On the $(p,h)$-convex function and some integral inequalities

Abstract

In this paper, we introduce a new class of $(p,h)$-convex functions which generalize P-functions and convex, $h,p,s$-convex, Godunova-Levin functions, and we give some properties of the functions. Moreover, we establish the corresponding Schur, Jensen, and Hadamard types of inequalities.

MSC:35K65, 35B33, 35B40.

1 Introduction

Let I and J be intervals in R. To motivate our work, let us recall the definitions of some special classes of functions.

Definition 1 [1]

A function $f:I\to R$ is said to be a Godunova-Levin function or belongs to the class $Q(I)$ if f is non-negative and

$$f(\alpha x+(1-\alpha )y)\le \frac{f(x)}{\alpha }+\frac{f(y)}{1-\alpha }$$

for all $x,y\in I$ and $\alpha \in (0,1)$.

The class $Q(I)$ was firstly described in [1] by Godunova and Levin. Some further properties of it are given in [2, 3]. It has been known that non-negative convex and monotone functions belong to this class of functions.

Definition 2 [4]

Let $s\in (0,1)$ be a fixed real number. A function $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is said to be an s-convex function (in the second sense) or belongs to the class $K_s^2$, if

$$f(\alpha x+(1-\alpha )y)\le {\alpha }^{s}f(x)+{(1-\alpha )}^{s}f(y)$$

for all $x,y\in I$ and $\alpha \in [0,1]$.

An s-convex function was introduced by Breckner [4] and a number of properties and connections with s-convexity (in the first sense) were discussed in [5]. Of course, s-convexity means just convexity when $s=1$.

Definition 3 [2]

A function $f:I\to R$ is said to be a P-function or belongs to the class $P(I)$, if f is non-negative and

$$f(\alpha x+(1-\alpha )y)\le f(x)+f(y)$$

for all $x,y\in I$ and $\alpha \in [0,1]$.

For some results on the class $P(I)$, see [6, 7].

Definition 4 [8]

Let I be a p-convex set. A function $f:I\to R$ is said to be a p-convex function or belongs to the class $PC(I)$, if

$$f\left({[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\right)\le \alpha f(x)+(1-\alpha )f(y)$$

for all $x,y\in I$ and $\alpha \in [0,1]$.

Remark 1 [8]

An interval I is said to be a p-convex set if ${[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\in I$ for all $x,y\in I$ and $\alpha \in [0,1]$, where $p=2k+1$ or $p=\frac{n}{m}$, $n=2r+1$, $m=2t+1$, and $k,r,t\in N$.

Definition 5 [9]

Let $h:J\to R$ be a non-negative and non-zero function. We say that $f:I\to R$ is an h-convex function or that f belongs to the class $SX(I)$, if f is non-negative and

$$f(\alpha x+(1-\alpha )y)\le h(\alpha )f(x)+h(1-\alpha )f(y)$$

for all $x,y\in I$ and $\alpha \in (0,1)$.

The h- and p-convex functions were introduced by Varšanec, Zhang and Wan, and a number of properties and Jensen’s inequalities of the functions were established (cf. [8]). As one can see, the definitions of the P-function, convex, $h,p,s$-convex, Godunova-Levin functions have similar forms. This observation leads us to generalize these varieties of convexity.

2 Definitions and basic results

In this section, we give new definitions and properties of the $(p,h)$-convex function. Throughout this paper, we assume that $(0,1)\subseteq J$, f and h are real non-negative functions defined on I and J, respectively, and the set I is p-convex when $f\in ghx(p,h,I)$ or $f\in ghv(p,h,I)$. We first give a definition of the new class of convex functions.

Definition 6 Let $h:J\to R$ be a non-negative and non-zero function. We say that $f:I\to R$ is a $(p,h)$-convex function or that f belongs to the class $ghx(h,p,I)$, if f is non-negative and

$$f\left({[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\right)\le h(\alpha )f(x)+h(1-\alpha )f(y)$$

(2.1)

for all $x,y\in I$ and $\alpha \in (0,1)$. Similarly, if the inequality sign in (2.1) is reversed, then f is said to be a $(p,h)$-concave function or belong to the class $ghv(h,p,I)$.

Remark 2 It can be obviously seen that if $h(\alpha )=\alpha$, then all non-negative p-convex and p-concave functions belong to $ghx(h,p,I)$ and $ghv(h,p,I)$, respectively; if $h(\alpha )=\alpha$ and $p=1$, then all non-negative convex functions belong to $ghx(h,p,I)$; if $h(\alpha )=\frac{1}{\alpha }$ and $p=1$, then $Q(I)=ghx(h,p,I)$; if $h(\alpha )={\alpha }^s$, $s\in (0,1)$, and $p=1$, then $K_s^2\subseteq ghx(h,p,I)$; if $h(\alpha )=1$ and $p=1$, then $P(I)\subseteq ghx(h,p,I)$, and if $p=1$, then $SX(I)\subseteq ghx(h,p,I)$.

Example 1 Let $h_k(\alpha )={\alpha }^k$, where $k\le 1$ and $\alpha >0$. If f is a function defined as $f(x)=x^p$, where p is an odd number and $x\ge 0$, we then have

$$f\left({[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\right)\le \alpha f(x)+(1-\alpha )f(y)\le h_k(\alpha )f(x)+h_k(1-\alpha )f(y),$$

and hence, f belongs to $ghx(h_k,p,I)$.

Next, we discuss some interesting properties of $(p,h)$-convex (concave) functions, which include linearity, product, composition properties, and an ordered property of h and p. In addition, we give some interesting properties of the $(p,h)$-convex function, when h is a super(sub)-multiplicative function.

Property 1 If $f,g\in ghx(h,p,I)$ and $\lambda >0$, then $f+g,\lambda f\in ghx(h,p,I)$. Similarly, if $f,g\in ghv(h,p,I)$ and $\lambda >0$, then $f+g,\lambda f\in ghv(h,p,I)$.

