Several integral inequalities and an upper bound for the bidimensional Hermite-Hadamard inequality

Abstract

In this paper we prove several integral inequalities and we find an upper bound of the Hermite-Hadamard inequality for a convex function on a bounded area from the plane in special cases.

1 Introduction

Let f be a convex function on $[a,b]$. Then we have the following inequality, which is called Hermite-Hadamard inequality:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2}.$$

(1.1)

There are many extensions, generalizations and similar results of inequality (1.1). In [1], Fejer established the following weighted generalization of inequality (1.1).

Theorem 1.1 If $f:[a,b]\to \mathbb{R}$ is a convex function, then the inequality

$$f\left(\frac{a+b}{2}\right){\int }_a^{b}w(x)dx\le {\int }_a^{b}f(x)w(x)dx\le \frac{f(a)+f(b)}{2}{\int }_a^{b}w(x)dx$$

(1.2)

holds, where $w:[a,b]\to \mathbb{R}$ is non-negative, integrable and symmetric about $\frac{a+b}{2}$.

In [2], Yang and Tseng proved the following theorem which refines inequality (1.2).

Theorem 1.2 Let f and w be defined as in Theorem  1.1. If $P:[a,b]\to \mathbb{R}$ is defined by

$$P(t)={\int }_a^{b}f[tx+(1-t)\frac{a+b}{2}]w(x)dx,$$

then P is convex, increasing on $[0,1]$ and for all $t\in [0,1]$,

$$f\left(\frac{a+b}{2}\right){\int }_a^{b}w(x)dx=P(0)\le P(t)\le P(1)={\int }_a^{b}f(x)w(x)dx.$$

In this paper, we find an upper bound for ${\int }_a^{b}f(x)g(x)dx$, where f is a convex function on $[a,b]$ and g is non-negative increasing (or decreasing) on $[a,b]$, and ${\int }_a^{b}g(t)dt=1$. Finally, in Section 3 we find an upper bound for the following integral:

$$\frac{1}{{\int }_a^b(g(x)-h(x))dx}{\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx.$$

2 Integral inequalities

Theorem 2.1 Let $f:[a,b]\to \mathbb{R}$ be a differentiable convex function and $g:[a,b]\to [0,\mathrm{\infty }]$ be a continuous function.

• (i) If g is decreasing on $[a,b]$, then

  $$\frac{1}{{\int }_a^{b}g(x)dx}{\int }_a^{b}f(x)g(x)dx\le \frac{f(a)+f(b)}{2}.$$

• (ii) If g is increasing on $[a,b]$, then

  $$f\left(\frac{a+b}{2}\right)\le \frac{1}{{\int }_a^{b}g(x)dx}{\int }_a^{b}f(x)g(x)dx.$$

Proof

• (i) Denote

  $$H(x)={\int }_a^{x}f(t)g(t)dt-\frac{1}{2}(f(a)+f(x)){\int }_a^{x}g(t)dt.$$

We will show that $H^{\prime }(x)\le 0$. We have

$$\begin{array}{rcl}{H}^{\prime }(x)& =& f(x)g(x)-\frac{1}{2}{f}^{\prime }(x){\int }_a^{x}g(t)dt-\frac{1}{2}(f(a)+f(x))g(x)\\ =& \frac{1}{2}[g(x)(f(x)-f(a))-f^{\prime }(x){\int }_a^{x}g(t)dt].\end{array}$$

By the extended mean value theorem (Cauchy’s theorem), we have

$$\frac{f(x)-f(a)}{{\int }_a^{x}g(t)dt}=\frac{f^{\prime }(\zeta )}{g(\zeta )}(a<\zeta <x).$$

On the other hand, by the convexity of f and decreasing of g, we obtain

$$\frac{f(x)-f(a)}{{\int }_a^{x}g(t)dt}=\frac{f^{\prime }(\zeta )}{g(\zeta )}\le \frac{f^{\prime }(x)}{g(x)}.$$

Since g is non-negative,

$$H^{\prime }(x)=\frac{1}{2}[(f(x)-f(a))g(x)-f^{\prime }(x){\int }_a^{x}g(t)dt]\le 0,$$

which implies that H is decreasing. Hence, $H(b)\le H(a)=0$. The proof is complete.

