Generalizations of Several Inequalities Related to Multivariate Geometric Means

Abstract

In the paper, by some methods in the theory of majorization, the authors generalize several inequalities related to multivariate geometric means.

1. Introduction

Let $\mathit{x}=(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}_{+}^n={(0,\infty )}^n$ and $n\ge 2$. Then, the arithmetic and geometric means of n positive numbers $x_1,x_2,\dots ,x_n$ are defined by

$$A(\mathit{x})=\frac{1}{n}\sum _{i=1}^n{x}_i\mathrm{and}G(\mathit{x})={\left(\prod _{i=1}^n{x}_i\right)}^{1/n}.$$

In [1] (p. 208, 3.2.34), it was stated that

$$\prod _{i=1}^n(x_i+1)\ge {[G(\mathit{x})+1]}^n.$$

(1)

In the paper [2], Wang and Chen established that the inequality

$$\prod _{i=1}^n[{(x_i+1)}^p-1]\ge {[{(G(\mathit{x})+1)}^p-1]}^n$$

(2)

is valid for $p>1$. In the paper [3], Wang extended Inequality (2) as follows:

• if $p\ge 1$, then Inequality (2) is valid;
• if $0<p\le 1$, then Inequality (2) is reversed;
• if $p>0$, then

  $$\prod _{i=1}^n[{(x_i+1)}^p+1]\ge {[{(G(\mathit{x})+1)}^p+1]}^n;$$

  (3)
• if $p<0$ and $0<x_i<\frac{1}{|p|}$ for $1\le i\le n$, then Inequality (3) is reversed.

In [4] (p. 11), it was proven that

$$\sqrt[n]{\prod _{k=1}^n(a_k+b_k)}\ge \sqrt[n]{\prod _{k=1}^n{a}_k}+\sqrt[n]{\prod _{k=1}^n{b}_k},$$

(4)

where $a_k,b_k\ge 0$. When $a_k=x_k$ and $b_k=1$, Inequality (4) becomes (1). Inequality (4) is the Minkowski inequality of the product form.

We observe that the inequalities in (1)–(4) can be rearranged as

$$G(\mathit{x}+\mathbf{1})\ge G(\mathit{x})+1,G({(\mathit{x}+\mathbf{1})}^p\pm \mathbf{1})\ge {[G(\mathit{x})+1]}^p\pm 1$$

(5)

and

$$G(\mathit{a}+\mathit{b})>G(\mathit{a})+G(\mathit{b}),$$

(6)

where

$$\mathit{x}+\mathbf{1}=(x_1+1,x_2+1,\dots ,x_n+1)$$

and

$${(\mathit{x}+\mathbf{1})}^p\pm \mathbf{1}=({(x_1+1)}^p\pm 1,{(x_2+1)}^p\pm 1,\dots ,{(x_n+1)}^p\pm 1).$$

Inequality (6) reveals that the geometric mean $G(\mathit{x})$ is sub-additive. For information about the sub-additivity, please refer to [5,6,7,8,9,10,11,12] and the closely related references therein. The sub-additive property of the geometric mean $G(\mathit{x})$ can also be derived from the property that the geometric mean $G(\mathit{x})$ is a Bernstein function; see [13,14,15,16,17,18,19] and the closely related references therein.

In this paper, by some methods in the theory of majorization, we will generalize the above inequalities in (1)–(3), (5), and (6).

2. Definitions and Lemmas

Now, we recall some definitions and lemmas.

It is well known that a function $f(x_1,x_2,\dots ,x_n)$ of n variables is said to be symmetric if its value is unchanged for any permutation of its n variables $x_1,x_2,\dots ,x_n$.

Definition 1

([20,21]). Let $\mathit{x}=(x_1,x_2,\dots ,x_n)$ and $\mathit{y}=(y_1,y_2,\dots ,y_n)\in {\mathbb{R}}^n$.

1. 

If

$$\sum _{i=1}^k{x}_{[i]}\le \sum _{i=1}^k{y}_{[i]}and\sum _{i=1}^n{x}_{[i]}=\sum _{i=1}^n{y}_{[i]}$$

for $1\le k\le n-1$, then $\mathit{x}$ is said to be majorized by $\mathit{y}$ (in symbol $\mathit{x}\prec \mathit{y}$), where $x_{[1]}\ge x_{[2]}\ge \cdots \ge x_{[n]}$ and $y_{[1]}\ge y_{[2]}\ge \cdots \ge y_{[n]}$ are rearrangements of $\mathit{x}$ and $\mathit{y}$ in descending order.

2. 

For $\mathit{x},\mathit{y}\in \mathsf{\Omega }$ and $\lambda \in [0,1]$, if

$$\lambda \mathit{x}+(1-\lambda )\mathit{y}=(\lambda x_1+(1-\lambda )y_1,\lambda x_2+(1-\lambda )y_2,\dots ,\lambda x_n+(1-\lambda )y_n)\in \mathsf{\Omega },$$

then $\mathsf{\Omega }\subseteq {\mathbb{R}}^n$ is said to be a convex set.

Definition 2

([20,21]). Let $\mathsf{\Omega }\subseteq {\mathbb{R}}^n$ be a symmetric and convex set.

1. 

If $\mathit{x}\prec \mathit{y}$ on Ω implies $\phi (\mathit{x})\le \phi (\mathit{y})$, then we say that the function $\phi :\mathsf{\Omega }\to \mathbb{R}$ is Schur convex on Ω.

2. 

If $-\phi$ is a Schur convex function on Ω, then we say that φ is Schur concave on Ω.

Lemma 1

([20,21]). Let $\mathsf{\Omega }\subseteq {\mathbb{R}}^n$ be a symmetric and convex set with nonempty interior $\mathsf{\Omega }°$, and let $\phi :\mathsf{\Omega }\to \mathbb{R}$ be continuous on Ω and continuously differentiable on $\mathsf{\Omega }°$. If and only if φ is symmetric on Ω and

$$(x_1-x_2)\left(\frac{\partial \phi }{\partial x_1}-\frac{\partial \phi }{\partial x_2}\right)\ge 0,\mathit{x}\in \mathsf{\Omega }°,$$

the function φ is Schur convex on Ω.