Proof The proof immediately follows from the definitions of the classes $ghx(h,p,I)$ and $ghv(h,p,I)$. □

Property 2 Let $h_1$ and $h_2$ be non-negative functions defined on an interval J with $h_2\le h_1$ in $(0,1)$. If $f\in ghx(h_2,p,I)$, then $f\in ghx(h_1,p,I)$. Similarly, if $f\in ghv(h_1,p,I)$, then $f\in ghv(h_2,p,I)$.

Proof If $f\in ghx(h_2,p,I)$, then for any $x,y\in I$ and $\alpha \in (0,1)$ we have

$$\begin{array}{rcl}f\left({[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\right)& \le & h_2(\alpha )f(x)+h_2(1-\alpha )f(y)\\ \le & h_1(\alpha )f(x)+h_1(1-\alpha )f(y),\end{array}$$

and hence, $f\in ghx(h_1,p,I)$. □

Property 3 Let $f\in ghx(h,p_1,I)$.

• (a) For $I\subseteq (0,1]$, if f is monotone increasing (monotone decreasing), and $p_2\ge p_1>0$ or $p_2\le p_1<0$, and ($p_1\ge p_2>0$ or $p_1\le p_2<0$), then $f\in ghx(h,p_2,I)$.

• (b) For $I\subseteq [1,\mathrm{\infty })$, if f is monotone increasing (monotone decreasing), and $p_1\ge p_2>0$ or $p_1\le p_2<0$, and ($p_2\ge p_1>0$ or $p_2\le p_1<0$), then $f\in ghx(h,p_2,I)$.

Let $f\in ghv(h,p_1,I)$.

• (c) For $I\subseteq (0,1]$, if f is monotone increasing (monotone decreasing), and $p_1\ge p_2>0$ or $p_1\le p_2<0$, and ($p_2\ge p_1>0$ or $p_2\le p_1<0$), then $f\in ghv(h,p_2,I)$.

• (d) For $I\subseteq [1,\mathrm{\infty })$, if f is monotone increasing (monotone decreasing), and $p_2\ge p_1>0$ or $p_2\le p_1<0$, and ($p_1\ge p_2>0$ or $p_1\le p_2<0$), then $f\in ghv(h,p_2,I)$.

Proof (a) Setting $g(p)={(\alpha x^p+(1-\alpha )y^p)}^{\frac{1}{p}}$, we have

$$g^{\prime }(p)=\frac{1}{p}{(\alpha x^p+(1-\alpha )y^p)}^{\frac{1}{p}-1}(\alpha x^{p}ln(x)+(1-\alpha )y^{p}ln(y)).$$

When $p>0$ and $x,y\in (0,1]$, we have $g^{\prime }(p)<0$, and so $g(p_2)\le g(p_1)$. We then obtain

$$f(g(p_2))\le f(g(p_1))\le h(\alpha )f(x)+(1-\alpha )f(y),$$

since f is monotone increasing and $f\in ghx(h,p_1,I)$. Therefore, we get $f\in ghx(h,p_2,I)$.

The results of (b), (c), and (d) follow by similar arguments as above. □

Property 4 Let f and g be similarly ordered functions on I, i.e.,

$$(f(x)-f(y))(g(x)-g(y))\ge 0$$

(2.2)

for all $x,y\in I$. If $f\in ghx(h_1,p,I)$, $g\in ghx(h_2,p,I)$, and $h(\alpha )+h(1-\alpha )\le c$ for all $\alpha \in (0,1)$, where $h(t)=max(h_1(t),h_2(t))$ and c is a fixed positive number, then the product fg belongs to $ghx(ch,p,I)$. Similarly, let f and g be oppositely ordered, i.e.,

$$(f(x)-f(y))(g(x)-g(y))\le 0$$

for all $x,y\in I$. If $f\in ghv(h_1,p,I)$, $g\in ghv(h_2,p,I)$, and $h(\alpha )+h(1-\alpha )\ge c$ for all $\alpha \in (0,1)$, where $h(t)=min(h_1(t),h_2(t))$ and c is a fixed positive number, then the product fg belongs to $ghv(ch,p,I)$.

Proof We only give a proof for the first part, since the result of the second part of this theorem follows by a similar argument. By (2.2), we have

$$f(x)g(x)+f(y)g(y)\ge f(x)g(y)+f(y)g(x).$$

Let α and β be positive numbers such that $\alpha +\beta =1$. We then obtain

$$\begin{array}{rcl}fg\left({[\alpha x^p+\beta y^p]}^{\frac{1}{p}}\right)& \le & (h_1(\alpha )f(x)+h_1(\beta )f(y))(h_2(\alpha )g(x)+h_2(\beta )g(y))\\ \le & h^2(\alpha )fg(x)+h(\alpha )h(\beta )f(x)g(y)+h(\alpha )h(\beta )f(y)g(x)+h^2(\beta )fg(y)\\ \le & h^2(\alpha )fg(x)+h(\alpha )h(\beta )f(x)g(x)+h(\alpha )h(\beta )f(y)g(y)+h^2(\beta )fg(y)\\ =& (h(\alpha )+h(\beta ))(h(\alpha )fg(x)+h(\beta )fg(y))\\ \le & ch(\alpha )fg(x)+ch(\beta )fg(y),\end{array}$$

which completes the proof. □

Definition 7 [9]

A function $h:I\to R$ is called a super-multiplicative function if

$$h(xy)\ge h(x)h(y)$$

(2.3)

for all $x,y\in J$.

If the inequality sign in (2.3) is reversed, then h is said to be a sub-multiplicative function, and if the equality holds in (2.3), then h is called a multiplicative function.

Example 2 Let $h(x)=c{e}^x$. If $c=1$, then h is a multiplicative function. If $c>1$, then h is a sub-multiplicative function, and if $0<c<1$, then h is a super-multiplicative function.

Property 5 Let I be an interval such that $0\in I$. We then have the following.

• (a) If $f\in ghx(h,p,I)$, $f(0)=0$, and h is super-multiplicative, then the inequality

  $$f\left({[\alpha x^p+\beta y^p]}^{\frac{1}{p}}\right)\le h(\alpha )f(x)+h(\beta )f(y)$$

  (2.4)

  holds for all $x,y\in I$ and all $\alpha ,\beta >0$ such that $\alpha +\beta \le 1$.

