Some inequalities for $(h,m)$-convex functions

Abstract

In the paper, the authors give some inequalities of Jensen type and Popoviciu type for $(h,m)$-convex functions and apply these inequalities to special means.

MSC: 26A51, 26D15, 26E60.

1 Introduction

The following definition is well known in the literature.

Definition 1 A function $f:I\subseteq \mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })\to \mathbb{R}$ is said to be convex if

$$f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$$

(1)

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

We cite the following inequalities for convex functions.

Theorem 1 ([[1], p.6])

If f is a convex function on I and $x_1,x_2,x_3\in I$, then

$$\begin{array}{r}f(x_1)+f(x_2)+f(x_3)+f\left(\frac{x_1+x_2+x_3}{3}\right)\\ \ge \frac{4}{3}[f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+f\left(\frac{x_3+x_1}{2}\right)].\end{array}$$

(2)

Theorem 2 ([[2], Popoviciu inequality])

If f is a convex function on I and $x_1,x_2,\dots ,x_n\in I$ with $n\ge 3$, then

$$\sum _{i=1}^{n}f(x_i)+\frac{n}{n-2}f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{2}{n-2}\sum _{i<j}f\left(\frac{x_i+x_j}{2}\right).$$

(3)

Theorem 3 ([[2], Generalized Popoviciu inequality])

If f is a convex function on I and $a_1,a_2,\dots ,a_n\in I$ for $n\ge 3$, then

$$(n-1)\sum _{i=1}^{n}f(b_i)\le n(n-2)f(a)+\sum _{i=1}^{n}f(a_i),$$

(4)

where $a=\frac{1}{n}{\sum }_{i=1}^n{a}_i$ and $b_i=\frac{na-a_i}{n-1}$ for $i=1,2,\dots ,n$.

The above inequalities were generalized as follows.

Theorem 4 ([3])

If f is a convex function on I and $x_1,x_2,\dots ,x_n\in I$ for $n\ge 3$, then

$$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{k=1}^n{x}_k\right)\ge \frac{n-1}{n}\sum _{i=1}^{n}f\left(\frac{x_i+x_{i+1}}{2}\right)$$

(5)

and

$$(n-1)\sum _{i=1}^{n}f(b_i)\le n[\sum _{i=1}^{n}f(a_i)-f(a)],$$

(6)

where $x_{n+1}=x_1$, $a=\frac{1}{n}{\sum }_{i=1}^n{a}_i$, and $b_i=\frac{na-a_i}{n-1}$ for $i=1,2,\dots ,n$.

Definition 2 ([4])

Let $s\in (0,1]$. A function $f:{\mathbb{R}}_0=[0,\mathrm{\infty })\to {\mathbb{R}}_0$ is said to be s-convex in the second sense if

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

(7)

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

The following inequalities for s-convex functions were established.

Theorem 5 ([[5], Theorem 4.2])

If f is nonnegative and s-convex in the second sense on I and if $x_1,x_2,\dots ,x_n\in I$ for $n\ge 3$, then

$$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{2^{s-1}(n^s-1)}{n}\sum _{i=1}^{n}f\left(\frac{x_i+x_{i+1}}{2}\right),$$

(8)

where $x_1=x_{n+1}$.

Theorem 6 ([[5], Theorem 4.4])

If f is nonnegative and s-convex in the second sense on I and $a_1,a_2,\dots ,a_n\in I$ for $n\ge 3$, then

$$(n^s-1)\sum _{i=1}^n{b}_i\le n^s[\sum _{i=1}^{n}f(a_i)-f(a)],$$

(9)

where $a=\frac{1}{n}{\sum }_{i=1}^n{a}_i$ and $b_i=\frac{na-a_i}{n-1}$ for $i=1,2,\dots ,n$.

The concept of h-convex functions below was innovated as follows.

Definition 3 ([[6], Definition 4])

Let $I,J\subseteq \mathbb{R}$ be intervals, $(0,1)\subseteq J$, and $h:J\to {\mathbb{R}}_0$ be a nonnegative function. A function $f:I\to {\mathbb{R}}_0$ is called h-convex, or as we say, f belongs to the class $SX(h,I)$, if f is nonnegative and

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

(10)

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

Definition 4 ([[6], Section 3])

A function $h:J\subseteq \mathbb{R}$ is said to be a super-multiplicative on an interval J if

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

(11)

is valid for all $x,y\in J$. If the inequality (11) reverses, then f is said to be a sub-multiplicative function on J.

