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GENERALIZATIONS OF SEVERAL INEQUALITIES
RELATED TO MULTIVARIATE GEOMETRIC
MEANS

Bo-Yan Xi, Ying Wu, Huan-Nan Shi, Feng Qi

► To cite this version:

Bo-Yan Xi, Ying Wu, Huan-Nan Shi, Feng Qi. GENERALIZATIONS OF SEVERAL INEQUALITIES RELATED TO MULTIVARIATE GEOMETRIC MEANS. Mathematics, 2019, 7 (6), 15 p. 10.3390/math7060552. hal-01865896v2

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Submitted on 19 Jan 2020

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Article

# Generalizations of Several Inequalities Related to Multivariate Geometric Means

Bo-Yan Xi<fnref n="1" />,*, Ying Wu<fnref n="1" />,*, Huan-Nan Shi<fnref n="2" />,*, and Feng Qi<fnref n="3" />,<fnref n="4" />,†

Received 16 May 2019; Accepted: 11 June 2019; Published: 17 June 2019

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

**Keywords:** Schur-convex function; Schur-geometrically convex function; elementary symmetric function; inequality; multivariate geometric mean; majorization

**MSC:** Primary 26E60; Secondary 26D15, 26A51

## 1. Introduction

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

$$A(\mathbf{x}) = \frac{1}{n} \sum_{i=1}^{n} x_i \quad \text{and} \quad G(\mathbf{x}) = \left( \prod_{i=1}^{n} x_i \right)^{1/n}.$$

In [12, p. 208, 3.2.34], it was stated that

$$\prod_{i=1}^{n} (x_i + 1) \ge [G(\mathbf{x}) + 1]^n. \qquad (1.1)$$

In the paper [26], Wang and Chen established that the inequality

$$\prod_{i=1}^{n} [(x_i + 1)^p - 1] \ge [(G(\mathbf{x}) + 1)^p - 1]^n \qquad (1.2)$$

is valid for $p > 1$. In the paper [24], Wang extended the inequality (1.2) as follows:

1. if $p \ge 1$, then the inequality (1.2) is valid;

2. if $0 < p \le 1$, then the inequality (1.2) is reversed;

3. if $p > 0$, then

$$\prod_{i=1}^{n} [(x_i + 1)^p + 1] \ge [(G(\mathbf{x}) + 1)^p + 1]^n; \qquad (1.3)$$

4. if $p < 0$ and $0 < x_i < \frac{1}{|p|}$ for $1 \le i \le n$, then the inequality (1.3) is reversed.

<footnote n="1">¹ College of Mathematics, Inner Mongolia University for Nationalities, Tongliao 028043, Inner Mongolia, China; baoyintu78@qq.com (Xi), wuying19800920@qq.com (Wu)</footnote>

<footnote n="2">² Department of Electronic Information, Teacher's College, Beijing Union University, Beijing 100011, China; sfthuannan@buu.edu.cn</footnote>

<footnote n="3">³ Institute of Mathematics, Henan Polytechnic University, Jiaozuo 454010, Henan, China</footnote>

<footnote n="4">⁴ School of Mathematical Sciences, Tianjin Polytechnic University, Tianjin 300387, China</footnote>

<footnote>* Correspondence: qifeng618@gmail.com</footnote>
In [8, p. 11], it was proved that

$$ \sqrt[n]{\prod_{k=1}^{n}(a_k + b_k)} \ge \sqrt[n]{\prod_{k=1}^{n}a_k} + \sqrt[n]{\prod_{k=1}^{n}b_k}, \qquad (1.4) $$

where $a_k, b_k \ge 0$. When $a_k = x_k$ and $b_k = 1$, the inequality (1.4) becomes (1.1). The inequality (1.4) is the Minkowski inequality of product form.

We observe that inequalities in (1.1), (1.2), (1.3), and (1.4) can be rearranged as

$$ G(\mathbf{x} + \mathbf{1}) \ge G(\mathbf{x}) + 1, \quad G((\mathbf{x} + \mathbf{1})^p \pm \mathbf{1}) \ge [G(\mathbf{x}) + 1]^p \pm 1, \qquad (1.5) $$

and

$$ G(a + b) > G(a) + G(b), \qquad (1.6) $$

where

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

and

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

The inequality (1.6) reveals that the geometric mean $G(\mathbf{x})$ is sub-additive. For information about the sub-additivity, please refer to [2,6,7,9,10,14,15,17] and closely related references therein. The sub-additive property of the geometric mean $G(\mathbf{x})$ can also be derived from the property that the geometric mean $G(\mathbf{x})$ is a Bernstein function, see [5,16,18–22] and closely related references therein.