• (b) Let h be a non-negative function with $h(\alpha )<\frac{1}{2}$ for some $\alpha \in (0,\frac{1}{2})$. If f is a non-negative function satisfying (2.4) for all $x,y\in I$ and all $\alpha ,\beta >0$ with $\alpha +\beta \le 1$, then $f(0)=0$.

• (c) If $f\in ghv(h,p,I)$, $f(0)=0$, and h is sub-multiplicative, then the inequality

  $$f\left({[\alpha x^p+\beta y^p]}^{\frac{1}{p}}\right)\ge h(\alpha )f(x)+h(\beta )f(y)$$

  (2.5)

  holds for all $x,y\in I$ and all $\alpha ,\beta >0$ such that $\alpha +\beta \le 1$.

• (d) Let h be a non-negative function with $h(\alpha )>\frac{1}{2}$ for some $\alpha \in (0,\frac{1}{2})$. If f is a non-negative function satisfying (2.5) for all $x,y\in I$ and all $\alpha ,\beta >0$ with $\alpha +\beta \le 1$, then $f(0)=0$.

Proof (a) Let $\alpha ,\beta >0$, $\alpha +\beta =\gamma <1$, and let a and b be numbers such that $a=\frac{\alpha }{\gamma }$ and $b=\frac{\beta }{\gamma }$. We then have $a+b=1$ and

$$\begin{array}{rcl}f\left({[\alpha x^p+\beta y^p]}^{\frac{1}{p}}\right)& =& f\left({[a\gamma x^p+b\gamma y^p]}^{\frac{1}{p}}\right)\\ \le & h(a)f\left({\gamma }^{\frac{1}{p}}x\right)+h(b)f\left({\gamma }^{\frac{1}{p}}y\right)\\ =& h(a)f\left({[\gamma x^p+(1-\gamma )0^p]}^{\frac{1}{p}}\right)+h(b)f\left({[\gamma y^p+(1-\gamma )0^p]}^{\frac{1}{p}}\right)\\ \le & h(a)h(\gamma )f(x)+h(a)h(1-\gamma )f(0)\\ +h(b)h(\gamma )f(y)+h(b)h(1-\gamma )f(0)\\ =& h(a)h(\gamma )f(x)+h(b)h(\gamma )f(y)\\ \le & h(a\gamma )f(x)+h(b\gamma )f(y)=h(\alpha )f(x)+h(\beta )f(y).\end{array}$$

(b) If $f(0)\ne 0$, then $f(0)>0$. Setting $x=y=0$ in (2.4), we get

$$f(0)\le h(\alpha )f(0)+h(\beta )f(0).$$

By setting $\alpha =\beta$, where $\alpha \in (0,\frac{1}{2})$, and dividing both sides of the inequality above by $f(0)$, we obtain $2h(\alpha )\ge 1$ for all $\alpha \in (0,\frac{1}{2})$, which is a contradiction to the assumption $h(\alpha )<\frac{1}{2}$ for some $\alpha \in (0,\frac{1}{2})$, and so $f(0)=0$.

The results of (c) and (d) follow by using similar arguments as above, and so we omit the proofs here. □

Corollary 1 Let $h_s(x)=x^s$, where $s,x>0$, and let $0\in I$. For all $f\in ghx(h_s,p,I)$, inequality (2.4) holds for all $\alpha ,\beta >0$ with $\alpha +\beta \le 1$ if and only if $f(0)=0$. For all $f\in ghv(h_s,p,I)$, inequality (2.5) holds for all $\alpha ,\beta >0$ with $\alpha +\beta \le 1$ if and only if $f(0)=0$.

Proof Let $\alpha ,\beta >0$, $\alpha +\beta =\gamma <1$, and let a and b be positive numbers such that $a=\frac{\alpha }{\gamma }$ and $b=\frac{\beta }{\gamma }$. We then have $a+b=1$ and

$$\begin{array}{rcl}f\left({[\alpha x^p+\beta y^p]}^{\frac{1}{p}}\right)& =& f\left({[a\gamma x^p+b\gamma y^p]}^{\frac{1}{p}}\right)\\ \le & a^{s}f\left({\gamma }^{\frac{1}{p}}x\right)+b^{s}f\left({\gamma }^{\frac{1}{p}}y\right)\\ =& a^{s}f\left({[\gamma x^p+(1-\gamma )0^p]}^{\frac{1}{p}}\right)+b^{s}f\left({[\gamma y^p+(1-\gamma )0^p]}^{\frac{1}{p}}\right)\\ \le & a^s{\gamma }^{s}f(x)+a^s{(1-\gamma )}^{s}f(0)+b^s{\gamma }^{s}f(y)+b^s{(1-\gamma )}^{s}f(0)\\ =& a^s{\gamma }^{s}f(x)+b^s{\gamma }^{s}f(y)\\ =& {\alpha }^{s}f(x)+{\beta }^{s}f(y).\end{array}$$

Setting $x=y=\alpha =\beta =0$ in (2.4), we get $f(0)\le 0$, while $f(0)\ge 0$ by the definition of the $(p,h)$-convex function, and hence $f(0)=0$. □

Property 6 Suppose that $h_i:J_i\to (0,\mathrm{\infty })$, $i=1,2$, are functions such that $h_2(J_2)\subseteq J_1$ and $h_2(\alpha )+h_2(1-\alpha )\le 1$ for all $\alpha \in (0,1)$, and that $f:I_1\to [0,\mathrm{\infty })$ and $g:I_2\to [0,\mathrm{\infty })$ are functions with $g(I_2)\subseteq I_1$, $0\in I_1$, and $f(0)=0$.

If $h_1$ is a super-multiplicative function, $f\in SX(h_1,I_1)$, and f is increasing (decreasing) and $g\in ghx(h_2,p,I_2)$ ($g\in ghv(h_2,p,I_2)$), then the composite function $f\circ g$ belongs to $ghx(h_1\circ h_2,p,I_2)$. If $h_1$ is a sub-multiplicative function, $f\in SV(h_1,I_1)$, and f is increasing (decreasing) and $g\in ghv(h_2,p,I_2)$ ($g\in ghx(h_2,p,I_2)$), then the composite function $f\circ g$ belongs to $ghv(h_1\circ h_2,p,I_2)$.