• (ii) Denote

  $$H(x)=f\left(\frac{a+x}{2}\right){\int }_a^{x}g(t)dt-{\int }_a^{x}f(t)g(t)dt.$$

Then we have

$$\begin{array}{rcl}{H}^{\prime }(x)& =& \frac{1}{2}{f}^{\prime }\left(\frac{a+x}{2}\right){\int }_a^{x}g(t)dt+f\left(\frac{a+x}{2}\right)g(x)-f(x)g(x)\\ =& \frac{1}{2}{f}^{\prime }\left(\frac{a+x}{2}\right){\int }_a^{x}g(t)dt-g(x)(f(x)-f\left(\frac{a+x}{2}\right)).\end{array}$$

By the mean value theorem (Lagrange’s theorem), there exist ${\zeta }_1\in (\frac{a+x}{2},x)$ and ${\zeta }_2\in (a,x)$ such that

$$\frac{f(x)-f(\frac{a+x}{2})}{x-\frac{a+x}{2}}=f^{\prime }({\zeta }_1)\text{and}\frac{{\int }_a^{x}g(t)dt-0}{x-a}=g({\zeta }_2).$$

Hence,

$$\frac{2(f(x)-f(\frac{a+x}{2}))}{{\int }_a^{x}g(t)dt}=\frac{f^{\prime }({\zeta }_1)}{g({\zeta }_2)}.$$

By the convexity of f and increasing of g, we obtain

$$\frac{2(f(x)-f(\frac{a+x}{2}))}{{\int }_a^{x}g(t)dt}=\frac{f^{\prime }({\zeta }_1)}{g({\zeta }_2)}\ge \frac{f^{\prime }(\frac{a+x}{2})}{g(x)}.$$

So,

$$H^{\prime }(x)=\frac{1}{2}{f}^{\prime }\left(\frac{a+x}{2}\right){\int }_a^{x}g(t)dt-g(x)(f(x)-f\left(\frac{a+x}{2}\right))\le 0.$$

Therefore, H is decreasing and $H(b)\le H(a)=0$. The proof is complete. □

Theorem 2.2 Let $f:[a,b]\to \mathbb{R}$ be a convex function and $P:[a,b]\to [0,\mathrm{\infty })$ be an integrable function such that ${\int }_a^{b}P(x)dx=1$. Then

$${\int }_a^{b}f(x)P(x)dx\le \frac{bf(a)-af(b)}{b-a}+\frac{f(b)-f(a)}{b-a}{\int }_a^{b}xP(x)dx.$$

Proof

We have

$$\begin{array}{rcl}\frac{1}{b-a}{\int }_a^{b}f(x)P(x)dx& =& {\int }_0^{1}f(tb+(1-t)a)P(tb+(1-t)a)dt\\ \le & f(b){\int }_0^{1}tP(bt+(1-t)a)dt+f(b){\int }_0^1(1-t)P(bt+(1-t)a)dt\\ =& f(b){\int }_a^b\frac{x-a}{b-a}P(x)\frac{dx}{b-a}+f(a){\int }_a^b\frac{b-x}{b-a}P(x)\frac{dx}{b-a}\\ =& \frac{f(b)}{{(b-a)}^2}[{\int }_a^{b}xP(x)-a{\int }_a^{b}P(x)dx]\\ +\frac{f(a)}{{(b-a)}^2}[b{\int }_a^{b}P(x)dx-{\int }_a^{b}xP(x)dx]\\ =& \frac{f(b)}{{(b-a)}^2}[{\int }_a^{b}xP(x)dx-a]+\frac{f(a)}{{(b-a)}^2}[b-{\int }_a^{b}xP(x)dx].\end{array}$$

So, we get

$${\int }_a^{b}f(x)P(x)dx\le \frac{bf(a)-af(b)}{b-a}+\frac{f(b)-f(a)}{b-a}{\int }_a^{b}xP(x)dx.$$

 □

Corollary 2.1 Let $f:[a,b]\to \mathbb{R}$ be a convex function and g be a non-negative integrable function. Then

$${\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2}$$

and

$$\frac{{\int }_a^{b}f(x)g(x)dx}{{\int }_a^{b}g(x)dx}\le \frac{bf(a)-af(b)}{b-a}+\frac{f(b)-f(a)}{b-a}\frac{{\int }_a^{b}xg(x)dx}{{\int }_a^{b}g(x)dx}.$$

The proof is similar to the proof of theorem.