Definition 3

([22] (pp. 64 and 107)). Let $\mathit{x},\mathit{y}\in \mathsf{\Omega }\subseteq {\mathbb{R}}_{+}^n$.

1. 

If $\left(x_1^{\lambda }{y}_1^{1-\lambda },x_2^{\lambda }{y}_2^{1-\lambda },\dots ,x_n^{\lambda }{y}_n^{1-\lambda }\right)\in \mathsf{\Omega }$ for $\mathit{x},\mathit{y}\in \mathsf{\Omega }$ and $\lambda \in [0,1]$, then Ω is called a geometrically-convex set.

2. 

If Ω is a geometrically-convex set and

$$(\ln x_1,ln{x}_2,\dots ,ln{x}_n)\prec (\ln y_1,ln{y}_2,\dots ,ln{y}_n)$$

implies $\phi (\mathit{x})\le \phi (\mathit{y})$ for any $\mathit{x},\mathit{y}\in \mathsf{\Omega }$, then $\phi :\mathsf{\Omega }\to {\mathbb{R}}_{+}$ is said to be a Schur geometrically-convex function on Ω.

3. 

If $-\phi$ is a Schur geometrically-convex function on Ω, then φ is said to be a Schur geometrically-concave function on Ω.

Lemma 2

([22] (p. 108)). Let $\mathsf{\Omega }\subseteq {\mathbb{R}}_{+}^n$ be a symmetric and geometrically-convex set with a nonempty interior $\mathsf{\Omega }°$, and let $\phi :\mathsf{\Omega }\to {\mathbb{R}}_{+}$ be continuous on Ω and differentiable in $\mathsf{\Omega }°$. If and only if φ is symmetric on Ω and

$$(ln{x}_1-ln{x}_2)\left(x_1\frac{\partial \phi }{\partial x_1}-x_2\frac{\partial \phi }{\partial x_2}\right)\ge 0,\mathit{x}\in \mathsf{\Omega }°,$$

the function φ is Schur geometrically-convex on Ω.

For more information on the Schur convexity and the Schur geometric convexity, please refer to the papers [23,24,25,26] and the monographs [20,22].

Lemma 3

(Bernoulli’s inequality [27,28]). For $0\ne x>-1$, if $\alpha \ge 1$ or $\alpha <0$, then

$${(1+x)}^{\alpha }\ge 1+\alpha x;$$

(7)

if $0<\alpha <1$, then Inequality (7) is reversed.

Lemma 4.

For $p\in \mathbb{R}\backslash \{0\}$ and $x\in {\mathbb{R}}_{+}$, define

$$h_1(x)=\frac{p{(x+1)}^{p-1}}{{(x+1)}^p+1},g_1(x)=\frac{px{(x+1)}^{p-1}}{{(x+1)}^p+1},h_2(x)=\frac{p{(x+1)}^{p-1}}{{(x+1)}^p-1},g_2(x)=\frac{px{(x+1)}^{p-1}}{{(x+1)}^p-1}.$$

1. 

If $p<0$, the function $h_1(x)$ is increasing on ${\mathbb{R}}_{+}$; if $0<p\le 2$, the function $h_1(x)$ is decreasing on ${\mathbb{R}}_{+}$; if $p>2$, the function $h_1(x)$ is increasing on $\left(0,{(p-1)}^{1/p}-1\right)$ and decreasing on $\left({(p-1)}^{1/p}-1,\infty \right)$.

2. 

If $p>0$, the function $g_1(x)$ is increasing on $\mathbb{R}$; if $p<0$, the function $g_1(x)$ is decreasing on $\left(0,\frac{1}{|p|}\right]$ and increasing on $\left[\frac{2}{|p|},\infty \right)$.

3. 

For $p\ne 0$, the function $h_2(x)$ is decreasing on ${\mathbb{R}}_{+}$.

4. 

If $p\ge 1$, the function $g_2(x)$ is increasing on ${\mathbb{R}}_{+}$; if $p<1$ and $p\ne 0$, the function $g_2(x)$ is decreasing on ${\mathbb{R}}_{+}$.

Proof. 

Straightforward computation gives

$$\begin{array}{cccc}\hfill {[h_1(x)]}^{\prime }=& \frac{p[p-1-{(x+1)}^p]}{{[{(x+1)}^p+1]}^2}{(x+1)}^{p-2},\hfill & \hfill {[g_1(x)]}^{\prime }& =\frac{p[1+px+{(x+1)}^p]}{{[{(x+1)}^p+1]}^2}{(x+1)}^{p-2},\hfill \\ \hfill [h_2{(x)]}^{\prime }=& \frac{p[1-p-{(x+1)}^p]}{{[{(x+1)}^p-1]}^2}{(x+1)}^{p-2},\hfill & \hfill {[g_2(x)]}^{\prime }& =\frac{p[{(x+1)}^p-1-px]}{{[{(x+1)}^p-1]}^2}{(x+1)}^{p-2}.\hfill \end{array}$$

If $p<0$ and $x\in {\mathbb{R}}_{+}$ or $p>2$ and $0<x\le {(p-1)}^{1/p}-1$, we have ${[h_1(x)]}^{\prime }\ge 0$; if $0<p\le 2$ and $x\in {\mathbb{R}}_{+}$ or $p>2$ and ${(p-1)}^{1/p}-1<x$, we obtain ${[h_1(x)]}^{\prime }\le 0$.

If $p>0$ and $x\in {\mathbb{R}}_{+}$, we acquire ${[g_1(x)]}^{\prime }\ge 0$; if $p<0$ and $0<x\le \frac{1}{|p|}$, we have ${[g_1(x)]}^{\prime }\le 0$; if $p<0$ and $x\ge \frac{2}{|p|}$, since ${(x+1)}^p<1$ and $1+px\le -1$, we acquire ${[g_1(x)]}^{\prime }\ge 0$.

If $p\in \mathbb{R}\backslash \{0\}$, we obtain ${[h_2(x)]}^{\prime }\le 0$ for $x\in {\mathbb{R}}_{+}$.