Proof If $g\in ghx(h_2,p,I_2)$ and f is an increasing function, then we have

$$(f\circ g)\left({[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\right)\le f(h_2(\alpha )g(x)+h_2(1-\alpha )g(y))$$

for all $x,y\in I_2$ and $\alpha \in (0,1)$. Using Property 5(a) with $p=1$, we obtain

$$f(h_2(\alpha )g(x)+h_2(1-\alpha )g(y))\le h_1(h_2(\alpha ))f(g(x))+h_1(h_2(1-\alpha ))f(g(y)),$$

which implies that $f\circ g$ belongs to $ghx(h_1\circ h_2,p,I_2)$. □

If f is a convex or concave function, then we may give a similar statement on the composite function of f and g.

Property 7 Let $f:I_1\to [0,\mathrm{\infty })$ and $g:I_2\to [0,\mathrm{\infty })$ be functions with $g(I_2)\subseteq I_1$. If the function f is convex and increasing (decreasing), and $g\in ghx(h,p,I_2)$ ($g\in ghv(h,p,I_2)$) with $h(\alpha )+h(1-\alpha )=1$ for $\alpha \in (0,1)$, then $f\circ g$ belongs to $ghx(h,p,I_2)$. If the function f is concave and increasing (decreasing), and $g\in ghv(h,p,I_2)$ ($g\in ghx(h,p,I_2)$) with $h(\alpha )+h(1-\alpha )=1$ for $\alpha \in (0,1)$, then $f\circ g$ belongs to $ghv(h,p,I_2)$.

Proof If $g\in ghx(h,p,I_2)$ and f is an increasing function, we then have

$$(f\circ g)\left({[\alpha x^p+(1-\alpha )y^p]}^{\frac{1}{p}}\right)\le f(h(\alpha )g(x)+h(1-\alpha )g(y))$$

for all $x,y\in I_2$ and $\alpha \in (0,1)$. Since $h(\alpha )+h(1-\alpha )=1$ and f is convex, we obtain

$$f(h(\alpha )g(x)+h(1-\alpha )g(y))\le h(\alpha )f(g(x))+h(1-\alpha )f(g(y)),$$

which implies that $f\circ g$ belongs to $ghx(h,p,I_2)$. □

3 Schur-type inequalities

In this section, we establish Schur-type inequalities of $(p,h)$-convex functions.

Theorem 1 Let $h:J\to R$ be a non-negative super-multiplicative function and let $f:I\to R$ be a function such that $f\in ghx(h,p,I)$. Then for all $x_1,x_2,x_3\in I$ such that $x_1<x_2<x_3$ and $x_3^p-x_1^p,x_3^p-x_2^p,x_2^p-x_1^p\in J$, the following inequality holds:

$$h(x_3^p-x_2^p)f(x_1)-h(x_3^p-x_1^p)f(x_2)+h(x_2^p-x_1^p)f(x_3)\ge 0.$$

(3.1)

If the function h is sub-multiplicative and $f\in ghv(h,p,I)$, then the inequality sign in (3.1) is reversed.

Proof Let $f\in ghx(h,p,I)$ and let $x_1,x_2,x_3\in I$ be the numbers stated in this theorem. Then one can easily see that

$$\frac{x_3^p-x_2^p}{x_3^p-x_1^p},\frac{x_2^p-x_1^p}{x_3^p-x_1^p}\in (0,1)\subseteq J\text{and}\frac{x_3^p-x_2^p}{x_3^p-x_1^p}+\frac{x_2^p-x_1^p}{x_3^p-x_1^p}=1.$$

We also have

$$h(x_3^p-x_2^p)=h\left(\frac{x_3^p-x_2^p}{x_3^p-x_1^p}(x_3^p-x_1^p)\right)\ge h\left(\frac{x_3^p-x_2^p}{x_3^p-x_1^p}\right)h(x_3^p-x_1^p)$$

and

$$h(x_2^p-x_1^p)\ge h\left(\frac{x_2^p-x_1^p}{x_3^p-x_1^p}\right)h(x_3^p-x_1^p).$$

Setting $\alpha =\frac{x_3^p-x_2^p}{x_3^p-x_1^p}$, $x=x_1$, and $y=x_3$ in (2.1), we have $x_2^p=\alpha x^p+(1-\alpha )y^p$ and

$$\begin{array}{rl}f(x_2)& \le h\left(\frac{x_3^p-x_2^p}{x_3^p-x_1^p}\right)f(x_1)+h\left(\frac{x_2^p-x_1^p}{x_3^p-x_1^p}\right)f(x_3)\\ \le \frac{h(x_3^p-x_2^p)}{h(x_3^p-x_1^p)}f(x_1)+\frac{h(x_2^p-x_1^p)}{h(x_3^p-x_1^p)}f(x_3).\end{array}$$

(3.2)

Assuming $h(x_3^p-x_1^p)>0$ and multiplying both sides of the inequality above by $h(x_3^p-x_1^p)$, we obtain inequality (3.1). □

Remark 3 In fact, if $f(x)=x^{\lambda }$, $\lambda \in R$, $h(x)=h_{-1}(x)=\frac{1}{x}$, $p=1$, and $x_1,x_2,x_3\in I=(0,1)$, then inequality (3.1) gives the Schur inequality, see [[10], p.177].

The following corollary gives a Schur-type inequality for the $(p,h)$-convex function.

Corollary 2 If $f:I=(0,1)\to I$ belongs to the class $ghx(h_{-k},p,I)$ and $h_{-k}=\frac{1}{x^k}$, then we have the inequality

$$\begin{array}{r}f(x_1){(x_3^p-x_1^p)}^k{(x_2^p-x_1^p)}^k-f(x_2){(x_3^p-x_2^p)}^k{(x_2^p-x_1^p)}^k\\ +f(x_3){(x_3^p-x_1^p)}^k{(x_3^p-x_2^p)}^k\ge 0\end{array}$$

(3.3)

for all $x_1,x_2,x_3\in I$ with $x_1<x_2<x_3$. If $f\in ghv(h_{-k},p,I)$, then the inequality sign in (3.3) is reversed. If $k=1$, $p=1$, and $f(x)=x^{\lambda }$, $\lambda \in R$, then $f\in ghx(h_{-1},1,I)$ and inequality (3.3) gives the Schur inequality.