3 Right bidimensional Hermite-Hadamard inequality

Let us consider the bidimensional interval $\mathrm{△}=[a,b]\times [c,d]$ in ${\mathbb{R}}^2$. Recall that the mapping $f:\mathrm{△}\to \mathbb{R}$ is convex on △ if

$$f(\lambda x+(1-\lambda )z,\lambda y+(1-\lambda )w)\le \lambda f(x,y)+(1-\lambda )f(z,w)$$

holds for all $(x,y),(z,w)\in \mathrm{△}$ and $\lambda \in [0,1]$. A function $f:\mathrm{△}\to \mathbb{R}$ is called co-ordinated convex on △ if the partial mappings $f_y:[a,b]\to \mathbb{R}$, $f_y(u)=f(u,y)$ and $f_x:[c,d]\to \mathbb{R}$, $f_x(v)=f(x,v)$ are convex for all $y\in [c,d]$ and $x\in [a,b]$. Note that every convex function $f:\mathrm{△}\to \mathbb{R}$ is co-ordinated convex, but the converse is not generally true; see [3].

Dragomir in [4] established the following similar inequality of the Hermite-Hadamard inequality for a co-ordinated convex function on a rectangle from the plane ${\mathbb{R}}^2$.

Theorem 3.1 Suppose that $f:\mathrm{△}=[a,b]\times [c,d]\to \mathbb{R}$ is co-ordinated convex on △. Then one has the inequalities

$$\begin{array}{rcl}f(\frac{a+b}{2},\frac{c+d}{2})& \le & \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dydx\\ \le & \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.\end{array}$$

Now, let △ be a convex area from the plane ${\mathbb{R}}^2$, bounded by a convex function $y=h(x)$ and a concave function $y=g(x)$ and $x=a$, $x=b$, such that for any $x\in [a,b]$, $g(x)\ge h(x)$. Also, let F be a two-variable convex function on △. In [5] and [6], the following inequality is proved:

In this paper, we want to find an upper bound for the integral

$$\frac{1}{{\int }_a^b(g(x)-h(x))dx}{\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx.$$

(3.1)

For this purpose, we reach to the following integral:

$$\frac{1}{{\int }_a^b(g(x)-h(x))dx}{\int }_a^b[F(x,g(x))+F(x,h(x))](g(x)-h(x))dx.$$

It is well known that if $F(x,y)$ is increasing relative to y and $y=h(x)$ is convex on $[a,b]$, then $F(x,h(x))$ is convex on $[a,b]$, but we have no information about the convexity of $F(x,h(x))$ generally. So, in special cases, we will find an upper bound for the integral (3.1).

Theorem 3.2 Let △ be a bounded area by a convex function $y=h(x)$ and a concave function $y=g(x)$ on $[a,b]$ such that for any $x\in [a,b]$, $g(x)\ge h(x)$ and $g-h$ is increasing on $[a,b]$. Also, let F be a two-variable convex function on △ such that $F(x,g(x))$ and $F(x,h(x))$ are convex on $[a,b]$. Then one has the inequality

Proof Since F is convex on △, hence F is co-ordinated convex on △. So, $F_x:[h(x),g(x)]\to \mathbb{R}$, $F_x(y)=F(x,y)$ is convex on $[h(x),g(x)]$ for all $x\in [a,b]$. By the right-hand side of Hermite-Hadamard inequality (1.1), we have

$${\int }_{h(x)}^{g(x)}F(x,y)dy\le (g(x)-h(x))\left[\frac{F(x,g(x))+F(x,h(x))}{2}\right].$$

Integrating this inequality on $[a,b]$, we obtain

Since $g-h$ is increasing and $F(x,g(x))$, $F(x,h(x))$ are convex on $[a,b]$, by Theorem 2.1(i), we have

The proof is complete. □

Theorem 3.3 Let △ be a bounded area by a convex function h and a concave function g on $[a,b]$ such that for any $x\in [a,b]$, $g(x)\ge h(x)$. Also, let F be a two-variable convex function on △ such that $F(x,g(x))$ and $F(x,h(x))$ are convex on $[a,b]$. Then one has the inequality

where $\alpha (b)=\frac{{\int }_a^{b}t(g(t)-h(t))dt}{{\int }_a^b(g(t)-h(t))dt}$.

Proof

By a similar way to the proof of Theorem 3.2, we have

Since $F(x,g(x))+F(x,h(x))$ is convex, by Theorem 2.2 $(P(x)=\frac{g(x)-h(x)}{{\int }_a^b(g(x)-h(x))dx})$, we obtain

The proof is complete. □

In the following theorem, we prove the assertion of Theorem 3.3 with weak conditions.