For $x\in {\mathbb{R}}_{+}$, by Lemma 3, if $p\ge 1$, we have ${(x+1)}^p\ge 1+px$, then ${[g_2(x)]}^{\prime }\ge 0$; if $0<p\le 1$, we have ${(x+1)}^p\le 1+px$, and so, ${[g_2(x)]}^{\prime }\le 0$; if $p<0$, we obtain ${(x+1)}^p\ge 1+px$, hence ${[g_2(x)]}^{\prime }\le 0$. The proof of Lemma 4 is complete. □

Let $\mathit{x}=(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}^n$ and $i_1,\dots ,i_k$ be positive integers. The elementary symmetric functions are defined by $E_0(\mathit{x})=1$,

$$E_k(\mathit{x})=E_k(x_1,x_2,\dots ,x_n)=\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k{x}_{i_j},1\le k\le n,$$

and $E_k(\mathit{x})=0$ for $k<0$ or $k>n$.

Lemma 5

(Newton’s inequality [20] (p. 134)). For $\mathit{x}=(x_1,x_2,\dots ,x_n)\in {\mathbb{R}}_{+}^n$ and $n\ge 2$, let ${\widehat{F}}_k(\mathit{x})=\frac{1}{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}{E}_k(\mathit{x})$ for $1\le k\le n-1$. Then

$${\left[{\widehat{F}}_k(\mathit{x})\right]}^2\ge {\widehat{F}}_{k-1}(\mathit{x}){\widehat{F}}_{k+1}(\mathit{x}).$$

3. Main Results

In this section, we will make use of the Schur convexity of the symmetric function

$$E_k({(\mathit{x}+\mathbf{1})}^p+\xi )=\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}+1)}^p+\xi ],2\le k\le n,n\ge 2$$

to generalize the inequalities in (1)–(3), (5) and (6), where $\mathit{x}\in {\mathbb{R}}_{+}^n$, the quantities $i_1,i_2,\dots ,i_k$ are positive integers, $\xi =0, \pm 1$, and $\xi =\mathbf{0},\pm \mathbf{1}$.

Our main results are Theorems 1–3 below.

Theorem 1.

Let $2\le k\le n$, $p\in \mathbb{R}\backslash \{0\}$, $i_1,\dots ,i_k\in \mathbb{N}$, and $a=\frac{(n-k)(p-1)}{k-1}$ for $p>1$.

1. 

If $2\le k\le n-1$, $0<p\le 1$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, or if $2\le k\le n-1$, $p>1$ with $a<1$, and $\mathit{x}\in {\left(0,a^{-1/p}-1\right]}^n$, or if $k=n$, $p>0$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[G(\mathit{x})+1]}^{kp}\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k{(x_{i_j}+1)}^p\le \left(\genfrac{}{}{0pt}{}{n}{k}\right){[A(\mathit{x})+1]}^{kp};$$

(8)

if $p<0$ and $\mathit{x}\in {\left(0,\frac{1}{|p|}\right]}^n$ or if $k=n$, $p<0$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then the double Inequality (8) is reversed.

2. 

If $2\le k\le n-1$, $p>1$ with $a>1$, and $\mathit{x}\in {\left(0,a^{1/p}-1\right]}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[A(\mathit{x})+1]}^{kp}\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k{(x_{i_j}+1)}^p.$$

3. 

If $2\le k\le n-1$, $p>0$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[G(\mathit{x})+1]}^{kp}\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k{(x_{i_j}+1)}^p;$$

(9)

if $p<0$ and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then Inequality (9) is reversed.

Proof. 

When $k=n$, we have

$$\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p)}{\partial x_1}=\frac{p}{(x_1+1)}{E}_n({(\mathit{x}+\mathbf{1})}^p).$$

From this, we obtain

$$(x_1-x_2)\left[\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p)}{\partial x_1}-\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p)}{\partial x_2}\right]=-\frac{p{(x_1-x_2)}^2}{(x_1+1)(x_2+1)}{E}_n({(\mathit{x}+\mathbf{1})}^p)\left\{\begin{array}{cc}\le 0,\hfill & p>0;\hfill \\ \ge 0,\hfill & p<0\hfill \end{array}\right.$$

and

$$(x_1-x_2)\left[x_1\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p)}{\partial x_1}-x_2\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p)}{\partial x_2}\right]=\frac{p{(x_1-x_2)}^2}{(x_1+1)(x_2+1)}{E}_n({(\mathit{x}+\mathbf{1})}^p)\left\{\begin{array}{cc}\ge 0,\hfill & p>0;\hfill \\ \le 0,\hfill & p<0.\hfill \end{array}\right.$$

Using Lemmas 1 and 2, we arrive at

• if $p>0$, then $E_n({(\mathit{x}+\mathbf{1})}^p)$ is Schur concave on ${\mathbb{R}}_{+}^n$;
• if $p<0$, then $E_n({(\mathit{x}+\mathbf{1})}^p)$ is Schur convex on ${\mathbb{R}}_{+}^n$;
• if $p>0$, then $E_n({(\mathit{x}+\mathbf{1})}^p)$ is Schur geometrically convex on ${\mathbb{R}}^n$;
• if $p<0$, the $E_n({(\mathit{x}+\mathbf{1})}^p)$ is Schur geometrically concave on ${\mathbb{R}}^n$.

Since

$$(A_n(\mathit{x}),A_n(\mathit{x}),\dots ,A_n(\mathit{x}))\prec \mathit{x}$$

(10)

and

$$(ln{G}_n(\mathit{x}),ln{G}_n(\mathit{x}),\dots ,ln{G}_n(\mathit{x}))\prec ln\mathit{x}$$

(11)

for $\mathit{x}\in {\mathbb{R}}_{+}^n$, applying Definitions 2 and 3, we obtain the double Inequality (8) for $k=n$.

