## Journal of Inequalities and Applications

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# On the Hermite-Hadamard type inequalities

Journal of Inequalities and Applications20132013:228

DOI: 10.1186/1029-242X-2013-228

Received: 20 October 2012

Accepted: 18 April 2013

Published: 7 May 2013

## Abstract

In the present paper, we establish some new Hermite-Hadamard type inequalities involving two functions. Our results in a special case yield recent results on Hermite-Hadamard type inequalities.

MSC:26D15.

### Keywords

Hermite-Hadamard inequality Barnes-Godunova-Levin inequality Minkowski integral inequality Hölder inequality

## 1 Introduction

The following inequality is well known in the literature as Hermite-Hadamard’s inequality [1].

Theorem 1.1 Let $f:\left[a,b\right]\subset \mathbb{R}\to \mathbb{R}$ be a convex function on an interval of real numbers. Then the following Hermite-Hadamard inequality for convex functions holds:
$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{2}.$
(1.1)
If the function f is concave, the inequality (1.1) can be written as follows:
$f\left(\frac{a+b}{2}\right)\ge \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\ge \frac{f\left(a\right)+f\left(b\right)}{2}.$
(1.2)

Recently, many generalizations, extensions and variants of this inequality have appeared in the literature (see, e.g., [210]) and the references given therein. In particular, in 2010, Özdemir and Dragomir [11] established some new Hermite-Hadamard inequalities and other integral inequalities involving two functions in . Following this work, the main purpose of the present paper is to establish some dual Hermite-Hadamard type inequalities involving two functions in ${\mathbb{R}}^{2}$. Our results provide some new estimates on such type of inequalities.

## 2 Preliminaries

A region $D\subset {\mathbb{R}}^{2}$ is called convex if it contains the close line segment joining any two of its points, or equivalently, if $\lambda {x}_{1}+\left(1-\lambda \right){x}_{2},\lambda {y}_{1}+\left(1-\lambda \right){y}_{2}\in D$ whenever $x\left({x}_{1},{y}_{1}\right),y\left({x}_{2},{y}_{2}\right)\in D$ and $0\le \lambda \le 1$.

Let $z=f\left(x,y\right)$ be a duality function on the convex region $D\subset {\mathbb{R}}^{2}$. $z=f\left(x,y\right)$ is called a duality convex function on the convex region D if
$f\left[\lambda {x}_{1}+\left(1-\lambda \right){x}_{2},\lambda {y}_{1}+\left(1-\lambda \right){y}_{2}\right]\le \lambda f\left({x}_{1},{y}_{1}\right)+\left(1-\lambda \right)f\left({x}_{2},{y}_{2}\right),$
(2.1)

whenever $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\in D$ and $0\le \lambda \le 1$.

If the function $f\left(x,y\right)$ is concave, the inequality (2.1) can be written as follows:
$f\left[\lambda {x}_{1}+\left(1-\lambda \right){x}_{2},\lambda {y}_{1}+\left(1-\lambda \right){y}_{2}\right]\ge \lambda f\left({x}_{1},{y}_{1}\right)+\left(1-\lambda \right)f\left({x}_{2},{y}_{2}\right).$
(2.2)
Let $x=\left({x}_{11},\dots ,{x}_{1n},\dots ,{x}_{m1},\dots ,{x}_{mn}\right)$ and $p=\left({p}_{11},\dots ,{p}_{1n},\dots ,{p}_{m1},\dots ,{p}_{mn}\right)$ be two positive nm-tuples, and let $r\in \mathbb{R}\cup \left\{+\mathrm{\infty },-\mathrm{\infty }\right\}$. Then, on putting ${P}_{mn}={\sum }_{{k}_{2}=1}^{n}{\sum }_{{k}_{1}=1}^{m}{p}_{{k}_{1}{k}_{2}}$, it easy follows that if $-\mathrm{\infty }\le r, then
${M}_{mn}^{\left[r\right]}\le {M}_{mn}^{\left[s\right]}$
(2.3)