The following inequalities were established for $f\in SX(h,I)$.

Theorem 7 ([[7], Theorem 6])

Let $w_1,\dots ,w_n$ for $n\ge 2$ be positive real numbers. If h is a nonnegative and super-multiplicative function and if $f\in SX(h,I)$ and $x_1,\dots ,x_n\in I$, then

$$f\left(\frac{1}{W_n}\sum _{i=1}^n{w}_i{x}_i\right)\le \sum _{i=1}^{n}h\left(\frac{w_i}{W_n}\right)f(x_i),$$

(12)

where $W_n={\sum }_{i=1}^n{w}_i$. If h is sub-multiplicative and $f\in SV(h,I)$, then the inequality (12) is reversed.

Theorem 8 ([[8], Theorem 11])

Let h be a nonnegative and super-multiplicative function. If $f\in SX(h,I)$ and $x_1,\dots ,x_n\in I$, then

$$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{2h(1/2)}\sum _{i=1}^{n}f\left(\frac{x_i+x_{i+1}}{2}\right),$$

(13)

where $x_{n+1}=x_1$. The inequality (13) is reversed if $f\in SV(h,I)$.

Theorem 9 ([[8], Theorem 12])

Let h be a nonnegative and super-multiplicative function. If $f\in SX(h,I)$ and $x_1,\dots ,x_n\in I$, then

$$[1-h\left(\frac{1}{n}\right)]\sum _{i=1}^{n}f(b_i)\le (n-1)h\left(\frac{1}{n-1}\right)[\sum _{i=1}^{n}f(a_i)-f(a)],$$

(14)

where $a=\frac{1}{n}{\sum }_{i=1}^n{a}_i$ and $b_i=\frac{na-a_i}{n-1}$ for $i=1,2,\dots ,n$ and $n\ge 3$. The inequality (14) is reversed if $f\in SV(h,I)$.

Two new kinds of convex functions were introduced as follows.

Definition 5 ([9])

For $f:[0,b]\to \mathbb{R}$ and $m\in (0,1]$, if

$$f(tx+m(1-t)y)\le tf(x)+m(1-t)f(y)$$

(15)

is valid for all $x,y\in [0,b]$ and $t\in [0,1]$, then we say that $f(x)$ is an m-convex function on $[0,b]$.

Definition 6 ([10])

Let $J\subseteq \mathbb{R}$ be an interval, $(0,1)\subseteq J$, $h:J\to \mathbb{R}$ be a nonnegative function. We say that $f:[0,b]\to \mathbb{R}$ is an $(h,m)$-convex function, or say, f belongs to the class $SMX((h,m),[0,b])$, if f is nonnegative and, for all $x,y\in [0,b]$ and $t\in [0,1]$ and for some $m\in (0,1]$, we have

$$f(tx+m(1-t)y)\le h(t)f(x)+mh(1-t)f(y).$$

(16)

If the inequality (16) is reversed, then f is said to be $(h,m)$-concave and denoted by $f\in SMV((h,m),[0,b])$.

Recently the h- and $(h,m)$-convex functions were generalized and some properties and inequalities for them were obtained in [11, 12].

The aim of this paper is to find some inequalities of Jensen type and Popoviciu type for $(h,m)$-convex functions.

2 Inequalities of Jensen type and Popoviciu type

Now we are in a position to establish some inequalities of Jensen type and Popoviciu type for $(h,m)$-convex functions.

Theorem 10 Let $h:[0,1]\to {\mathbb{R}}_0$ be a super-multiplicative function and $m\in (0,1]$. If $f\in SMX((h,m),[0,b])$, then for all $x_i\in [0,b]$ and $w_i>0$ with $i=1,2,\dots ,n$ and $n\ge 2$, we have

$$f\left(\frac{1}{W_n}\sum _{i=1}^n{m}^{i-1}{w}_i{x}_i\right)\le \sum _{i=1}^n{m}^{i-1}h\left(\frac{w_i}{W_n}\right)f(x_i),$$

(17)

where $W_n={\sum }_{i=1}^n{w}_i$.