4 Jensen-type inequalities

In this section, we introduce some Jensen-type inequalities of $(p,h)$-convex functions.

Theorem 2 Let $w_1,\dots ,w_n$ be positive real numbers with $n\ge 2$. If h is a non-negative super-multiplicative function and if $f\in ghx(h,p,I)$ and $x_1,\dots ,x_n\in I$, then we have the inequality

$$f\left({\left[\frac{1}{W_n}\sum _{i=1}^n{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)\le \sum _{i=1}^{n}h\left(\frac{w_i}{W_n}\right)f(x_i),\mathit{\text{where }}{W}_n=\sum _{i=1}^n{w}_i.$$

(4.1)

If h is sub-multiplicative and $f\in ghv(h,p,I)$, then the inequality sign in (4.1) is reversed.

Proof When $n=2$, inequality (4.1) holds by (2.1) with $\alpha =\frac{w_1}{W_2}$. Assuming inequality (4.1) holds for $n-1$, we obtain

$$\begin{array}{rcl}f\left({\left[\frac{1}{W_n}\sum _{i=1}^n{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)& =& f\left({[\frac{w_n}{W_n}{x}_n^p+\sum _{i=1}^{n-1}\frac{w_i}{W_n}{x}_i^p]}^{\frac{1}{p}}\right)\\ =& f\left({[\frac{w_n}{W_n}{x}_n^p+\frac{W_{n-1}}{W_n}\sum _{i=1}^{n-1}\frac{w_i}{W_{n-1}}{x}_i^p]}^{\frac{1}{p}}\right)\\ \le & h\left(\frac{w_n}{W_n}\right)f(x_n)+h\left(\frac{W_{n-1}}{W_n}\right)f\left({\left[\sum _{i=1}^{n-1}\frac{w_i}{W_{n-1}}{x}_i^p\right]}^{\frac{1}{p}}\right)\\ \le & h\left(\frac{w_n}{W_n}\right)f(x_n)+h\left(\frac{W_{n-1}}{W_n}\right)\sum _{i=1}^{n-1}h\left(\frac{w_i}{W_{n-1}}\right)f(x_i)\\ \le & \sum _{i=1}^{n}h\left(\frac{w_i}{W_n}\right)f(x_i),\end{array}$$

and, hence, the result follows by mathematical induction. □

Remark 4 For $h(\alpha )=\alpha$ and $p=1$, inequality (4.1) becomes the classical Jensen inequality.

Theorem 3 Let $w_1,\dots ,w_n$ be positive real numbers and let $(m,M)$ be an interval in I. If $h:(0,\mathrm{\infty })\to R$ is a non-negative super-multiplicative function and $f\in ghx(h,p,I)$, then for all $x_1,\dots ,x_n\in (m,M)$ we have the inequality

$$\begin{array}{rl}\sum _{i=1}^{n}h\left(\frac{w_i}{W_n}\right)f(x_i)\le & f(m)\sum _{i=1}^{n}h\left(\frac{w_i}{W_n}\right)h\left(\frac{M^p-x_i^p}{M^p-m^p}\right)\\ +f(M)\sum _{i=1}^{n}h\left(\frac{w_i}{W_n}\right)h\left(\frac{x_i^p-m^p}{M^p-m^p}\right).\end{array}$$

(4.2)

If h is a non-negative sub-multiplicative function and $f\in ghv(h,p,I)$, then the inequality sign in (4.2) is reversed.

Proof Setting $x_1=m$, $x_2=x_i$, and $x_3=M$ in (3.2), we get the inequalities

$$f(x_i)\le h\left(\frac{M^p-x_i^p}{M^p-m^p}\right)f(m)+h\left(\frac{x_i^p-m^p}{M^p-m^p}\right)f(M),i=1,\dots ,n.$$

Multiplying both sides of the above inequality with $h(\frac{w_i}{W_n})$ and adding all inequalities side by side for $i=1,\dots ,n$, we obtain (4.2). □

Let K be a finite nonempty set of positive integers and let F be an index set function defined by

$$F(K)=h(W_K)f\left({\left[\frac{1}{W_K}\sum _{i\in K}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)-\sum _{i\in K}h(w_i)f(x_i),\text{where }{W}_K=\sum _{i\in K}{w}_i.$$

Theorem 4 Let $h:(0,\mathrm{\infty })\to R$ be a non-negative function, and let M and K be finite nonempty sets of positive integers such that $M\cap K=\mathrm{\varnothing }$. If h is super-multiplicative and $f:I\to R$ belongs to the class $ghx(h,p,I)$, then for $w_i>0$, $x_i\in I$, $i\in M\cup K$ we have the inequality

$$F(M\cup K)\le F(M)+F(K).$$

(4.3)

If h is sub-multiplicative and $f\in ghv(h,p,I)$, then the inequality sign in (4.3) is reversed.

Proof Setting $x={[\frac{1}{W_M}{\sum }_{i\in M}{w}_i{x}_i^p]}^{\frac{1}{p}}$, $y={[\frac{1}{W_K}{\sum }_{i\in K}{w}_i{x}_i^p]}^{\frac{1}{p}}$, and $\alpha =\frac{W_M}{W_{M\cup K}}$ in (2.1), we obtain the inequality

$$\begin{array}{r}f\left({\left[\frac{1}{W_{M\cup K}}\sum _{i\in M\cup K}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)\\ \le h\left(\frac{W_M}{W_{M\cup K}}\right)f\left({\left[\frac{1}{W_M}\sum _{i\in M}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)+h\left(\frac{W_K}{W_{M\cup K}}\right)f\left({\left[\frac{1}{W_K}\sum _{i\in K}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right).\end{array}$$

Multiplying both sides of the above inequality with $h(W_{M\cup K})$, we get the inequality

$$\begin{array}{r}h(W_{M\cup K})f\left({\left[\frac{1}{W_{M\cup K}}\sum _{i\in M\cup K}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)\\ \le h(W_M)f\left({\left[\frac{1}{W_M}\sum _{i\in M}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right)+h(W_K)f\left({\left[\frac{1}{W_K}\sum _{i\in K}{w}_i{x}_i^p\right]}^{\frac{1}{p}}\right).\end{array}$$

Subtracting ${\sum }_{i\in M\cup K}h(w_i)f(x_i)$ from both sides of the inequality above and using the identity ${\sum }_{i\in M\cup K}h(w_i)f(x_i)={\sum }_{i\in M}h(w_i)f(x_i)+{\sum }_{i\in K}h(w_i)f(x_i)$, we obtain (4.3). □

A simple consequence of Theorem 4 is stated in the following corollary without proof.

