑎𝑠/21+𝑎𝑏𝑠/21=(𝑏𝑎)|𝑠|4𝑎𝐴(𝑎,𝑏)𝐴2(𝑠1),𝑏2(𝑠1)(𝜔+2)𝐺2𝑠+𝑎(𝑎,𝑏)𝜔𝐴(𝑎,𝑏)𝐴𝑠2,𝑏𝑠22(𝜔+2)𝐺𝑠.(𝑎,𝑏)(4.9) The proof is complete.

Remark 4.4. Some inequalities of Hermite-Hadamard type were also obtained in [69] by the authors.

Acknowledgment

The first author was supported in part by the National Natural Science Foundation of China under Grant no. 10962004.