In this paper, by some methods in the theory of majorization, we will generalize the above inequalities in (1.1), (1.2), (1.3), (1.5), and (1.6).

## 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 ([11,25]).** Let $\mathbf{x} = (x_1, x_2, \dots, x_n)$ and $\mathbf{y} = (y_1, y_2, \dots, y_n) \in \mathbb{R}^n$.

1. If

$$ \sum_{i=1}^{k} x_{[i]} \le \sum_{i=1}^{k} y_{[i]} \quad \text{and} \quad \sum_{i=1}^{n} x_{[i]} = \sum_{i=1}^{n} y_{[i]} $$

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

2. For $\mathbf{x}, \mathbf{y} \in \Omega$ and $\lambda \in [0, 1]$, if

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

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

**Definition 2 ([11,25]).** Let $\Omega \subseteq \mathbb{R}^n$ be a symmetric and convex set.

1. If $\mathbf{x} \prec \mathbf{y}$ on $\Omega$ implies $\varphi(\mathbf{x}) \le \varphi(\mathbf{y})$, then we say that the function $\varphi: \Omega \to \mathbb{R}$ is Schur-convex on $\Omega$.

2. If $-\varphi$ is a Schur-convex function on $\Omega$, then we say that $\varphi$ is Schur-concave on $\Omega$.

**Lemma 1 ([11,25]).** Let $\Omega \subseteq \mathbb{R}^n$ be a symmetric and convex set with nonempty interior $\Omega^\circ$ and let $\varphi: \Omega \to \mathbb{R}$ is continuous on $\Omega$ and continuously differentiable on $\Omega^\circ$. If and only if $\varphi$ is symmetric on $\Omega$ and

$$ (x_1 - x_2) \left( \frac{\partial \varphi}{\partial x_1} - \frac{\partial \varphi}{\partial x_2} \right) \ge 0, \quad \mathbf{x} \in \Omega^\circ, $$
the function $\varphi$ is Schur-convex on $\Omega$.

**Definition 3 ([29, pp. 64 and 107]).** Let $x, y \in \Omega \subseteq \mathbb{R}_+^n$.

1. If $(x_1^\lambda y_1^{1-\lambda}, x_2^\lambda y_2^{1-\lambda}, \dots, x_n^\lambda y_n^{1-\lambda}) \in \Omega$ for $x, y \in \Omega$ and $\lambda \in [0, 1]$, then $\Omega$ is called a geometrically convex set.

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

3. If $-\varphi$ is a Schur-geometrically convex function on $\Omega$, then $\varphi$ is said to be a Schur-geometrically concave function on $\Omega$.

**Lemma 2 ([29, p. 108]).** Let $\Omega \subseteq \mathbb{R}_+^n$ be a symmetric and geometrically convex set with a nonempty interior $\Omega^\circ$ and let $\varphi: \Omega \to \mathbb{R}_+$ be continuous on $\Omega$ and differentiable in $\Omega^\circ$. If and only if $\varphi$ is symmetric on $\Omega$ and

$$ (\ln x_1 - \ln x_2) \left( x_1 \frac{\partial \varphi}{\partial x_1} - x_2 \frac{\partial \varphi}{\partial x_2} \right) \ge 0, \quad x \in \Omega^\circ, $$

the function $\varphi$ is Schur-geometrically convex on $\Omega$.

For more information on the Schur-convexity and the Schur-geometric convexity, please refer to the papers [3,4,23,27] and the monographs [11,29].

**Lemma 3 (Bernoulli's inequality [1,13]).** For $0 \ne x > -1$, if $\alpha \ge 1$ or $\alpha < 0$, then

$$ (1+x)^\alpha \ge 1+\alpha x; \qquad (2.1) $$

if $0 < \alpha < 1$, then the inequality (2.1) is reversed.

**Lemma 4.** For $p \in \mathbb{R} \setminus \{0\}$ and $x \in \mathbb{R}_+$, define

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

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 $(0, (p-1)^{1/p} - 1)$ and decreasing on $((p-1)^{1/p} - 1, \infty)$.

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 $(0, \frac{1}{|p|}]$ and increasing on $[\frac{2}{|p|}, \infty)$.