Corollary 3 Let $h:(0,\mathrm{\infty })\to R$ be a non-negative super-multiplicative function. If $w_i>0$, $i=1,\dots ,n$, and $M_k=\{1,\dots ,K\}$, then for $f\in ghx(h,p,I)$ we have

$$F(M_n)\le F(M_{n-1})\le \cdots \le F(M_2)\le 0$$

(4.4)

and

$$F(M_n)\le \underset{1\le i<j\le n}{min}\{h(w_i+w_j)f\left({\left[\frac{w_i{x}_i^p+w_j{x}_j^p}{w_i+w_j}\right]}^{\frac{1}{p}}\right)-h(w_i)f(x_i)-h(w_j)f(x_j)\}.$$

(4.5)

If h is sub-multiplicative and $f\in ghv(h,p,I)$, then the inequality signs in (4.4) and (4.5) are reversed, and min is replaced with max.

Remark 5 Some results obtained from Theorem 4 and Corollary 3 are given in [[11], p.7], when $h(\alpha )=\alpha$, $p=1$, and h is a convex or concave function.

5 Hadamard-type inequalities

In this section, we give some Hadamard-type inequalities of $(p,h)$-convex functions.

Theorem 5 If $f\in ghx(h,p,I)\cap L_1([a,b])$ for $a,b\in I$ with $a<b$, then we have

$$\frac{1}{2h(\frac{1}{2})}f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)\le \frac{p}{b^p-a^p}{\int }_a^b{x}^{p-1}f(x)dx\le (f(a)+f(b)){\int }_0^{1}h(t)dt.$$

(5.1)

Proof Setting $x^p=\frac{y-a}{b-a}{b}^p+\frac{b-y}{b-a}{a}^p$, we get

$$\frac{p}{b^p-a^p}{\int }_a^b{x}^{p-1}f(x)dx=\frac{1}{b-a}{\int }_a^{b}f\left({[\frac{y-a}{b-a}{b}^p+\frac{b-y}{b-a}{a}^p]}^{\frac{1}{p}}\right)dy.$$

By using inequality (2.1) we obtain

$$f\left({[\frac{y-a}{b-a}{b}^p+\frac{b-y}{b-a}{a}^p]}^{\frac{1}{p}}\right)\le h\left(\frac{y-a}{b-a}\right)f(b)+h\left(\frac{b-y}{b-a}\right)f(a),$$

and hence, by integrating the above inequality over $[a,b]$, we have

$$\begin{array}{rcl}{\int }_a^{b}f\left({[\frac{y-a}{b-a}{b}^p+\frac{b-y}{b-a}{a}^p]}^{\frac{1}{p}}\right)dy& \le & f(b){\int }_a^{b}h\left(\frac{y-a}{b-a}\right)dy+f(a){\int }_a^{b}h\left(\frac{b-y}{b-a}\right)dy\\ =& (b-a)(f(a)+f(b)){\int }_0^{1}h(t)dt,\end{array}$$

which gives the second inequality.

Setting $y=\frac{1}{2}(a+b)+t$, we obtain

$$\begin{array}{r}{\int }_{-\frac{1}{2}(b-a)}^{\frac{1}{2}(b-a)}f\left({[\frac{1}{2}(a^p+b^p)+\frac{b^p-a^p}{b-a}t]}^{\frac{1}{p}}\right)dt\\ ={\int }_0^{\frac{1}{2}(b-a)}f\left({[\frac{1}{2}(a^p+b^p)+\frac{b^p-a^p}{b-a}t]}^{\frac{1}{p}}\right)dt\\ +{\int }_0^{\frac{1}{2}(b-a)}f\left({[\frac{1}{2}(a^p+b^p)-\frac{b^p-a^p}{b-a}t]}^{\frac{1}{p}}\right)dt\\ \ge \frac{1}{h(1/2)}{\int }_0^{\frac{1}{2}(b-a)}f\left({\left[\frac{1}{2}(a^p+b^p)\right]}^{\frac{1}{p}}\right)dt=\frac{b-a}{2h(1/2)}f\left({\left[\frac{1}{2}(a^p+b^p)\right]}^{\frac{1}{p}}\right),\end{array}$$

and, hence, the first inequality follows. □

Remark 6 If $h(\alpha )=\alpha$ and $p=1$, then inequality (5.1) gives the classical Hadamard inequality.

Theorem 6 Suppose that f and g are functions such that $f\in ghx(h_1,p,I)$, $g\in ghx(h_2,p,I)$, $fg\in L_1([a,b])$, and $h_1{h}_2\in L_1([0,1])$ with $a,b\in I$ and $a<b$. We then have

$$\begin{array}{rl}\frac{p}{b^p-a^p}{\int }_a^b{x}^{p-1}f(x)g(x)dx\le & M(a,b){\int }_0^1{h}_1(t)h_2(t)dt\\ +N(a,b){\int }_0^1{h}_1(t)h_2(1-t)dt,\end{array}$$

(5.2)

where $M(a,b)=f(a)g(a)+f(b)g(b)$ and $N(a,b)=f(a)g(b)+f(b)g(a)$.