Theorem 3.4 Let △, g and h be defined as in Theorem  3.3. Also, let F be a two-variable convex function on △ such that

$$\frac{\partial F(x,g(x))}{\partial g}(\frac{g(x)-g(a)}{x-a}-g^{\prime }(x))+\frac{\partial F(x,h(x))}{\partial h}(\frac{h(x)-h(a)}{x-a}-h^{\prime }(x))\le 0,$$

then we have

where $\alpha (b)=\frac{{\int }_a^{b}t(g(t)-h(t))dt}{{\int }_a^b(g(t)-h(t))}$.

Proof

Denote

$$H(x)={\int }_a^x{\int }_{h(x)}^{g(x)}f(t,y)dydt-\frac{1}{2}K(x){\int }_a^x(g(t)-h(t))dt,$$

where

$$K(x)=\left(\frac{x-\alpha (x)}{x-a}\right)[F(a,g(a))+F(a,h(a))]+\left(\frac{\alpha (x)-a}{x-a}\right)[F(x,g(x))+F(x,h(x))].$$

Then we have

$$H^{\prime }(x)={\int }_{h(x)}^{g(x)}F(x,y)dy-\frac{1}{2}K(x)(g(x)-h(x))-\frac{1}{2}{K}^{\prime }(x){\int }_a^x(g(t)-h(t))dt.$$

Since F is convex, so it is co-ordinated convex. Hence, by the right-hand side of the Hermite-Hadamard inequality, we obtain

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{1}{2}(g(x)-h(x))(F(x,g(x))+F(x,h(x)))\\ -\frac{1}{2}K(x)(g(x)-h(x))-\frac{1}{2}{K}^{\prime }(x){\int }_a^x(g(t)-h(t))dt.\end{array}$$

So,

$$H^{\prime }(x)\le \frac{1}{2}[(g(x)-h(x))(F(x,g(x))+F(x,h(x))-K(x))-K^{\prime }(x){\int }_a^x(g(t)-h(t))dt].$$

On the other hand, we have

Now, multiplying each term by

$${\int }_a^x(g(t)-h(t))dt$$

and using the fact

$${\int }_a^x(g(t)-h(t))dt\alpha (x)={\int }_a^{x}t(g(t)-h(t))dt,$$

we obtain

$${\int }_a^x(g(t)-h(t))dt{\alpha }^{\prime }(x)=(g(x)-h(x))(x-\alpha (x)).$$

Therefore,

By a similar way, we obtain

Thus,

So,

$$\begin{array}{lll}{H}^{\prime }(x)& \le & \frac{1}{2}[(g(x)-h(x))(F(x,g(x))+F(x,h(x))-K(x))-K^{\prime }(x){\int }_a^x(g(t)-h(t))dt]\\ =& \frac{1}{2}(g(x)-h(x))\frac{x-\alpha (x)}{x-a}[F(x,g(x))+F(x,h(x))-F(a,g(a))-F(a,h(a))]\\ -\frac{1}{2}[\frac{(g(x)-h(x))(x-\alpha (x))}{x-a}-\frac{\alpha (x)-a}{{(x-a)}^2}{\int }_a^x(g(t)-h(t))dt]\\ \times [F(x,g(x))+F(x,h(x))-F(a,g(a))-F(a,h(a))]\\ -\frac{1}{2}{\int }_a^x(g(t)-h(t))dt\left(\frac{\alpha (x)-a}{x-a}\right)(F^{\prime }(x,g(x))+F^{\prime }(x,h(x))).\end{array}$$

Then it follows that

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{1}{2}[F(x,g(x))+F(x,h(x))-F(a,g(a))-F(a,h(a))]\\ \times [(g(x)-h(x))\frac{x-\alpha (x)}{x-a}-\frac{(g(x)-h(x))(x-\alpha (x))}{x-a}\\ +\frac{\alpha (x)-a}{{(x-a)}^2}{\int }_a^x(g(t)-h(t))dt]\\ -\frac{1}{2}{\int }_a^x(g(t)-h(t))dt\left(\frac{\alpha (x)-a}{x-a}\right)(F^{\prime }(x,g(x))+F^{\prime }(x,h(x))).\end{array}$$

Thus,

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{1}{2}\left(\frac{\alpha (x)-a}{x-a}\right){\int }_a^x(g(t)-h(t))dt\\ \times [\frac{F(x,g(x))-F(a,g(a))}{x-a}+\frac{F(x,h(x))-F(x,g(x))}{x-a}-F^{\prime }(x,g(x))-F^{\prime }(x,h(x))].\end{array}$$