(also see, e.g., [[1], p.15]). Here, the r th power mean of x with weights p is the following: ${M}_{mn}^{\left[r\right]}={\left(\frac{1}{{P}_{mn}}{\sum }_{{k}_{2}=1}^{n}{\sum }_{{k}_{1}=1}^{m}{p}_{{k}_{1}{k}_{2}}{x}_{{k}_{1}{k}_{2}}^{r}\right)}^{1/r}$ if $r\ne +\mathrm{\infty },0,-\mathrm{\infty }$; ${M}_{mn}^{\left[r\right]}={\left({\prod }_{{k}_{2}=1}^{n}{\prod }_{{k}_{1}=1}^{m}{x}_{{k}_{1}{k}_{2}}^{{p}_{{k}_{1}{k}_{2}}}\right)}^{{P}_{mn}}$ if $r=0$; ${M}_{mn}^{\left[r\right]}=min\left({x}_{11},\dots ,{x}_{1n},\dots ,{x}_{n1},\dots ,{x}_{mn}\right)$ if $r=-\mathrm{\infty }$ and ${M}_{mn}^{\left[r\right]}=max\left({x}_{11},\dots ,{x}_{1n},\dots ,{x}_{n1},\dots ,{x}_{mn}\right)$ if $r=+\mathrm{\infty }$.

Let $f\left(x,y\right):\left[a,b\right]×\left[c,d\right]\to \mathbb{R}$, and $p\ge 1$. Now, we define the p-norm of the function $f\left(x,y\right)$ on $\left[a,b\right]×\left[c,d\right]$ as follows:
${\parallel f\left(x,y\right)\parallel }_{p}={\left({\int }_{a}^{b}{\int }_{c}^{d}{|f\left(x,y\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p},\phantom{\rule{1em}{0ex}}1\le p<\mathrm{\infty },$
and
${\parallel f\left(x,y\right)\parallel }_{p}=sup|f\left(x,y\right)|,\phantom{\rule{1em}{0ex}}p=\mathrm{\infty },$

and ${L}^{p}\left(\left[a,b\right]×\left[c,d\right]\right)$ is the set of all functions $f\left(x,y\right):\left[a,b\right]×\left[c,d\right]\to \mathbb{R}$ such that ${\parallel f\left(x,y\right)\parallel }_{p}<\mathrm{\infty }$.

Lemma 2.1 (see [12]) (Barnes-Godunova-Levin inequality)

Let $f\left(x,y\right)$, $g\left(x,y\right)$ be nonnegative concave functions on $\left[a,b\right]×\left[c,d\right]$, then for $p,q>1$ we have
${\parallel f\left(x,y\right)\parallel }_{p}{\parallel g\left(x,y\right)\parallel }_{q}\le B\left(p,q\right){\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy,$
(2.4)
where
$B\left(p,q\right)=\frac{6{\left[\left(b-a\right)\left(d-c\right)\right]}^{1/p+1/q-1}}{{\left(p+1\right)}^{1/p}{\left(q+1\right)}^{1/q}}.$

Lemma 2.2 (see [1]) (Hermite-Hadamard inequality)

Let $f\left(x,y\right):\left[a,b\right]×\left[c,d\right]\subset {\mathbb{R}}^{2}\to \mathbb{R}$ be a convex function. Then the following dual Hermite-Hadamard inequality for convex functions holds:
$f\left(\frac{a+c}{2},\frac{b+d}{2}\right)\le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\le \frac{f\left(a,b\right)+f\left(c,d\right)}{2}.$
(2.5)

The inequality is reversed if the function $f\left(x,y\right)$ is concave.