When $2\le k\le n-1$, a direct differentiation yields

$$\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p)}{\partial x_1}=p{(x_1+1)}^{p-1}{(x_2+1)}^p{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)+p{(x_1+1)}^{p-1}{E}_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p),$$

where $\widehat{\mathit{x}}=(x_3,\dots ,x_n)$. We clearly see that

$$\begin{array}{cc}\hfill {\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p))& \triangleq (x_1-x_2)\left[\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p)}{\partial x_1}-\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p)}{\partial x_2}\right]\hfill \\ & =-p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\hfill \\ & +p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\hfill \end{array}$$

(12)

and

$$\begin{array}{c}(ln{x}_1-ln{x}_2)\left[x_1\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p)}{\partial x_1}-x_2\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p)}{\partial x_2}\right]\hfill \\ =p(ln{x}_1-ln{x}_2)(x_1-x_2){(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\hfill \\ +p(ln{x}_1-ln{x}_2)\left[x_1{(x_1+1)}^{p-1}-x_2{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p).\hfill \end{array}$$

(13)

Using Equation (13) and Lemma 2, we obtain that

• if $p>0$, then $E_k({(\mathit{x}+\mathbf{1})}^p)$ is Schur geometrically convex on ${\mathbb{R}}_{+}^n$;
• if $p<0$, then $E_k({(\mathit{x}+\mathbf{1})}^p)$ is Schur geometrically concave on ${\left(0,\frac{1}{|p|}\right]}^n$.

By Equation (12) and Lemma 1, we reveal that

• if $p<0$, then $E_k({(\mathit{x}+\mathbf{1})}^p)$ is Schur convex on ${\mathbb{R}}_{+}^n$;
• if $0<p\le 1$, then $E_k({(\mathit{x}+\mathbf{1})}^p)$ is Schur concave on ${\mathbb{R}}_{+}^n$;
• if $p>1$ and $\mathit{x}\in {\mathbb{R}}_{+}^n$,

  (a)

  when $x_1=x_2$, we have ${\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p))=0$;

  (b)

  when $x_1\ne x_2$,

  • by Lagrange’s mean value theorem, we have

    $$\frac{1}{{(x_2+1)}^{p-1}}-\frac{1}{{(x_1+1)}^{p-1}}=\frac{(p-1)(x_1-x_2)}{{(\xi +1)}^p}$$

    (14)

    for at least one interior point $\xi \in (min\{x_1,x_2\},max\{x_1,x_2\})$;
  • from Newton’s inequality, we obtain

    $$\begin{array}{cc}\hfill \frac{(n-k)(n-2)}{(k-1){\sum }_{i=3}^n{(x_i+1)}^{-p}}& =\frac{(n-k){\widehat{F}}_{n-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}{(k-1){\widehat{F}}_{n-3}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}\le \frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}\hfill \\ & \le \frac{(n-k){\widehat{F}}_1({(\widehat{\mathit{x}}+\mathbf{1})}^p)}{(k-1){\widehat{F}}_0({(\widehat{\mathit{x}}+\mathbf{1})}^p)}=\frac{n-k}{(k-1)(n-2)}\sum _{i=3}^n{(x_i+1)}^p.\hfill \end{array}$$

    (15)

Substituting (14) and (15) into Inequality (12) yields

$$\begin{array}{c}{\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p))=-p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\\ +p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\\ =p(x_1-x_2){(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\\ \times \left[-(x_1-x_2)+\left(\frac{1}{{(x_2+1)}^{p-1}}-\frac{1}{{(x_1+1)}^{p-1}}\right)\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}\right]\\ =p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\left[\frac{p-1}{{(\xi +1)}^p}\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}-1\right].\end{array}$$

(16)

If $a>1$ and $\mathit{x}\in {(0,a^{1/p}-1]}^n$, using $min\{x_1,x_2\}<\xi <max\{x_1,x_2\}$, we have

$$\frac{n-2}{{\sum }_{i=3}^n{(x_i+1)}^{-p}}>\frac{n-2}{{\sum }_{i=3}^n{(0+1)}^{-p}}=1$$

and

$$\frac{1}{{(\xi +1)}^p}\ge \frac{k-1}{(n-k)(p-1)}.$$

(17)

Consequently, the inequalities from (16)–(17) imply $\Delta (E_k({(\mathit{x}+\mathbf{1})}^p))\ge 0$ for $\mathit{x}\in {(0,a^{1/p}-1]}^n$.

If $a<1$ and $\mathit{x}\in {\left(0,a^{-1/p}-1\right]}^n$, we obtain

$$\begin{array}{c}{\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p))=p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\\ \times \left[-1+\frac{p-1}{{(\xi +1)}^p}\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)}\right]\\ \le p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\left[-1+\frac{(n-k)(p-1)}{(k-1)(n-2){(\xi +1)}^p}\sum _{i=3}^n{(x_i+1)}^p\right]\\ \le p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\\ \times \left\{-1+\frac{(n-k)(p-1)}{(k-1){(\xi +1)}^p}{\left[\left({\left(\frac{k-1}{(n-k)(p-1)}\right)}^{1/p}-1\right)+1\right]}^p\right\}\\ =p{(x_1-x_2)}^2{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p)\left[\frac{1}{{(\xi +1)}^p}-1\right]<0.\end{array}$$

It is easy to obtain that, if $p>1$ and $a>1$, then $E_k({(\mathit{x}+\mathbf{1})}^p)$ is Schur convex on ${\left(0,a^{1/p}-1\right]}^n$; if $p>1$ and $a<1$, then $E_k({(\mathit{x}+\mathbf{1})}^p)$ is Schur concave on ${\left(0,a^{-1/p}-1\right]}^n$.

Using (10) and by Definitions 2 and 3, the inequalities in (8) and (9) hold. The proof of Theorem 1 is complete. □

Theorem 2.

Let $2\le k\le n$ and $p\in \mathbb{R}\backslash \{0\}$, $i_1,\dots ,i_k\in \mathbb{N}$, and

$$b={\left[\frac{1}{2}\left(\sqrt{\frac{4(4n-3k-1)(p-1)}{k-1}+1}-1\right)\right]}^{1/p},1<p\le 2.$$

1. 