If h is sub-multiplicative and $f\in SMV((h,m),[0,b])$, then the inequality (17) is reversed.

Proof Assume that $w_i^{\prime }=\frac{w_i}{W_n}$ for $i=1,2,\dots ,n$.

When $n=2$, taking $t=w_1^{\prime }$ and $1-t=w_2^{\prime }$ in Definition 6 gives the inequality (17) clearly.

Suppose that the inequality (17) holds for $n=k$, i.e.,

$$f\left(\sum _{i=1}^k{m}^{i-1}{w}_i^{\prime }{x}_i\right)\le \sum _{i=1}^k{m}^{i-1}h\left(w_i^{\prime }\right)f(x_i).$$

(18)

When $n=k+1$, letting ${\mathrm{\Delta }}_k={\sum }_{i=2}^{k+1}{w}_i^{\prime }$ and making use of (18) result in

$$\begin{array}{rl}f\left(\sum _{i=1}^{k+1}{m}^{i-1}{w}_i^{\prime }{x}_i\right)& =f(w_1^{\prime }{x}_1+m{\mathrm{\Delta }}_k\sum _{i=2}^{k+1}{m}^{i-2}\frac{w_i^{\prime }}{{\mathrm{\Delta }}_k}{x}_i)\\ \le h\left(w_1^{\prime }\right)f(x_1)+mh({\mathrm{\Delta }}_k)f\left(\sum _{i=2}^{k+1}{m}^{i-2}\frac{w_i^{\prime }}{{\mathrm{\Delta }}_k}{x}_i\right)\\ \le h\left(w_1^{\prime }\right)f(x_1)+mh({\mathrm{\Delta }}_k)\sum _{i=2}^{k+1}{m}^{i-2}h\left(\frac{w_i^{\prime }}{{\mathrm{\Delta }}_k}\right)f(x_i).\end{array}$$

Since h is a super-multiplicative function, it follows that

$$h({\mathrm{\Delta }}_k)h\left(\frac{w_i^{\prime }}{{\mathrm{\Delta }}_k}\right)\le h\left(w_i^{\prime }\right)$$

for $i=1,2,\dots ,n$. Namely, when $n=k+1$, the inequality (17) holds. By induction, Theorem 10 is proved. □

Corollary 1 Under the conditions of Theorem  10,

• 1. if $W_n=1$, we have

  $$f\left(\sum _{i=1}^n{m}^{i-1}{w}_i{x}_i\right)\le \sum _{i=1}^n{m}^{i-1}h(w_i)f(x_i);$$

  (19)

• 2. if $w_1=w_2=\cdots =w_n$, we have

  $$f\left(\frac{1}{n}\sum _{i=1}^n{m}^{i-1}{w}_i{x}_i\right)\le h\left(\frac{1}{n}\right)\sum _{i=1}^n{m}^{i-1}f(x_i);$$

  (20)

• 3. if h is sub-multiplicative and $f\in SMV((h,m),[0,b])$, then the inequalities (19) and (20) are reversed.

Corollary 2 For $m\in (0,1]$ and $s\in (0,1]$, the assertion $f\in SMX((t^s,m),[0,b])$ is valid if and only if for all $x_i\in [0,b]$ and $w_i>0$ with $i=1,2,\dots ,n$ and $n\ge 2$

$$f\left(\frac{1}{W_n}\sum _{i=1}^n{m}^{i-1}{x}_i\right)\le \sum _{i=1}^n{m}^{i-1}{\left(\frac{w_i}{W_n}\right)}^{s}f(x_i),$$

(21)

where $W_n={\sum }_{i=1}^n{w}_i$.

Corollary 3 Under the conditions of Corollary  1, if $h(t)=t^s$ for $s\in (0,1]$, then

$$f\left(\frac{1}{n}\sum _{i=1}^n{m}^{i-1}{x}_i\right)\le \frac{1}{n^s}\sum _{i=1}^n{m}^{i-1}f(x_i).$$

(22)

If $f\in SMV((h,m),[0,b])$, then the inequality (22) is reversed.