Now, notice that if $F(x,g(x))$, $F(x,h(x))$ were convex on $[a,b]$, we can deduce the assertion of Theorem 3.3. Since F is convex on △, we have

$$F(x,g(x))-F(a,g(a))\le \frac{\partial F(x,g(x))}{\partial x}(x-a)+\frac{\partial F(x,g(x))}{\partial g}(g(x)-g(a))$$

or

By a similar way, we have

Note that

$$F^{\prime }(x,g(x))=\frac{\partial F(x,g(x))}{\partial x}+\frac{\partial F(x,g(x))}{\partial g}{g}^{\prime }(x)$$

and

$$F^{\prime }(x,h(x))=\frac{\partial F(x,h(x))}{\partial x}+\frac{\partial F(x,h(x))}{\partial h}{h}^{\prime }(x).$$

So,

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{1}{2}\frac{\alpha (x)-a}{x-a}{\int }_a^x(g(t)-h(t))dt[\frac{\partial F(x,g(x))}{\partial x}\\ +\frac{\partial F(x,g(x))}{\partial g}\frac{g(x)-g(a)}{x-a}+\frac{\partial F(x,h(x))}{\partial x}\\ +\frac{\partial F(x,h(x))}{\partial h}\frac{h(x)-h(a)}{x-a}-\frac{\partial F(x,g(x))}{\partial x}\\ -\frac{\partial F(x,g(x))}{\partial g}{g}^{\prime }(x)-\frac{\partial F(x,h(x))}{\partial x}-\frac{\partial F(x,h(x))}{\partial h}{h}^{\prime }(x)].\end{array}$$

Thus,

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{1}{2}\frac{\alpha (x)-a}{x-a}{\int }_a^x(g(t)-h(t))dt[\frac{\partial F(x,g(x))}{\partial g}(\frac{g(x)-g(a)}{x-a}-g^{\prime }(x))\\ +\frac{\partial F(x,h(x))}{\partial x}(\frac{h(x)-h(a)}{x-a}-h^{\prime }(x))]\le 0.\end{array}$$

Note that $\alpha (x)\ge a$. Therefore, H is decreasing and

$$H(b)\le H(a)=0.$$

The proof is complete. □

Remark 3.1 Notice that since g is concave and h is convex on $[a,b]$, so $g^{\prime }$ is decreasing and $h^{\prime }$ is increasing on $[a,b]$. By the mean value theorem, we have

$$\frac{g(x)-g(a)}{x-a}-g^{\prime }(x)\ge 0\text{and}\frac{h(x)-h(a)}{x-a}-h^{\prime }(x)\le 0.$$

In particular, if we have $g(x)=mx+n$, then $\frac{g(x)-g(a)}{x-a}-g^{\prime }(x)=0$. So, if $\frac{\partial F(x,h(x))}{\partial h}\ge 0$, then

In the following theorem, we find an upper bound of the Hermite-Hadamard inequality for a co-ordinated convex function.

Theorem 3.5 Let △, g and h be defined as in Theorem  3.3. Also, let F be a convex function only relative to y, that is, $F_x:[h(x),g(x)]\to \mathbb{R}$, $F_x(v)=F(x,v)$ is convex for all $x\in [a,b]$. If $F^{\prime }(x,g(x))+F^{\prime }(x,h(x))\ge 0$, then

$$\frac{1}{{\int }_a^b(g(t)-h(t))dt}{\int }_a^b{\int }_{h(x)}^{g(x)}F(x,y)dydx\le \frac{1}{2}[F(b,g(b))+F(b,h(b))].$$

Proof

Denote

$$H(x)={\int }_a^x{\int }_{h(x)}^{g(x)}F(t,y)dydt-\frac{1}{2}{\int }_a^x(g(t)-h(t))dt[F(x,g(x))+F(x,h(x))].$$

We have

$$\begin{array}{rcl}{H}^{\prime }(x)& =& {\int }_{h(x)}^{g(x)}F(x,y)dy-\frac{1}{2}(g(x)-h(x))(F(x,g(x))+F(x,h(x)))\\ -\frac{1}{2}{\int }_a^x(g(t)-h(t))dt(F^{\prime }(x,g(x))+F^{\prime }(x,h(x))).\end{array}$$