Lemma 2.3 (see [13]) (A reversed Minkowski integral inequality)

Let $f\left(x,y\right)$ and $g\left(x,y\right)$ be positive functions satisfying
$0
(2.6)
Then
${\parallel f\left(x,y\right)\parallel }_{p}+{\parallel g\left(x,y\right)\parallel }_{p}\le c{\parallel f\left(x,y\right)+g\left(x,y\right)\parallel }_{p},$
(2.7)

where $c=\left[M\left(m+1\right)+\left(M+1\right)\right]/\left[\left(m+1\right)\left(M+1\right)\right]$.

## 3 Main results

Our main results are established in the following theorems.

Theorem 3.1 Let $p,q>1$ and let $f\left(x,y\right),g\left(x,y\right):\left[a,b\right]×\left[c,d\right]\to \mathbb{R}$ be nonnegative functions such that $f{\left(x,y\right)}^{p}$ and $g{\left(x,y\right)}^{q}$ are concave on $\left[a,b\right]×\left[c,d\right]$. Then
$\begin{array}{r}\frac{f\left(a,b\right)+f\left(c,d\right)}{2}×\frac{g\left(a,b\right)+g\left(c,d\right)}{2}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{1/p+1/q}}B\left(p,q\right){\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy,\end{array}$
(3.1)

where $B\left(p,q\right)$ is the Barnes-Godunova-Levin constant given by (2.4).

Proof Observe that whenever ${f}^{p}\left(x,y\right)$ is concave on $\left[a,b\right]×\left[c,d\right]$, the nonnegative function $f\left(x,y\right)$ is also concave on $\left[a,b\right]×\left[c,d\right]$. Namely,
$f{\left[\lambda a+\left(1-\lambda \right)c,\lambda b+\left(1-\lambda \right)d\right]}^{p}\ge \lambda f{\left(a,b\right)}^{p}+\left(1-\lambda \right)f{\left(c,d\right)}^{p},$
that is,
$f\left[\lambda a+\left(1-\lambda \right)c,\lambda b+\left(1-\lambda \right)d\right]\ge {\left(\left(\lambda f{\left(a,b\right)}^{p}+\left(1-\lambda \right)f{\left(c,d\right)}^{p}\right)\right)}^{1/p},$
and $p>1$, using the power-mean inequality (2.3), we obtain
$f\left[\lambda a+\left(1-\lambda \right)c,\lambda b+\left(1-\lambda \right)d\right]\ge \lambda f\left(a,b\right)+\left(1-\lambda \right)f\left(c,d\right).$

For $q>1$, similarly, if ${g}^{q}\left(x,y\right)$ is concave on $\left[a,b\right]×\left[c,d\right]$, the nonnegative function $g\left(x,y\right)$ is concave on $\left[a,b\right]×\left[c,d\right]$.

In view that ${f}^{p}\left(x,y\right)$ and ${g}^{q}\left(x,y\right)$ are concave functions on $\left[a,b\right]×\left[c,d\right]$, from Lemma 2.2, we get
$\begin{array}{rl}{\left(\frac{f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}}{2}\right)}^{1/p}& \le \frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{1/p}}{\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}\\ \le f\left(\frac{a+c}{2},\frac{b+d}{2}\right),\end{array}$
(3.2)
and
$\begin{array}{rl}{\left(\frac{g{\left(a,b\right)}^{p}+g{\left(c,d\right)}^{q}}{2}\right)}^{1/q}& \le \frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{1/q}}{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{q}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/q}\\ \le g\left(\frac{a+c}{2},\frac{b+d}{2}\right).\end{array}$
(3.3)
By multiplying the above inequalities, we obtain
$\begin{array}{r}{\left(\frac{f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}}{2}\right)}^{1/p}{\left(\frac{g{\left(a,b\right)}^{p}+g{\left(c,d\right)}^{q}}{2}\right)}^{1/q}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{1/p+1/q}}{\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{q}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/q}.\end{array}$
(3.4)
If $p,q>1$, then it is easy to show that
${\left(\frac{f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}}{2}\right)}^{1/p}\ge \frac{f\left(a,b\right)+f\left(c,d\right)}{2},$
(3.5)
and
${\left(\frac{g{\left(a,b\right)}^{q}+g{\left(c,d\right)}^{q}}{2}\right)}^{1/q}\ge \frac{g\left(a,b\right)+g\left(c,d\right)}{2}.$
(3.6)

Thus, by applying Barnes-Godunova-Levin inequality to the right-hand side of (3.4) with (3.5), (3.6), we get (3.1).