If $k=n$, $0<p\le 2$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, or if $k=n$, $p>2$, and $\mathit{x}\in {\left({(p-1)}^{1/p}-1,\infty \right)}^n$, or if $2\le k\le n-1$, $0<p\le 1$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(G(\mathit{x})+1)}^p+1]}^k\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}+1)}^p+1]\le \left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(A(\mathit{x})+1)}^p+1]}^k.$$

(18)

If $2\le k\le n$, $p<0$, and $\mathit{x}\in {\left(0,\frac{1}{|p|}\right]}^n$, then the double Inequality (18) is reversed.

2. 

If $2\le k\le n$, $p>2$, and $\mathit{x}\in {\left(0,{(p-1)}^{1/p}-1\right]}^n$, or if $2\le k\le n-1$, $1<p\le 2$, and $\mathit{x}\in {(0,b]}^n$, or if $2\le k\le n$, $p<0$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(A(\mathit{x})+1)}^p+1]}^k\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}+1)}^p+1].$$

(19)

3. 

If $2\le k\le n-1$, $p>0$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$ or if $2\le k\le n$, $p<0$, and $\mathit{x}\in {\left[\frac{2}{|p|},\infty \right)}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(G(\mathit{x})+1)}^p+1]}^k\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}+1)}^p+1].$$

(20)

4. 

If $k=n$, $p>2$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then

$${[{(G(\mathit{x})+1)}^p+1]}^n\le \prod _{i=1}^n\left[{(x_i+1)}^p+1\right].$$

(21)

Proof. 

When $k=n$, a direct differentiation yields

$$\begin{array}{cc}\hfill \frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_1}=& \frac{p{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}{E}_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1}).\hfill \end{array}$$

(22)

From Lemma 4, it follows that

$$\begin{array}{c}(x_1-x_2)\left[\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_1}-\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_2}\right]\\ =p(x_1-x_2)\left[\frac{{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}-\frac{{(x_2+1)}^{p-1}}{{(x_2+1)}^p+1}\right]E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})\\ \left\{\begin{array}{cc}\le 0,\hfill & 0<p\le 2,\mathit{x}\in {\mathbb{R}}_{+}^n;\hfill \\ \ge 0,\hfill & p>2,\mathit{x}\in {\left(0,{(p-1)}^{1/p}-1\right]}^n;\hfill \\ \le 0,\hfill & p>2,\mathit{x}\in {\left({(p-1)}^{1/p}-1,\infty \right)}^n;\hfill \\ \ge 0,\hfill & p<0,\mathit{x}\in {\mathbb{R}}_{+}^n\hfill \end{array}\right.\end{array}$$

and

$$\begin{array}{c}(ln{x}_1-ln{x}_2)\left[x_1\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_1}-x_2\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_2}\right]\\ =p(ln{x}_1-ln{x}_2)\left[\frac{x_1{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}-\frac{x_2{(x_2+1)}^{p-1}}{{(x_2+1)}^p+1}\right]E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})\\ \left\{\begin{array}{cc}\ge 0,\hfill & p>0,\mathit{x}\in {\mathbb{R}}_{+}^n,\hfill \\ \le 0,\hfill & p<0,\mathit{x}\in {\left(0,\frac{1}{|p|}\right]}^n,\hfill \\ \ge 0,\hfill & p<0,\mathit{x}\in {\left[\frac{2}{|p|},\infty \right)}^n.\hfill \end{array}\right.\end{array}$$

Therefore, from Lemmas 1 and 2, we acquire

• if $0<p\le 2$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur concave on ${\mathbb{R}}_{+}^n$;
• if $p>2$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur convex on ${\left(0,{(p-1)}^{1/p}-1\right]}^n$;
• if $p>2$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur concave on ${\left({(p-1)}^{1/p}-1,\infty \right)}^n$;
• if $p<0$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur convex on ${\mathbb{R}}_{+}^n$;
• if $p>0$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur geometrically convex on ${\mathbb{R}}_{+}^n$;
• if $p<0$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur geometrically concave on ${\left(0,\frac{1}{|p|}\right]}^n$;
• if $p<0$, then $E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur geometrically convex on ${\left[\frac{2}{|p|},\infty \right)}^n$.

From Relations (10) and (11), employing Definitions 2 and 3, we conclude that inequalities in (18)–(21) for $k=n$.

If $2\le k\le n-1$, since

$$\begin{array}{c}\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_1}=p{(x_1+1)}^{p-1}[{(x_2+1)}^p+1]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\hfill \\ +p{(x_1+1)}^{p-1}{E}_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1).\hfill \end{array}$$

Therefore, we have

$$\begin{array}{c}{\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1}))\triangleq (x_1-x_2)\left[\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_1}-\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_2}\right]\\ =p(x_1-x_2)\left[\frac{{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}-\frac{{(x_2+1)}^{p-1}}{{(x_2+1)}^p+1}\right]{\displaystyle \prod _{i=1}^2}[{(x_i+1)}^p+1]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ +p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1).\end{array}$$

(23)

By Equation (23) and Lemma 1, it is easy to obtain that

• if $p<0$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur convex on ${\mathbb{R}}_{+}^n$;
• if $0<p\le 1$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur concave on ${\mathbb{R}}_{+}^n$;
• if $p>1$ and $\mathit{x}\in {\mathbb{R}}_{+}^n$,

  (a)

  when $x_1=x_2$, we have ${\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})=0$;

  (b)

  when $x_1\ne x_2$, using Cauchy’s mean value theorem, we have

  $$\begin{array}{c}\hfill \frac{\frac{{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}-\frac{{(x_2+1)}^{p-1}}{{(x_2+1)}^p+1}}{{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}}=\frac{p-1-{(\xi +1)}^p}{(p-1){[{(\xi +1)}^p+1]}^2}\end{array}$$

  for some point $\xi \in (min\{x_1,x_2\},max\{x_1,x_2\})$ such that

  $$\begin{array}{c}{\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1}))=p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{\frac{\frac{{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}-\frac{{(x_2+1)}^{p-1}}{{(x_2+1)}^p+1}}{{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}}\prod _{i=1}^2[{(x_i+1)}^p+1]+\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)}\right\}\\ =p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{\frac{p-1-{(\xi +1)}^p}{(p-1){[{(\xi +1)}^p+1]}^2}\prod _{i=1}^2[{(x_i+1)}^p+1]+\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)}\right\}.\end{array}$$

If $p>2$ and $x\in \left(0,{(p-1)}^{1/p}-1\right]$, then $p-1-{(x+1)}^p\ge 0$, so ${\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1}))\ge 0$ for $\mathit{x}\in {\left(0,{(p-1)}^{1/p}-1\right]}^n$.