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

$$ [h_1(x)]' = \frac{p[p-1-(x+1)^p]}{[(x+1)^p+1]^2} (x+1)^{p-2}, \quad [g_1(x)]' = \frac{p[1+px+(x+1)^p]}{[(x+1)^p+1]^2} (x+1)^{p-2}, $$

$$ [h_2(x)]' = \frac{p[1-p-(x+1)^p]}{[(x+1)^p-1]^2} (x+1)^{p-2}, \quad [g_2(x)]' = \frac{p[(x+1)^p-1-px]}{[(x+1)^p-1]^2} (x+1)^{p-2}. $$

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

If $p > 0$ and $x \in \mathbb{R}_+$, we acquire $[g_1(x)]' \ge 0$; if $p < 0$ and $0 < x \le \frac{1}{|p|}$, we have $[g_1(x)]' \le 0$; if $p < 0$ and $x \ge \frac{2}{|p|}$, since $(x+1)^p < 1$ and $1+px \le -1$, we acquire $[g_1(x)]' \ge 0$.
If $p \in \mathbb{R} \setminus \{0\}$, we obtain $[h_2(x)]' \le 0$ 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 $[g_2(x)]' \ge 0$; if $0 < p \le 1$, we have $(x+1)^p \le 1+px$, and so $[g_2(x)]' \le 0$; if $p < 0$, we obtain $(x+1)^p \ge 1+px$, hence $[g_2(x)]' \le 0$.
The proof of Lemma 4 is complete. □

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

$$E_k(\mathbf{x}) = E_k(x_1, x_2, \dots, x_n) = \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} x_{i_j}, \quad 1 \le k \le n,$$

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

**Lemma 5 (Newton's inequality [11, p. 134]).** For $\mathbf{x} = (x_1, x_2, \dots, x_n) \in \mathbb{R}_+^n$ and $n \ge 2$, let $\hat{F}_k(\mathbf{x}) = \frac{1}{(n/k)} E_k(\mathbf{x})$ for $1 \le k \le n-1$. Then

$$[\hat{F}_k(\mathbf{x})]^2 \ge \hat{F}_{k-1}(\mathbf{x})\hat{F}_{k+1}(\mathbf{x}).$$

### 3. Main results

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

$$E_k((\mathbf{x} + \mathbf{1})^p + \xi) = \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} [(x_{i_j} + 1)^p + \xi], \quad 2 \le k \le n, \quad n \ge 2$$

to generalize inequalities in (1.1), (1.2), (1.3), (1.5), and (1.6), where $\mathbf{x} \in \mathbb{R}_+^n$, the quantities $i_1, i_2, \dots, i_k$ are positive integers, $\xi = 0, \pm 1$, and $\xi = 0, \pm 1$.

Our main results are Theorems 1 to 3 below.

**Theorem 1.** Let $2 \le k \le n$, $p \in \mathbb{R} \setminus \{0\}$, $i_1, \dots, i_k \in \mathbb{N}$, 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 $\mathbf{x} \in \mathbb{R}_+^n$, or if $2 \le k \le n-1$, $p > 1$ with $a < 1$, and $\mathbf{x} \in (0, a^{-1/p} - 1]^n$, or if $k = n$, $p > 0$, and $\mathbf{x} \in \mathbb{R}_+^n$, then

$$\binom{n}{k} [G(\mathbf{x}) + 1]^{kp} \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} (x_{i_j} + 1)^p \le \binom{n}{k} [A(\mathbf{x}) + 1]^{kp}; \quad (3.1)$$

if $p < 0$ and $\mathbf{x} \in (0, \frac{1}{|p|}]^n$, or if $k = n$, $p < 0$, and $\mathbf{x} \in \mathbb{R}_+^n$, then the double inequality (3.1) is reversed.

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

$$\binom{n}{k} [A(\mathbf{x}) + 1]^{kp} \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} (x_{i_j} + 1)^p.$$

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

$$\binom{n}{k} [G(\mathbf{x}) + 1]^{kp} \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} (x_{i_j} + 1)^p; \quad (3.2)$$

if $p < 0$ and $\mathbf{x} \in \mathbb{R}_+^n$, then the inequality (3.2) is reversed.

**Proof.** When $k = n$, we have

$$\frac{\partial E_n((\mathbf{x} + \mathbf{1})^p)}{\partial x_1} = \frac{p}{(x_1 + 1)} E_n((\mathbf{x} + \mathbf{1})^p).$$
From this, we obtain

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

and

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

Using Lemmas 1 and 2, we arrive at

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

2. if $p < 0$, then $E_n((\mathbf{x} + \mathbf{1})^p)$ is Schur-convex on $\mathbb{R}_+^n$;

3. if $p > 0$, then $E_n((\mathbf{x} + \mathbf{1})^p)$ is Schur-geometrically convex on $\mathbb{R}^n$;

4. if $p < 0$, the $E_n((\mathbf{x} + \mathbf{1})^p)$ is Schur-geometrically concave on $\mathbb{R}^n$.