The proof is complete. □

Remark 3.1 By multiplying inequalities (3.2), (3.3), we obtain
$\begin{array}{r}\frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{1/p+1/q}}{\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{q}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/q}\\ \phantom{\rule{1em}{0ex}}\le f\left(\frac{a+c}{2},\frac{b+d}{2}\right)g\left(\frac{a+c}{2},\frac{b+d}{2}\right).\end{array}$
(3.7)
By applying the Hölder inequality to the left-hand side of (3.7) with $\left(1/p\right)+\left(1/q\right)=1$, we get
$\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\le f\left(\frac{a+c}{2},\frac{b+d}{2}\right)g\left(\frac{a+c}{2},\frac{b+d}{2}\right).$
(3.8)

Remark 3.2 Let $f\left(x,y\right)$ and $g\left(x,y\right)$ change to $f\left(x\right)$ and $g\left(x\right)$, respectively, and with suitable changes in Theorem 3.1 and Remark 3.1, we have the following.

Corollary 3.1 Let $p,q>1$ and let $f\left(x\right),g\left(x\right):\left[a,b\right]\to \mathbb{R}$, $a, be nonnegative functions such that $f{\left(x\right)}^{p}$ and $g{\left(x\right)}^{q}$ are concave on $\left[a,b\right]$. Then
$\frac{f\left(a\right)+f\left(b\right)}{2}\cdot \frac{g\left(a\right)+g\left(b\right)}{2}\le \frac{1}{{\left(b-a\right)}^{1/p+1/q}}B\left(p,q\right){\int }_{a}^{b}f\left(x\right)g\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,$
and if $\left(1/p\right)+\left(1/q\right)=1$, then one has
$\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)g\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le f\left(\frac{a+b}{2}\right)g\left(\frac{a+b}{2}\right).$

This is just Theorem 2.1 established by Özdemir and Dragomir [11].

Theorem 3.2 Let $p\ge 1$ and let ${\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy<\mathrm{\infty }$ and ${\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy<\mathrm{\infty }$, and let $f\left(x,y\right),g\left(x,y\right):\left[a,b\right]×\left[c,d\right]\to \mathbb{R}$ be positive functions with
$0
Then
${\parallel f\left(x,y\right)\parallel }_{p}^{2}+{\parallel g\left(x,y\right)\parallel }_{p}^{2}\ge \left(\frac{\left(M+1\right)\left(m+1\right)}{M}-2\right){\parallel f\left(x,y\right)\parallel }_{p}{\parallel g\left(x,y\right)\parallel }_{p}.$
(3.9)
Proof Since $f\left(x,y\right)$, $g\left(x,y\right)$ are positive, as in the proof of Lemma 2.3 (see [[13], p.2]), we have
${\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}\le \frac{M}{M+1}{\left({\int }_{a}^{b}{\int }_{c}^{d}{\left(f\left(x,y\right)+g\left(x,y\right)\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}$
and
${\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}\le \frac{1}{m+1}{\left({\int }_{a}^{b}{\int }_{c}^{d}{\left(f\left(x,y\right)+g\left(x,y\right)\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}.$
By multiplying the above inequalities and in view of the Minkowski inequality, we get
$\begin{array}{r}{\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}\\ \phantom{\rule{1em}{0ex}}\le \frac{M}{\left(M+1\right)\left(m+1\right)}{\left({\int }_{a}^{b}{\int }_{c}^{d}{\left(f\left(x,y\right)+g\left(x,y\right)\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{2/p}\\ \phantom{\rule{1em}{0ex}}\le \frac{M}{\left(M+1\right)\left(m+1\right)}\left({\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}\\ \phantom{\rule{2em}{0ex}}{+{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}\right)}^{2}.\end{array}$
(3.10)
Hence
$\begin{array}{c}{\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{2/p}+{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{2/p}\hfill \\ \phantom{\rule{1em}{0ex}}\ge \left(\frac{\left(M+1\right)\left(m+1\right)}{M}-2\right){\left({\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}{\left({\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\right)}^{1/p}.\hfill \end{array}$