If $1<p\le 2$ and $\mathit{x}\in (0,b]$, we derive $p-1-{(x+1)}^p\le 0$. Using

$${(min\{x_1,x_2\}+1)}^p+1<{(\xi +1)}^p+1<{(max\{x_1,x_2\}+1)}^p+1$$

and Newton’s inequality leads to

$$\begin{array}{c}{\Delta }_A(E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1}))=p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{\frac{p-1-{(\xi +1)}^p}{(p-1){[{(\xi +1)}^p+1]}^2}\prod _{i=1}^2[{(x_i+1)}^p+1]+\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)}\right\}\\ \ge p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{\frac{p-1-{(\xi +1)}^p}{(p-1){[{(\xi +1)}^p+1]}^2}\prod _{i=1}^2[{(x_i+1)}^p+1]+\frac{n-k}{k-1}\frac{n-2}{{\sum }_{i=3}^n{[{(x_i+1)}^p+1]}^{-1}}\right\}\\ \ge p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{\frac{p-1-{(\xi +1)}^p}{(p-1)[{(\xi +1)}^p+1]}[{(max\{x_1,x_2\}+1)}^p+1]+\frac{n-k}{k-1}\frac{n-2}{{\sum }_{i=3}^n{[{(x_i+1)}^p+1]}^{-1}}\right\}\\ \ge p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{\frac{p-1-{(b+1)}^p}{2(p-1)}[{(b+1)}^p+1]+\frac{2(n-k)}{k-1}\right\}\\ =\frac{p}{2(p-1)}(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ \times \left\{-{(b+1)}^p[{(b+1)}^p+1]+\frac{(4n-3k-1)(p-1)}{k-1}\right\}\ge 0.\end{array}$$

By Lemma 1, if $1<p\le 2$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur convex with respect to $\mathit{x}\in {(0,b]}^n$; if $p>2$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur convex on ${\left(0,{(p-1)}^{1/p}-1\right]}^n$.

When $2\le k\le n-1$, from (22), we obtain

$$\begin{array}{c}(ln{x}_1-ln{x}_2)\left[x_1\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_1}-x_2\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_2}\right]=p(ln{x}_1-ln{x}_2)\\ \times \left[\frac{x_1{(x_1+1)}^{p-1}}{{(x_1+1)}^p+1}-\frac{x_2{(x_2+1)}^{p-1}}{{(x_2+1)}^p+1}\right]{\displaystyle \prod _{i=1}^2}[{(x_i+1)}^p+1]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1)\\ +p(x_1-x_2)\left[x_1{(x_1+1)}^{p-1}-x_2{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p+1).\end{array}$$

(24)

Using Equation (24) and Lemma 2, we obtain that

• if $p>0$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur geometrically convex on ${\mathbb{R}}_{+}^n$;
• if $p<0$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur geometrically concave on ${\left(0,\frac{1}{|p|}\right]}^n$;
• if $p<0$, then $E_k({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})$ is Schur geometrically convex on ${\left[\frac{2}{|p|},\infty \right)}^n$.

By (10) and Lemmas 1 and 2, we arrive at Inequalities (18)–(21). The proof of Theorem 2 is complete. □

Theorem 3.

Let $2\le k\le n$, $p\in \mathbb{R}\backslash \{0\}$, $i_1,\dots ,i_k\in \mathbb{N}$,

$$c={\left[\frac{n-1+\sqrt{{(n-1)}^2+4(n-k)(k-1)/(p-1)}}{2(n-k)}\right]}^{1/p}-1,p>1,$$

and

$$d={\left(\frac{2n+k-3}{k-1}\right)}^{1/p}-1,0<p<1.$$

1. 

If $k=n$, $p\ge 1$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, or if $2\le k\le n-1$, $p>1$, and $\mathit{x}\in {(0,c]}^n$, or if $2\le k\le n-1$, $p=1$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, then

$$\left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(G(\mathit{x})+1)}^p-1]}^k\le \sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}+1)}^p-1]\le \left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(A(\mathit{x})+1)}^p-1]}^k.$$

(25)

2. 

If $k=n$, $0<p<1$, and $\mathit{x}\in {\mathbb{R}}_{+}^n$ or f $2\le k\le n-1$, $0<p<1$, and $\mathit{x}\in {(0,d]}^n$, then

$$\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}-1)}^p+1]\le \left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(G(\mathit{x})+1)}^p-1]}^k.$$

3. 

If $p<0$, $k=n$ is an even integer, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, or if $2\le k\le n-1$, k is an even integer, $p<0$, and $\mathit{x}\in {\left(0,\frac{1}{|p|}\right]}^n$, then

$$\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k[{(x_{i_j}-1)}^p+1]\le \left(\genfrac{}{}{0pt}{}{n}{k}\right){[{(A(\mathit{x})+1)}^p-1]}^k.$$

(26)

If $p<0$, $k=n$ is an odd integer, and $\mathit{x}\in {\mathbb{R}}_{+}^n$, or if $2\le k\le n-1$, k is an odd integer, $p<0$, and $\mathit{x}\in {\left(0,\frac{1}{|p|}\right]}^n$, then Inequality (26) is reversed.

Proof. 

The proof is divided into three cases.