Proof Since $f\in ghx(h_1,p,I)$ and $g\in ghx(h_2,p,I)$, we have

$$\begin{array}{c}f\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)\le h_1(t)f(a)+h_1(1-t)f(b),\hfill \\ g\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)\le h_2(t)g(a)+h_2(1-t)g(b)\hfill \end{array}$$

for all $t\in [0,1]$. Because f and g are non-negative, we get the inequality

$$\begin{array}{r}f\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)g\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)\\ \le h_1(t)h_2(t)f(a)g(a)+h_1(1-t)h_2(t)f(b)g(a)+h_1(t)h_2(1-t)f(a)g(b)\\ +h_1(1-t)h_2(1-t)f(b)g(b).\end{array}$$

Integrating both sides of the above inequality over $(0,1)$, we obtain the inequality

$$\begin{array}{r}{\int }_0^{1}f\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)g\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)dt\\ \le f(a)g(a){\int }_0^1{h}_1(t)h_2(t)dt+f(b)g(a){\int }_0^1{h}_1(1-t)h_2(t)dt\\ +f(a)g(b){\int }_0^1{h}_1(t)h_2(1-t)dt+f(b)g(b){\int }_0^1{h}_1(1-t)h_2(1-t)dt.\end{array}$$

Setting $x={[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}$, we get

$$\frac{p}{b^p-a^p}{\int }_a^b{x}^{p-1}f(x)g(x)dx\le M(a,b){\int }_0^1{h}_1(t)h_2(t)dt+N(a,b){\int }_0^1{h}_1(t)h_2(1-t)dt.$$

 □

Theorem 7 Let $f\in ghx(h_1,p,I)$, $g\in ghx(h_2,p,I)$ be functions such that $fg\in L_1([a,b])$ and $h_1{h}_2\in L_1([0,1])$, and let $a,b\in I$ with $a<b$. We then have

$$\begin{array}{r}\frac{1}{2h_1(\frac{1}{2})h_2(\frac{1}{2})}f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)-\frac{p}{b^p-a^p}{\int }_a^b{x}^{p-1}f(x)g(x)dx\\ \le M(a,b){\int }_0^1{h}_1(t)h_2(1-t)dt+N(a,b){\int }_0^1{h}_1(t)h_2(t)dt.\end{array}$$

(5.3)

Proof Since $\frac{a^p+b^p}{2}=\frac{t{a}^p+(1-t)b^p}{2}+\frac{(1-t)a^p+t{b}^p}{2}$, we have

$$\begin{array}{r}f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)\\ =f\left({[\frac{t{a}^p+(1-t)b^p}{2}+\frac{(1-t)a^p+t{b}^p}{2}]}^{\frac{1}{p}}\right)g\left({[\frac{t{a}^p+(1-t)b^p}{2}+\frac{(1-t)a^p+t{b}^p}{2}]}^{\frac{1}{p}}\right)\\ =h_1\left(\frac{1}{2}\right)[f\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)+f\left({[(1-t)a^p+t{b}^p]}^{\frac{1}{p}}\right)]\\ \times h_2\left(\frac{1}{2}\right)[g\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)+g\left({[(1-t)a^p+t{b}^p]}^{\frac{1}{p}}\right)]\\ \le h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\left[f\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)g\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)\right]\\ +h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\left[f\left({[(1-t)a^p+t{b}^p]}^{\frac{1}{p}}\right)g\left({[(1-t)a^p+t{b}^p]}^{\frac{1}{p}}\right)\right]\\ +h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)[h_1(t)f(a)+h_1(1-t)f(b)][h_2(1-t)g(a)+h_2(t)g(b)]\\ +h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)[h_1(1-t)f(a)+h_1(t)f(b)][h_2(t)g(a)+h_2(1-t)g(b)]\\ =h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\left[f\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)g\left({[t{a}^p+(1-t)b^p]}^{\frac{1}{p}}\right)\right]\\ +h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\left[f\left({[(1-t)a^p+t{b}^p]}^{\frac{1}{p}}\right)g\left({[(1-t)a^p+t{b}^p]}^{\frac{1}{p}}\right)\right]\\ +h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)[(h_1(t)h_2(1-t)+h_1(1-t)h_2(t))M(a,b)]\\ +h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)[(h_1(t)h_2(t)+h_1(1-t)h_2(1-t))N(a,b)].\end{array}$$

Integrating the above inequality over $[0,1]$, we obtain

$$\begin{array}{r}\frac{1}{2h_1(\frac{1}{2})h_2(\frac{1}{2})}f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)-\frac{p}{b^p-a^p}{\int }_a^b{x}^{p-1}f(x)g(x)dx\\ \le M(a,b){\int }_0^1{h}_1(t)h_2(1-t)dt+N(a,b){\int }_0^1{h}_1(t)h_2(t)dt.\end{array}$$

 □

Theorem 8 Let $f\in ghx(h_1,p,I)$ and $g\in ghx(h_2,p,I)$ be functions such that $fg\in L_1([a,b])$, $h_1{h}_2\in L_1([0,1])$, and let $a,b\in I$ with $a<b$. We then have the inequality

$$\begin{array}{r}\frac{p^2}{2{(b^p-a^p)}^2}{\int }_a^b{\int }_a^b{\int }_0^1{x}^{p-1}{y}^{p-1}f\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)dxdydt\\ \le \frac{p}{b^p-a^p}{\int }_0^1{h}_1(t)h_2(t)dt{\int }_a^b{x}^{p-1}f(x)g(x)dx\\ +{\int }_0^1{h}_1(t)dt{\int }_0^1{h}_2(t)dt{\int }_0^1{h}_1(t)h_2(1-t)dt[M(a,b)+N(a,b)].\end{array}$$

(5.4)

Proof Since $f\in ghx(h_1,p,I)$ and $g\in ghx(h_2,p,I)$, we have

$$\begin{array}{c}f\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)\le h_1(t)f(x)+h_1(1-t)f(y),\hfill \\ g\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)\le h_2(t)g(x)+h_2(1-t)g(y)\hfill \end{array}$$

for all $t\in [0,1]$. Because f and g are non-negative, we get the inequality

$$\begin{array}{r}f\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)\\ \le h_1(t)h_2(t)f(x)g(x)+h_1(1-t)h_2(t)f(y)g(x)+h_1(t)h_2(1-t)f(x)g(y)\\ +h_1(1-t)h_2(1-t)f(y)g(y).\end{array}$$