This proof is complete. □

Remark 3.3 Let $f\left(x,y\right)$ and $g\left(x,y\right)$ change to $f\left(x\right)$ and $g\left(x\right)$, respectively, and with suitable changes in (3.9), (3.9) reduces to an inequality established by Özdemir and Dragomir [11].

Theorem 3.3 If ${f}^{p}\left(x,y\right)$ and ${g}^{q}\left(x,y\right)$ are as in Theorem  3.1, then the following inequality holds:
$\frac{1}{\left(b-a\right)\left(d-c\right)}{\parallel f\left(x,y\right)\parallel }_{p}^{p}\cdot {\parallel g\left(x,y\right)\parallel }_{q}^{q}\ge \frac{{\left(f\left(a,b\right)+f\left(c,d\right)\right)}^{p}{\left(g\left(a,b\right)+g\left(c,d\right)\right)}^{q}}{{2}^{p+q}}.$
(3.11)
Proof If ${f}^{p}\left(x,y\right)$ and ${g}^{q}\left(x,y\right)$ are concave on $\left[a,b\right]×\left[c,d\right]$, then from Lemma 2.2, we get
$\frac{f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}}{2}\le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy$
and
$\frac{g{\left(a,b\right)}^{q}+g{\left(c,d\right)}^{q}}{2}\le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{q}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy,$
which imply that
$\begin{array}{r}\frac{\left[f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}\right]\left[g{\left(a,b\right)}^{q}+g{\left(c,d\right)}^{q}\right]}{4}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{2}}{\int }_{a}^{b}{\int }_{c}^{d}f{\left(x,y\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy{\int }_{a}^{b}{\int }_{c}^{d}g{\left(x,y\right)}^{q}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy.\end{array}$
(3.12)
On the other hand, if $p,q\ge 1$, from (2.3) we get
$\frac{f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}}{2}\le {2}^{-p}{\left[f\left(a,b\right)+f\left(c,d\right)\right]}^{p}$
and
$\frac{g{\left(a,b\right)}^{q}+g{\left(c,d\right)}^{q}}{2}\le {2}^{-q}{\left[g\left(a,b\right)+g\left(c,d\right)\right]}^{q},$
which imply that
$\begin{array}{r}\frac{\left[f{\left(a,b\right)}^{p}+f{\left(c,d\right)}^{p}\right]\left[g{\left(a,b\right)}^{p}+g{\left(c,d\right)}^{q}\right]}{4}\\ \phantom{\rule{1em}{0ex}}\ge {2}^{-p-q}{\left[f\left(a,b\right)+f\left(c,d\right)\right]}^{p}{\left[g\left(a,b\right)+g\left(c,d\right)\right]}^{q}.\end{array}$
(3.13)
Combining (3.12) and (3.13), we obtain the desired inequality as
$\begin{array}{c}{2}^{-p-q}{\left[f\left(a,b\right)+f\left(c,d\right)\right]}^{p}{\left[g\left(a,b\right)+g\left(c,d\right)\right]}^{q}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\left[\left(b-a\right)\left(d-c\right)\right]}^{2}}{\parallel f\left(x,y\right)\parallel }_{p}^{p}\cdot {\parallel g\left(x,y\right)\parallel }_{q}^{q}.\hfill \end{array}$

This proof is complete. □

Remark 3.4 Let $f\left(x,y\right)$ and $g\left(x,y\right)$ change to $f\left(x\right)$ and $g\left(x\right)$, respectively, and with suitable changes in (3.11), (3.11) reduces to an inequality established by Özdemir and Dragomir [11].