Case 1. If $k=n$, a direct differentiation yields

$$\begin{array}{cc}\hfill \frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_1}=& \frac{p{(x_1+1)}^{p-1}}{{(x_1+1)}^p-1}{E}_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}).\hfill \end{array}$$

From Lemma 4, it follows that

$$\begin{array}{c}(x_1-x_2)\left[\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_1}-\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_2}\right]\\ =p(x_1-x_2)\left[\frac{{(x_1+1)}^{p-1}}{{(x_1+1)}^p-1}-\frac{{(x_2+1)}^{p-1}}{{(x_2+1)}^p-1}\right]E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})\\ \left\{\begin{array}{cc}\le 0,\hfill & \mathrm{if}p>0;\hfill \\ \le 0,\hfill & \mathrm{if}p<0\mathrm{and}n\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{integer};\hfill \\ \ge 0,\hfill & \mathrm{if}p<0\mathrm{and}n\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{integer}\hfill \end{array}\right.\end{array}$$

and

$$\begin{array}{c}(ln{x}_1-ln{x}_2)\left[x_1\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_1}-x_2\frac{\partial E_n({(\mathit{x}+\mathbf{1})}^p+\mathbf{1})}{\partial x_2}\right]\\ =p(ln{x}_1-ln{x}_2)\left[\frac{x_1{(x_1+1)}^{p-1}}{{(x_1+1)}^p-1}-\frac{x_2{(x_2+1)}^{p-1}}{{(x_2+1)}^p-1}\right]E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})\\ \left\{\begin{array}{cc}\ge 0,\hfill & \mathrm{if}p\ge 1;\hfill \\ \le 0,\hfill & \mathrm{if}0<p<1;\hfill \\ \le 0,\hfill & \mathrm{if}p<0\mathrm{and}n\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{integer};\hfill \\ \ge 0,\hfill & \mathrm{if}p<0\mathrm{and}n\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{integer}.\hfill \end{array}\right.\end{array}$$

Therefore, from Lemmas 1 and 2, we have

• if $p>0$, then $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur concave on ${\mathbb{R}}_{+}^n$; if $p<0$ and n is an even (or odd, respectively) integer, then $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur concave (or Schur convex, respectively) on ${\mathbb{R}}_{+}^n$;
• if $p\ge 1$, then $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur geometrically convex on ${\mathbb{R}}_{+}^n$; if $0<p<1$, then $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur geometrically concave on ${\mathbb{R}}_{+}^n$; if $p<0$ and n is an even (or odd, respectively) integer, then $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur geometrically concave (or convex, respectively) on ${\mathbb{R}}_{+}^n$.

For $k=n$, by Relations (10) and (11) and by Definitions 2 and 3, the inequalities in (25) and (26) hold.

Case 2. When $2\le k\le n-1$, since

$$\begin{array}{c}\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_1}=p{(x_1+1)}^{p-1}[{(x_2+1)}^p-1]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\hfill \\ +p{(x_1+1)}^{p-1}{E}_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1}),\hfill \end{array}$$

(27)

we have

$$\begin{array}{c}\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}))\triangleq (x_1-x_2)\left[\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_1}-\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_2}\right]\\ =p(x_1-x_2)\{{(x_1+1)}^{p-1}[{(x_2+1)}^p-1]-[{(x_1+1)}^p-1]{(x_2+1)}^{p-1}\}\\ \times E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})+p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ =p(x_1-x_2)\left[\frac{{(x_1+1)}^{p-1}}{{(x_1+1)}^p-1}-\frac{{(x_2+1)}^{p-1}}{{(x_2+1)}^p-1}\right]{\displaystyle \prod _{i=1}^2}[{(x_i+1)}^p-1]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ +p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1}).\end{array}$$

(28)

Utilizing the monotonicity of $p{(x+1)}^{p-1}$ and Lemma 4, we obtain that

• if $0<p\le 1$, then $\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}))\le 0$ for $\mathit{x}\in {\mathbb{R}}_{+}^n$, so the function $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is a Schur-concave function on ${\mathbb{R}}_{+}^n$;
• if $p<0$ and k is an even (or odd, respectively) integer, then $\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}))⋚0$ for $\mathit{x}\in {\mathbb{R}}_{+}^n$. This shows from Lemma 1 that, if $p<0$ and n is an even (or odd, respectively) integer, the function $E_n({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur concave (or Schur convex, respectively) on ${\mathbb{R}}_{+}^n$;
• if $p>1$ and $\mathit{x}\in {(0,c]}^n$, from (28), it follows that

  $$\begin{array}{c}\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}))=p(x_1-x_2)\{-(x_1-x_2){(x_1+1)}^{p-1}{(x_2+1)}^{p-1}\\ -\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]\}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ +p(x_1-x_2)\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ =p(x_1-x_2)E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\{-(x_1-x_2){(x_1+1)}^{p-1}{(x_2+1)}^{p-1}\\ +\left[{(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}\right]\left[\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}-1\right]\}.\end{array}$$

  (29)

Suppose $x_1\ne x_2$ and $\mathit{x}\in {(0,c]}^n$. By Lagrange’s mean value theorem, we have

$$\begin{array}{c}\hfill {(x_1+1)}^{p-1}-{(x_2+1)}^{p-1}=(p-1)(x_1-x_2){(\xi +1)}^{p-2}\end{array}$$

(30)

for some $\xi$ in the open interval $(min\{x_1,x_2\},max\{x_1,x_2\})$ and

$$\frac{{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}}{(p-1){(\xi +1)}^{p-2}}\ge \frac{1}{(p-1){(\xi +1)}^p}\ge \frac{1}{(p-1){(c+1)}^p}.$$

For $k=2,\dots ,n$ and $\mathit{x}\in {(0,c]}^n$, using Lemma 5 yields

$$\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}\le \frac{(n-k){\widehat{F}}_1({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{(k-1){\widehat{F}}_0({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}\le \frac{n-k}{k-1}{(c+1)}^p-\frac{n-k}{k-1}.$$

(31)

