# More About Hermite-Hadamard Inequalities, Cauchy's Means, and Superquadracity

- S Abramovich
^{1}, - G Farid
^{2}Email author and - J Pečarić
^{2, 3}

**2010**:102467

https://doi.org/10.1155/2010/102467

© S. Abramovich et al. 2010

**Received: **26 March 2010

**Accepted: **4 August 2010

**Published: **16 August 2010

## Abstract

New results associated with Hermite-Hadamard inequalities for superquadratic functions are given. A set of Cauchy's type means is derived from these Hermite-Hadamard-type inequalities, and its log-convexity and monotonicity are proved.

## Keywords

## 1. Introduction

is holding for any convex function, that is, well known in the literature as the Hermite-Hadamard inequality (see [1, page 137]). In many areas of analysis applications of Hermite-Hadamard inequality appear for different classes of functions with and without weights; see for convex functions, for example, [2, 3]. Also some useful mappings are defined connected to this inequality see in [4–6]. Here we focus on a class of functions which are superquadratic and analogs and refinements of (1.1) are applied to obtain results useful in analysis.

Now we present definitions, theorems, and results that we use in this paper.

The following definition is given in [7].

for all . One says that is subquadratic if is a superquadratic function.

The followings theorem is given in [8] and is used in our main results:

Theorem 1.1.

for all and all choices and such that

Proposition 1.2 (see [9, Proposition ]).

Let . The following are equivalent:

Remark 1.4.

The next two sections are about mean value theorems, positive semidefiniteness, exponential convexity, log-convexity, Cauchy means, and their monotonicity, that are associated with Hermite-Hadamard inequalities for superquadratic functions.

## 2. Mean Value Theorems

It is clear from (1.3) Theorem 1.1 of that; if is superquadratic function; then .

In [7] we have the following Lemma.

Lemma 2.1.

Suppose that is continuously differentiable and . If is superadditive or is increasing, then is superquadratic.

Lemma 2.2 (see [12, Lemma ]).

Then and are increasing functions. If also then they are superquadratic functions.

Theorem 2.3.

Proof.

By combining the above two inequalities we get that there exists such that (2.4) holds. Moreover if (for example) is bounded from above we have that (2.5) is valid. Also (2.4) holds when is not bounded.

We omit the proofs of Theorems 2.4 and 2.6 as they are similar to the proofs in [9, 13–16].

Theorem 2.4.

is a new mean.

It is easy to check that the set of functions , satisfies Lemma 2.1. Therefore if we put and in (2.8), we have a new mean defined as follows.

If we put and in (2.8), then by the substitution, , we have a new mean defined as

It is clear from (1.4) Theorem 1.1 of that if is superquadratic function, then .

Theorem 2.5.

Proof.

By combining the above two inequalities we get that there exist such that (2.23) holds. Moreover if (for example) is bounded from above we have that (2.24) is valid. Also (2.23) holds when is not bounded.

Theorem 2.6.

is a new mean.

If we put and in (2.27) we have new mean defined as follows.

If we put and in (2.27), then by the substitution we have new mean defined as follows.

## 3. Positive Semidefiniteness, Exponential Convexity, and Log-Convexity

Lemma 3.1 (see [12, Lemma ]).

Then, with the convention , is superquadratic.

Theorem 3.2.

For defined in (2.1) one has the following.

that is, is log-convex in the Jensen sense.

(c)The function is exponentially convex.

(d) is log-convex, that is, for where one has

Corollary 3.3.

One has the following

Proof.

By applying Theorem 3.2(b) with and respectively, we get the result.

Similar to Theorem 3.2 we get the following.

Theorem 3.4.

For defined in (2.22) one has the following.

(a)The matrix , , is a positive-semidefinite matrix, that is,

that is, is log-convex in the Jensen sense.

(c)The function is exponentially convex.

Proof.

The proof is the same as the proof of Theorem 3.2.

In the next results we use the continuity of and .

When log f is convex we see that (also see [13])

Lemma 3.5.

Theorem 3.6.

Proof.

From the continuity of we get our result for , , and for we can consider limiting case.

Theorem 3.7.

Proof.

By substituting , and , such that , , we get the result, and for , we can consider the limiting case.

Theorem 3.8.

Proof.

From the continuity of we get our result for , ; and for we can consider limiting case.

Theorem 3.9.

Proof.

By substituting , and , such that , , we get the result, and for , we can consider the limiting case.

## Declarations

### Acknowledgment

This research was partially funded by Higher Education Commission, Pakistan. The research of the third author was supported by the Croatian Ministry of Science, Education and Sports under the Research grant no. 117-1170889-0888.

## Authors’ Affiliations

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