Theorem 11 Let $h:[0,1]\to {\mathbb{R}}_0$ be a super-multiplicative function, $m\in (0,1]$, and $n\ge 2$. If $f\in SMX((h,m),[0,\frac{b}{m^{n-1}}])$, then for all $x_i\in [0,b]$ and $w_i>0$ with $i=1,2,\dots ,n$,

$$f\left(\frac{1}{W_n}\sum _{i=1}^n{w}_i{x}_i\right)\le \sum _{i=1}^n{m}^{i-1}h\left(\frac{w_i}{W_n}\right)f\left(\frac{x_i}{m^{i-1}}\right),$$

(23)

where $W_n={\sum }_{i=1}^n{w}_i$.

If h is sub-multiplicative and $f\in SMV((h,m),[0,\frac{b}{m^{n-1}}])$, then the inequality (23) is reversed.

Proof Putting $y_i=\frac{x_i}{m^{i-1}}$ for $i=1,2,\dots ,n$, then from inequality (17), we have

$$\begin{array}{rl}f\left(\frac{1}{W_n}\sum _{i=1}^n{w}_i{x}_i\right)& =f\left(\frac{1}{W_n}\sum _{i=1}^n{m}^{i-1}{w}_i{y}_i\right)\\ \le \sum _{i=1}^n{m}^{i-1}h\left(\frac{w_i}{W_n}\right)f(y_i)=\sum _{i=1}^n{m}^{i-1}h\left(\frac{w_i}{W_n}\right)f\left(\frac{x_i}{m^{i-1}}\right).\end{array}$$

The proof of Theorem 11 is complete. □

Corollary 4 For $m\in (0,1]$, $s\in (0,1]$, and $n\ge 2$, the assertion $f\in SMX((t^s,m),[0,\frac{b}{m^{n-1}}])$ is valid if and only if for all $x_i\in [0,b]$ and $w_i>0$ with $i=1,2,\dots ,n$ the inequality

$$f\left(\frac{1}{W_n}\sum _{i=1}^n{w}_i{x}_i\right)\le \sum _{i=1}^n{m}^{i-1}{\left(\frac{w_i}{W_n}\right)}^{s}f\left(\frac{x_i}{m^{i-1}}\right)$$

(24)

is valid, where $W_n={\sum }_{i=1}^n{w}_i$.

Corollary 5 Under the conditions of Theorem  11,

• 1. if $W_n=1$, then

  $$f\left(\sum _{i=1}^n{w}_i{x}_i\right)\le \sum _{i=1}^n{m}^{i-1}h(w_i)f\left(\frac{x_i}{m^{i-1}}\right);$$

  (25)

• 2. if $w_1=w_2=\cdots =w_n$, then

  $$f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\le h\left(\frac{1}{n}\right)\sum _{i=1}^n{m}^{i-1}f\left(\frac{x_i}{m^{i-1}}\right);$$

  (26)

• 3. if h is sub-multiplicative and $f\in SMV((h,m),[0,\frac{b}{m^{n-1}}])$, then the inequalities (25) and (26) are reversed.

Corollary 6 Under the conditions of Corollary  5,

• 1. if $h(t)=t^s$ for $s\in (0,1]$, then

  $$f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\le \frac{1}{n^s}\sum _{i=1}^n{m}^{i-1}f\left(\frac{x_i}{m^{i-1}}\right);$$

  (27)

• 2. if $f\in SMV((h,m),[0,\frac{b}{m^{n-1}}])$, then the inequality (27) is reversed.

Theorem 12 Let $h:[0,1]\to [0,1]$ be a super-multiplicative function and let $m\in (0,1]$ and $n\ge 3$. If $f\in SMX((h,m),[0,b])$, then for all $x_i\in [0,b]$ with $i=1,2,\dots ,n$ and $2\le k\le n$, we have

$$\begin{array}{r}\sum _{i=1}^{n}f(x_i)-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)\\ \ge \frac{1-h(1/n)}{h(1/k)}{\left(\sum _{j=0}^{k-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_j\right),\end{array}$$

(28)

where $x_{n+1}=x_1$, …, $x_{2n-1}=x_{n-1}$.