For $\mathit{x}\in {(0,c]}^n$, by Equation (29) and the inequalities in (30) and (31), we obtain

$$\begin{array}{c}\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}))=p{(x_1-x_2)}^2{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ \times \left\{-{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}+(p-1){(\xi +1)}^{p-2}\left[\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}-1\right]\right\}\\ \le p{(x_1-x_2)}^2(p-1){(\xi +1)}^{p-2}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ \times \left[-\frac{{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}}{(p-1){(\xi +1)}^{p-2}}+\frac{n-k}{(k-1)(n-2)}\sum _{i=3}^n{(x_i+1)}^p-\frac{n-1}{k-1}\right]\\ \le p{(x_1-x_2)}^2(p-1){(\xi +1)}^{p-2}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ \times \left[-\frac{1}{(p-1){(c+1)}^p}+\frac{n-k}{k-1}{(c+1)}^p-\frac{n-1}{k-1}\right]=0.\end{array}$$

Therefore, if $p>1$, then $\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1}))\le 0$ for $\mathit{x}\in {(0,c]}^n$. Therefore, from Lemma 1, it follows that

• if $0<p\le 1$, the function $E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur concave on ${\mathbb{R}}_{+}^n$;
• if $p<0$ and k is an even (or odd, respectively) integer, the function $E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur concave (or Schur convex, respectively) on ${\mathbb{R}}_{+}^n$;
• if $p>1$, the function $E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur concave on ${(0,c]}^n$.

Case 3. If $2\le k\le n-1$, then from (27), it follows that

$$\begin{array}{c}{\Delta }_G(E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})\triangleq (ln{x}_1-ln{x}_2)\left[x_1\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_1}-x_2\frac{\partial E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})}{\partial x_2}\right]\\ =p(ln{x}_1-ln{x}_2)\left[\frac{x_1{(x_1+1)}^{p-1}}{{(x_1+1)}^p-1}-\frac{x_2{(x_2+1)}^{p-1}}{{(x_2+1)}^p-1}\right]\prod _{i=1}^2[{(x_i+1)}^p-1]E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ +p(ln{x}_1-ln{x}_2)\left[x_1{(x_1+1)}^{p-1}-x_2{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1}).\end{array}$$

Using the monotonicity of function $px{(x+1)}^{p-1}$ and Lemma 4 results in

• if $p\ge 1$, then $\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})\ge 0$ for $\mathit{x}\in {\mathbb{R}}_{+}^n$;
• if $p<0$ and k is an even (or odd, respectively) integer, then $\Delta (E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})⋚0$ for $\mathit{x}\in {\left(0,\frac{1}{|p|}\right]}^n$;
• if $0<p<1$, then

  $$\begin{array}{c}{\Delta }_G(E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})=p(ln{x}_1-ln{x}_2)\{x_1{(x_1+1)}^{p-1}[{(x_2+1)}^p-1]\\ -[{(x_2+1)}^p-1]x_2{(x_2+1)}^{p-1}\}{E}_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ +p(ln{x}_1-ln{x}_2)\left[x_1{(x_1+1)}^{p-1}-x_2{(x_2+1)}^{p-1}\right]E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ =p(ln{x}_1-ln{x}_2)E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\{(x_1-x_2){(x_1+1)}^{p-1}{(x_2+1)}^{p-1}\\ +\left[x_1{(x_1+1)}^{p-1}-x_2{(x_2+1)}^{p-1}\right]\left[\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}-1\right]\}.\end{array}$$

  (32)

Suppose $x_1\ne x_2$, by Lagrange’s mean value theorem, we have

$$x_1{(x_1+1)}^{p-1}-x_2{(x_2+1)}^{p-1}=(x_1-x_2){(\xi +1)}^{p-2}(1+p\xi )$$

(33)

for some $\xi \in (min\{x_1,x_2\},max\{x_1,x_2\})$. Therefore, using Lemma 5 leads to

$$\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}\le \frac{n-k}{(k-1)(n-2)}\sum _{i=3}^n{(x_i+1)}^p-\frac{n-k}{k-1},\mathit{x}\in {\mathbb{R}}_{+}^n.$$

(34)

For $\mathit{x}\in {(0,d]}^n$, substituting (33) and (34) into (32) yields

$$\begin{array}{c}{\Delta }_G(E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})=p(ln{x}_1-ln{x}_2)(x_1-x_2)E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ \times \left\{{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}+{(\xi +1)}^{p-2}(1+p\xi )\left[\frac{E_{k-1}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}{E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})}-1\right]\right\}\\ \le p(ln{x}_1-ln{x}_2)(x_1-x_2)E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\{{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}\\ +{(\xi +1)}^{p-2}(1+p\xi )\left[\frac{n-k}{(k-1)(n-2)}\sum _{i=3}^n{(x_i+1)}^p-\frac{n-1}{k-1}\right]\}\\ =p(ln{x}_1-ln{x}_2)(x_1-x_2){(\xi +1)}^{p-2}(1+p\xi )E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ \times \left[\frac{{(x_1+1)}^{p-1}{(x_2+1)}^{p-1}}{{(\xi +1)}^{p-2}(1+p\xi )}+\frac{n-k}{(k-1)(n-2)}\sum _{i=3}^n{(x_i+1)}^p-\frac{n-1}{k-1}\right]\\ \le p(ln{x}_1-ln{x}_2)(x_1-x_2){(\xi +1)}^{p-2}(1+p\xi )E_{k-2}({(\widehat{\mathit{x}}+\mathbf{1})}^p-\mathbf{1})\\ \times \left[{(d+1)}^{2p}+\frac{n-k}{k-1}{(d+1)}^p-\frac{n-1}{k-1}\right]=0.\end{array}$$

Therefore, using Lemma 2, we obtain that

• if $p\ge 1$, the function $E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur geometrically convex on ${\mathbb{R}}_{+}^n$;
• if $p<0$ and k is an even (or odd, respectively) integer, the function $E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is a Schur geometrically-concave (or convex, respectively) function on ${\left(0,\frac{1}{|p|}\right]}^n$;
• if $0<p<1$, the function $E_k({(\mathit{x}+\mathbf{1})}^p-\mathbf{1})$ is Schur geometrically concave on ${(0,d]}^n$.

For $k=n$, by Relations (10) and (11) and by Definitions 2 and 3, the inequalities in (25) and (26) hold. Theorem 3 is thus proven. □

Remark 1.

This paper is a corrected and revised version of the preprint [29].