Multiplying both sides of the above inequality with $\frac{p^2{x}^{p-1}{y}^{p-1}}{{(b^p-a^p)}^2}$ and integrating the result over $[a,b]$ and $[0,1]$, we obtain the inequality

$$\begin{array}{r}\frac{p^2}{{(b^p-a^p)}^2}{\int }_a^b{\int }_a^b{\int }_0^1{x}^{p-1}{y}^{p-1}f\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)dxdydt\\ \le {\int }_0^1{h}_1(t)h_2(t)dt\left[\frac{p^2}{{(b^p-a^p)}^2}\right({\int }_a^b{x}^{p-1}f(x)g(x)dx{\int }_a^b{y}^{p-1}dy\\ +{\int }_a^b{y}^{p-1}f(y)g(y)dy{\int }_a^b{x}^{p-1}dx\left)\right]\\ +2{\int }_0^1{h}_1(t)h_2(1-t)dt[\frac{p^2}{{(b^p-a^p)}^2}{\int }_a^b{x}^{p-1}f(x)dx{\int }_a^b{y}^{p-1}f(y)dy].\end{array}$$

By (5.1), we have the inequality

$$\begin{array}{r}\frac{p^2}{2{(b^p-a^p)}^2}{\int }_a^b{\int }_a^b{\int }_0^1{x}^{p-1}{y}^{p-1}f\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)y^p]}^{\frac{1}{p}}\right)dxdydt\\ \le \frac{p}{b^p-a^p}{\int }_0^1{h}_1(t)h_2(t)dt{\int }_a^b{x}^{p-1}f(x)g(x)dx\\ +{\int }_0^1{h}_1(t)dt{\int }_0^1{h}_2(t)dt{\int }_0^1{h}_1(t)h_2(1-t)dt[M(a,b)+N(a,b)].\end{array}$$

 □

Theorem 9 Let $f\in ghx(h_1,p,I)$, $g\in ghx(h_2,p,I)$ be functions such that $fg\in L_1([a,b])$, $h_1{h}_2\in L_1([0,1])$, and let $a,b\in I$ with $a<b$. We then have the inequality

$$\begin{array}{r}{\int }_a^b{\int }_0^1{x}^{p-1}f\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)dtdx\\ \le {\int }_0^1{h}_1(t)h_2(t)dt{\int }_a^b{x}^{p-1}f(x)g(x)dx+\frac{b^p-a^p}{p}[M(a,b)+N(a,b)]\\ \times [h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right){\int }_0^1{h}_1(t)h_2(t)dt\\ +[h_1\left(\frac{1}{2}\right){\int }_0^1{h}_2(t)dt+h_2\left(\frac{1}{2}\right){\int }_0^1{h}_1(t)dt]{\int }_0^1{h}_1(t)h_2(1-t)dt].\end{array}$$

(5.5)

Proof Since $f\in ghx(h_1,p,I)$ and $g\in ghx(h_2,p,I)$, we have the inequalities

$$\begin{array}{c}f\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)\le h_1(t)f(x)+h_1(1-t)f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right),\hfill \\ g\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)\le h_2(t)g(x)+h_2(1-t)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)\hfill \end{array}$$

for all $t\in [0,1]$. Because f and g are non-negative, we get the inequality

$$\begin{array}{r}f\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)\\ \le h_1(t)h_2(t)f(x)g(x)+h_1(1-t)h_2(t)f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)g(x)\\ +h_1(t)h_2(1-t)f(x)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)\\ +h_1(1-t)h_2(1-t)f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right).\end{array}$$

Multiplying both sides of the inequality above with $x^{p-1}$ and integrating the result over $[a,b]$ and $[0,1]$, we obtain

$$\begin{array}{r}{\int }_a^b{\int }_0^1{x}^{p-1}f\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)dtdx\\ \le {\int }_0^1{h}_1(t)h_2(t)dt[{\int }_a^b{x}^{p-1}f(x)g(x)dx+\frac{b^p-a^p}{p}f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right)]\\ +{\int }_0^1{h}_1(t)h_2(1-t)dt[g\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right){\int }_a^b{x}^{p-1}f(x)dx\\ +f\left({\left[\frac{a^p+b^p}{2}\right]}^{\frac{1}{p}}\right){\int }_a^b{x}^{p-1}g(x)dx].\end{array}$$

By inequality (5.1), we have

$$\begin{array}{r}{\int }_a^b{\int }_0^1{x}^{p-1}f\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)g\left({[t{x}^p+(1-t)\frac{a^p+b^p}{2}]}^{\frac{1}{p}}\right)dtdx\\ \le {\int }_0^1{h}_1(t)h_2(t)dt[{\int }_a^b{x}^{p-1}f(x)g(x)dx\\ +\frac{b^p-a^p}{p}{h}_1\left(\frac{1}{2}\right)(f(a)+f(b))h_2\left(\frac{1}{2}\right)(g(a)+g(b))]\\ +{\int }_0^1{h}_1(t)h_2(1-t)dt\frac{b^p-a^p}{p}{h}_2\left(\frac{1}{2}\right)(f(a)+f(b))(g(a)+g(b)){\int }_0^1{h}_1(t)dt\\ +{\int }_0^1{h}_1(t)h_2(1-t)dt\frac{b^p-a^p}{p}{h}_1\left(\frac{1}{2}\right)(f(a)+f(b))(g(a)+g(b)){\int }_0^1{h}_2(t)dt\\ ={\int }_0^1{h}_1(t)h_2(t)dt{\int }_a^b{x}^{p-1}f(x)g(x)dx+\frac{b^p-a^p}{p}[M(a,b)+N(a,b)]\\ \times [h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right){\int }_0^1{h}_1(t)h_2(t)dt\\ +[h_1\left(\frac{1}{2}\right){\int }_0^1{h}_2(t)dt+h_2\left(\frac{1}{2}\right){\int }_0^1{h}_1(t)dt]{\int }_0^1{h}_1(t)h_2(1-t)dt].\end{array}$$

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