If h is sub-multiplicative and $f\in SMV((h,m),[0,b])$, then the inequality (28) is reversed.

Proof By using the inequality (20), we have

$$\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_j\right)\le h\left(\frac{1}{k}\right)\sum _{i=1}^n\sum _{j=i}^{k+i-1}{m}^{j-i}f(x_j)=h\left(\frac{1}{k}\right)\left(\sum _{j=0}^{k-1}{m}^j\right)\sum _{i=1}^{n}f(x_i)$$

(29)

and

$$\begin{array}{rl}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)& \le h\left(\frac{1}{n}\right)\sum _{i=1}^n\sum _{j=i}^{n+i-1}{m}^{j-i}f(x_j)\\ =h\left(\frac{1}{n}\right)\left(\sum _{j=0}^{n-1}{m}^j\right)\sum _{i=1}^{n}f(x_i).\end{array}$$

(30)

If $h(\frac{1}{n})=1$, then, from the inequality (30), the inequality (28) holds. If $h(\frac{1}{n})\le 1$, it is easy to see that

$$\begin{array}{r}\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_j\right)\\ \le h\left(\frac{1}{k}\right)\left(\sum _{j=0}^{k-1}{m}^j\right)\sum _{i=1}^{n}f(x_i)\\ =\frac{h(1/k)}{1-h(1/n)}\left(\sum _{j=0}^{k-1}{m}^j\right)[\sum _{i=1}^{n}f(x_i)-h\left(\frac{1}{n}\right)\sum _{i=1}^{n}f(x_i)]\\ \le \frac{h(1/k)}{1-h(1/n)}\left(\sum _{j=0}^{k-1}{m}^j\right)[\sum _{i=1}^{n}f(x_i)-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)].\end{array}$$

The proof of Theorem 12 is complete. □

Corollary 7 Under the conditions of Theorem  12, let ${\overline{x}}_n=\frac{1}{n}{\sum }_{i=1}^n{x}_i$.

• 1. When $m=1$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{kh(1/k)}\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{x}_j\right).$$

  (31)

• 2. When $m=1$ and $k=2$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{2h(1/2)}\sum _{i=1}^{n}f\left(\frac{x_i+x_{i+1}}{2}\right).$$

  (32)

• 3. When $m=1$ and $k=n-1$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{(n-1)h(1/(n-1))}\sum _{i=1}^{n}f\left(\frac{n{\overline{x}}_n-x_i}{n-1}\right).$$

  (33)

• 4. If h is sub-multiplicative and $f\in SMV((h,m),[0,b])$, then the inequalities (31) to (33) are reversed.

Remark 1 The inequality (14) can be deduced from applying (33) to $a_i=x_i$ for $i=1,2,\dots ,n$, $a=\frac{1}{n}{\sum }_{i=1}^n{a}_i$, and $b_i=\frac{na-a_i}{n-1}$ for $i=1,2,\dots ,n$.

Corollary 8 Under the conditions of Theorem  12,

• 1. if $h(t)=t^s$ for $s\in (0,1]$, then

  $$\begin{array}{r}\sum _{i=1}^{n}f(x_i)-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)\\ \ge \frac{k^s(n^s-1)}{n^s}{\left(\sum _{j=0}^{k-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_j\right);\end{array}$$

  (34)

• 2. if $h(t)=t^s$ for $s\in (0,1]$ and $m=1$, then

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{k^{s-1}(n^s-1)}{n^s}\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{x}_j\right);$$

  (35)

• 3. if $h(t)=t$ and $m=1$, then

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{n-1}{n}\sum _{i=1}^{n}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{x}_j\right);$$

  (36)

• 4. if $f\in SMV((h,m),[0,b])$, then the inequalities (34) to (36) are reversed.

Theorem 13 Let $h:[0,1]\to [0,1]$ be a super-multiplicative function and let $m\in (0,1]$ and $n\ge 3$. If $f\in SMX((h,m),[0,\frac{b}{m^{n-1}}])$, then for all $x_i\in [0,b]$ with $i=1,2,\dots ,n$ and $2\le k\le n$ and for ${\ell }_1,\dots ,{\ell }_k\in \mathbb{N}$, we have

$$\begin{array}{r}\sum _{i=1}^{n}f(x_i)-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)\\ \ge \frac{1-h(1/n)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)h(1/k)}{\left(\sum _{j=0}^{k-1}{m}^j\right)}^{-1}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^{k}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_{{\ell }_j}\right),\end{array}$$

(37)

where ${\ell }_{k+1}={\ell }_1$, …, ${\ell }_{2k-1}={\ell }_{k-1}$.