Since F is convex relative to y, by the right-hand side of the Hermite-Hadamard inequality, we obtain

$$\begin{array}{rcl}{H}^{\prime }(x)& \le & \frac{1}{2}(g(x)-h(x))(F(x,g(x))+F(x,h(x)))\\ -\frac{1}{2}(g(x)-h(x))(F(x,g(x))+F(x,h(x)))\\ -\frac{1}{2}{\int }_a^x(g(t)-h(t))dt(F^{\prime }(x,g(x))+F^{\prime }(x,h(x)))\\ =& -\frac{1}{2}{\int }_a^x(g(t)-h(t))dt(F^{\prime }(x,g(x))+F^{\prime }(x,h(x)))\\ \le & 0.\end{array}$$

So, H is decreasing on $[a,b]$. That is, $H(b)\le H(a)=0$. □

4 Examples

Example 4.1 Let $F(x,y)=x^2+y^2$ and △ be bounded by $g(x)=\sqrt{1-x^2}$, $h(x)=x-1$ on $[0,1]$. Then $g(x)-h(x)=\sqrt{1-x^2}-x+1$ is decreasing on $[0,1]$ and $F(x,g(x))=1$, $F(x,h(x))=x^2+{(x-1)}^2$ are convex on $[0,1]$. By Theorem 3.2, we have

By easy calculation, we see that

$${\int }_0^1(\sqrt{1-x^2}-x+1)dx=\frac{\pi }{4}+\frac{1}{2}=\frac{\pi +2}{4}$$

and

$$\frac{4}{\pi +2}{\int }_0^1{\int }_{x-1}^{\sqrt{1-x^2}}(x^2+y^2)dydx\le 1.$$

Example 4.2 Let F, g and h be defined as in Example 4.1. By Theorem 3.3, we have

$$\begin{array}{rcl}\alpha (1)& =& \frac{{\int }_0^{1}t(\sqrt{1-t^2}-t+1)dt}{{\int }_0^1(\sqrt{1-t^2}-t+1)dt}=\frac{\frac{5}{6}}{\frac{\pi }{4}+\frac{1}{2}}\\ =& \frac{10}{3(\pi +2)}\frac{1}{{\int }_0^1(\sqrt{1-t^2}-t+1)dt}{\int }_0^1{\int }_{x-1}^{\sqrt{1-x^2}}(x^2+y^2)dydx\\ \le & \frac{1}{2}[\frac{3\pi -4}{3(\pi +2)}(F(0,g(0))+F(0,h(0)))+\frac{10}{3(\pi +2)}(F(1,g(1))+F(1,h(1)))].\end{array}$$

So,

$$\frac{4}{\pi +2}{\int }_0^1{\int }_{x-1}^{\sqrt{1-x^2}}(x^2+y^2)dydx\le 1.$$

Example 4.3 Let $F(x,y)=x^2+y^2$ and △ be bounded by $g(x)=x+2$, $h(x)=x^2$ on $[-1,2]$. Then $g-h$ is not decreasing on $[-1,2]$ and also $F(x,h(x))=x^2+x^4$ is not convex on $[-1,2]$. So, g, h and F do not hold in the hypothesis of Theorems 3.2 and 3.3. But we have

$$\frac{g(x)-g(-1)}{x+1}-g^{\prime }(x)=\frac{x+1}{x+1}-1=0,\frac{h(x)-h(-1)}{x+1}-h^{\prime }(x)=-x-1\le 0$$

and

$$\frac{\partial F(x,g(x))}{\partial g}=2(x+2),\frac{\partial F(x,h(x))}{\partial h}=2x^2.$$

So,

Thus, we can apply Theorem 3.4

Hence,

$$\frac{2}{9}{\int }_{-1}^2{\int }_{x^2}^{x+2}(x^2+y^2)dydx\le 11.$$

Example 4.4 Let $F(x,y)=xy$ and △ be bounded by $g(x)=x+2$, and $h(x)=x^2$ on $[-1,2]$. Then F is not convex on △, but it is convex relative to y, we have

$$F(x,g(x))=x^2+2x\text{and}F(x,h(x))=x^3.$$

So,

$$F^{\prime }(x,g(x))+F^{\prime }(x,h(x))=2x+2+3x^2>0.$$

Hence, by Theorem 3.5, we have

$$\frac{1}{{\int }_{-1}^2(x+2-x^2)dx}{\int }_{-1}^2{\int }_{x^2}^{x+2}xydydx\le \frac{1}{2}[F(2,g(2))+F(2,h(2))].$$

Hence,

$$\frac{2}{9}{\int }_{-1}^2{\int }_{x^2}^{x+2}xydydx\le 8.$$