Theorem 3.4 Let $f\left(x,y\right),g\left(x,y\right):\left[a,b\right]×\left[c,d\right]\to {\mathbb{R}}^{+}$ be functions such that $f{\left(x,y\right)}^{p}$, $g{\left(x,y\right)}^{q}$ and $f\left(x,y\right)g\left(x,y\right)$ are in ${L}_{1}\left(\left[a,b\right]×\left[c,d\right]\right)$, and
$0
Then
$\begin{array}{r}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\\ \phantom{\rule{1em}{0ex}}\le {c}_{1}\left(\frac{{\parallel f\left(x,y\right)\parallel }_{p}^{p}+{\parallel g\left(x,y\right)\parallel }_{p}^{p}}{2}\right)+{c}_{2}\left(\frac{{\parallel f\left(x,y\right)\parallel }_{q}^{q}+{\parallel g\left(x,y\right)\parallel }_{q}^{q}}{2}\right),\end{array}$
(3.14)
where
${c}_{1}=\frac{{2}^{p}}{p}{\left(\frac{M}{M+1}\right)}^{p},\phantom{\rule{2em}{0ex}}{c}_{2}=\frac{{2}^{q}}{q}{\left(\frac{1}{m+1}\right)}^{q},$

and $\left(1/p\right)+\left(1/q\right)=1$ with $p>1$.

Proof Since $0, $\mathrm{\forall }\left(x,y\right)\in \left[a,b\right]×\left[c,d\right]$, we have
$f\left(x,y\right)\le \frac{M}{M+1}\left(f\left(x,y\right)+g\left(x,y\right)\right)$
and
$g\left(x,y\right)\le \frac{1}{m+1}\left(f\left(x,y\right)+g\left(x,y\right)\right).$
In view of the Young-type inequality and using the elementary inequality
${\left(a+b\right)}^{p}\le {2}^{p-1}\left({a}^{p}+{b}^{p}\right),\phantom{\rule{1em}{0ex}}p>1,a,b\in {\mathbb{R}}^{+},$
we have
$\begin{array}{c}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{p}{\left(\frac{M}{M+1}\right)}^{p}{\int }_{a}^{b}{\int }_{c}^{d}{\left(f\left(x,y\right)+g\left(x,y\right)\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{q}{\left(\frac{1}{m+1}\right)}^{q}{\int }_{a}^{b}{\int }_{c}^{d}{\left(f\left(x,y\right)+g\left(x,y\right)\right)}^{q}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{p}{\left(\frac{M}{M+1}\right)}^{p}{2}^{p-1}{\int }_{a}^{b}{\int }_{c}^{d}\left[f{\left(x,y\right)}^{p}+g{\left(x,y\right)}^{p}\right]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{q}{\left(\frac{1}{m+1}\right)}^{q}{2}^{q-1}{\int }_{a}^{b}{\int }_{c}^{d}\left[f{\left(x,y\right)}^{q}+g{\left(x,y\right)}^{q}\right]\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy.\hfill \end{array}$

This completes the proof. □

Remark 3.5 Let $f\left(x,y\right)$ and $g\left(x,y\right)$ change to $f\left(x\right)$ and $g\left(x\right)$, respectively, and with suitable changes in (3.14), (3.14) reduces to an inequality established by Özdemir and Dragomir [11].

## Declarations

### Acknowledgements

The first author’s research is supported by Natural Science Foundation of China (10971205). The second author’s research is partially supported by a HKU Seed Grant for Basic Research.

## Authors’ Affiliations

(1)
Department of Mathematics, China Jiliang University
(2)
Department of Mathematics, The University of Hong Kong
(3)
Department of Mathematics, Hunan Normal University

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