If h is sub-multiplicative and $f\in SMV((h,m),[0,b])$, then the inequality (37) is reversed.

Proof By the inequality (20), we have

$$\begin{array}{r}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^{k}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_{{\ell }_j}\right)\\ \le h\left(\frac{1}{k}\right)\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^k\sum _{j=i}^{k+i-1}{m}^{j-i}f(x_{{\ell }_j})\\ =h\left(\frac{1}{k}\right)\left(\sum _{j=0}^{k-1}{m}^j\right)\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^{k}f(x_{{\ell }_j})\\ =\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)h\left(\frac{1}{k}\right)\left(\sum _{j=0}^{k-1}{m}^j\right)\sum _{i=1}^{n}f(x_i).\end{array}$$

(38)

If $h(\frac{1}{n})=1$, then, from the inequality (30), the inequality (28) holds. If $h(\frac{1}{n})\le 1$, using (38) and (30), we have

$$\begin{array}{r}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^{k}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_{{\ell }_j}\right)\\ \le \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)h\left(\frac{1}{k}\right)\left(\sum _{j=0}^{k-1}{m}^j\right)\sum _{i=1}^{n}f(x_i)\\ =\frac{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)h(1/k)}{1-h(1/n)}\left(\sum _{j=0}^{k-1}{m}^j\right)[\sum _{i=1}^{n}f(x_i)-h\left(\frac{1}{n}\right)\sum _{i=1}^{n}f(x_i)]\\ \le \frac{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)h(1/k)}{1-h(1/n)}\left(\sum _{j=0}^{k-1}{m}^j\right)[\sum _{i=1}^{n}f(x_i)-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)].\end{array}$$

The proof of Theorem 13 is complete. □

Corollary 9 Under the conditions of Theorem  13, let ${\overline{x}}_n=\frac{1}{n}{\sum }_{i=1}^n{x}_i$.

• 1. When $m=1$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)h(1/k)}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}f\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right).$$

  (39)

• 2. When $m=1$ and $k=2$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{(n-1)h(1/2)}\sum _{1\le i<j\le n}f\left(\frac{x_i+x_j}{2}\right).$$

  (40)

• 3. When $m=1$ and $k=n-1$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{1-h(1/n)}{(n-1)h(1/(n-1))}\sum _{i=1}^n\left(\frac{n{\overline{x}}_n-x_i}{n-1}\right).$$

  (41)

• 4. If h is sub-multiplicative and $f\in SMV((h,m),[0,b])$, then the inequalities (39) to (41) are reversed.

Corollary 10 Under the conditions of Theorem  13,

• 1. if $h(t)=t^s$ for $s\in (0,1]$, then

  $$\begin{array}{r}\sum _{i=1}^{n}f(x_i)-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^{n}f\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)\\ \ge \frac{k^s(n^s-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n^s}{\left(\sum _{j=0}^{k-1}{m}^j\right)}^{-1}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^{k}f\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_{{\ell }_j}\right);\end{array}$$

  (42)

• 2. if $m=1$ and $h(t)=t^s$ for $s\in (0,1]$, we have

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{k^s(n^s-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n^s}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}f\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right);$$

  (43)

• 3. if $m=1$ and $h(t)=t$, then

  $$\sum _{i=1}^{n}f(x_i)-f\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\ge \frac{k(n-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}f\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right);$$

  (44)

• 4. if $f\in SMV((h,m),[0,b])$, then the inequalities (42) to (44) are reversed.

3 Applications to means

In what follows we will apply the theorems and corollaries in the above section to establish inequalities for some special means.

For $r\in \mathbb{R}$, $r\ne 0$, and $m,s\in (0,1]$, let $f(x)=x^r$ for $x\in {\mathbb{R}}_{+}$ and $h(t)=t^s$ for $t\in [0,1]$. Then

• 1. if $r\ge 1$ and $0<m\le 1$, or if $r<0$ and $m=1$, we have

  $${(tx+m(1-t)y)}^r\le t{x}^r+(1-t){(my)}^r\le t^s{x}^r+m{(1-t)}^s{y}^r$$

for $x,y\in {\mathbb{R}}_{+}$;

• 2. if $0<r\le 1$, $0<m\le 1$, and $s=1$, we have

  $${(tx+m(1-t)y)}^r\ge t{x}^r+(1-t){(my)}^r\ge t{x}^r+m(1-t)y^r$$

for $x,y\in {\mathbb{R}}_{+}$.

Using Definition 6 yields the following:

• 1. if $r\ge 1$ and $0<m\le 1$, or if $r<0$ and $m=1$, the function $f(x)=x^r\in SMX((t^s,m),{\mathbb{R}}_{+})$;

• 2. if $0<r\le 1$, $0<m\le 1$, and $s=1$, the function $f(x)=x^r\in SMV((t,m),{\mathbb{R}}_{+})$.

By virtue of Corollary 10, we obtain the following results.

Theorem 14 Let $n\ge 3$ and $x_i\in {\mathbb{R}}_{+}$ for $i=1,2,\dots ,n$, let $r\in \mathbb{R}$ with $r\ne 0$ and $m,s\in (0,1]$, and let ${\ell }_1,\dots ,{\ell }_k\in \mathbb{N}$ for $2\le k\le n$ and ${\ell }_{k+1}={\ell }_1$, …, ${\ell }_{2k-1}={\ell }_{k-1}$.

• 1. If $r\ge 1$ and $0<m\le 1$, or if $r<0$ and $m=1$, then we have

  $$\begin{array}{r}\sum _{i=1}^n{x}_i^r-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^n{\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)}^r\\ \ge \frac{k^s(n^s-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n^s}{\left(\sum _{j=0}^{k-1}{m}^j\right)}^{-1}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^k{\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_{{\ell }_j}\right)}^r;\end{array}$$

  (45)

• 2. if $r\ge 1$ or $r<0$ and if $m=1$, we have

  $$\sum _{i=1}^n{x}_i^r-{\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^r\ge \frac{k^s(n^s-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n^s}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}{\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right)}^r;$$

  (46)

• 3. if $r\ge 1$ or $r<0$ and if $m=s=1$, then

  $$\sum _{i=1}^n{x}_i^r-{\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^r\ge \frac{k(n-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}{\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right)}^r;$$

  (47)

• 4. if $0<r\le 1$, $0<m\le 1$, and $s=1$, then the inequality (47) are reversed.

Corollary 11 Under the conditions of Theorem  14, when ${\ell }_{k+1}={\ell }_1$, …, ${\ell }_{2k-1}={\ell }_{k-1}$, we have the following conclusions.

• 1. If $r=2$, we have

  $$\begin{array}{r}\sum _{i=1}^n{x}_i^2-{\left(\sum _{j=0}^{n-1}{m}^j\right)}^{-1}\sum _{i=1}^n{\left(\frac{1}{n}\sum _{j=i}^{n+i-1}{m}^{j-i}{x}_j\right)}^2\\ \ge \frac{k^s(n^s-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n^s}{\left(\sum _{j=0}^{k-1}{m}^j\right)}^{-1}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}\sum _{i=1}^k{\left(\frac{1}{k}\sum _{j=i}^{k+i-1}{m}^{j-i}{x}_{{\ell }_j}\right)}^2;\end{array}$$

  (48)

• 2. if $r=2$ and $m=1$, we have

  $$\sum _{i=1}^n{x}_i^2-{\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^2\ge \frac{k^s(n^s-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n^s}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}{\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right)}^2;$$

  (49)

• 3. if $r=2$ and $m=s=1$, then

  $$\sum _{i=1}^n{x}_i^2-{\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)}^2\ge \frac{k(n-1)}{\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)n}\sum _{1\le {\ell }_1<\cdots <{\ell }_k\le n}{\left(\frac{1}{k}\sum _{j=1}^k{x}_{{\ell }_j}\right)}^2.$$

  (50)