Since

$$ (A_n(\mathbf{x}), A_n(\mathbf{x}), \dots, A_n(\mathbf{x})) \prec \mathbf{x} \qquad (3.3) $$

and

$$ (\ln G_n(\mathbf{x}), \ln G_n(\mathbf{x}), \dots, \ln G_n(\mathbf{x})) \prec \ln \mathbf{x} \qquad (3.4) $$

for $\mathbf{x} \in \mathbb{R}_+^n$, applying Definitions 2 and 3, we obtain the double inequality (3.1) for $k = n$. When $2 \le k \le n - 1$, a direct differentiation yields

$$ \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p)}{\partial x_1} = p(x_1 + 1)^{p-1}(x_2 + 1)^p E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p) + p(x_1 + 1)^{p-1} E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p), $$

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

$$ \begin{aligned} \Delta_A(E_k((\mathbf{x} + \mathbf{1})^p)) &\triangleq (x_1 - x_2) \left[ \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p)}{\partial x_1} - \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p)}{\partial x_2} \right] \\ &= -p(x_1 - x_2)^2 (x_1 + 1)^{p-1} (x_2 + 1)^{p-1} E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p) \\ &\quad + p(x_1 - x_2) [(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}] E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p) \end{aligned} \qquad (3.5) $$

and

$$ \begin{aligned} & (\ln x_1 - \ln x_2) \left[ x_1 \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p)}{\partial x_1} - x_2 \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p)}{\partial x_2} \right] \\ &= p(\ln x_1 - \ln x_2)(x_1 - x_2)(x_1 + 1)^{p-1}(x_2 + 1)^{p-1} E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p) \\ &\quad + p(\ln x_1 - \ln x_2) [x_1(x_1 + 1)^{p-1} - x_2(x_2 + 1)^{p-1}] E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p). \end{aligned} \qquad (3.6) $$

Using the equation (3.6) and Lemma 2, we obtain that

1. if $p > 0$, then $E_k((\mathbf{x} + \mathbf{1})^p)$ is Schur-geometrically convex on $\mathbb{R}_+^n$;

2. if $p < 0$, then $E_k((\mathbf{x} + \mathbf{1})^p)$ is Schur-geometrically concave on $(0, \frac{1}{|p|}]^n$.

By the equation (3.5) and Lemma 1, we reveal that

1. if $p < 0$, then $E_k((\mathbf{x} + \mathbf{1})^p)$ is Schur-convex on $\mathbb{R}_+^n$;

2. if $0 < p \le 1$, then $E_k((\mathbf{x} + \mathbf{1})^p)$ is Schur-concave on $\mathbb{R}_+^n$;

3. if $p > 1$ and $\mathbf{x} \in \mathbb{R}_+^n$,

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

(b) when $x_1 \neq x_2$,
i. by Lagrange's mean value theorem, we have

$$ \frac{1}{(x_2+1)^{p-1}} - \frac{1}{(x_1+1)^{p-1}} = \frac{(p-1)(x_1-x_2)}{(\zeta+1)^p} \quad (3.7) $$

for at least one interior point $\zeta \in (\min\{x_1, x_2\}, \max\{x_1, x_2\})$;

ii. from Newton's inequality, we obtain

$$ \begin{aligned} \frac{(n-k)(n-2)}{(k-1) \sum_{i=3}^{n} (x_i+1)^{-p}} &= \frac{(n-k) \hat{F}_{n-2}((\hat{\mathbf{x}}+\mathbf{1})^p)}{(k-1) \hat{F}_{n-3}((\hat{\mathbf{x}}+\mathbf{1})^p)} \le \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p)}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p)} \\ &\le \frac{(n-k) \hat{F}_1((\hat{\mathbf{x}}+\mathbf{1})^p)}{(k-1) \hat{F}_0((\hat{\mathbf{x}}+\mathbf{1})^p)} = \frac{n-k}{(k-1)(n-2)} \sum_{i=3}^{n} (x_i+1)^p. \end{aligned} \quad (3.8) $$

Substituting (3.7) and (3.8) into the inequality (3.5) yields

$$ \begin{aligned} \Delta_A(E_k((\mathbf{x}+\mathbf{1})^p)) &= -p(x_1-x_2)^2(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \\ &\quad +p(x_1-x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p) \\ &= p(x_1-x_2)(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \\ &\quad \times \left[ -(x_1-x_2) + \left( \frac{1}{(x_2+1)^{p-1}} - \frac{1}{(x_1+1)^{p-1}} \right) \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p)}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p)} \right] \end{aligned} \quad (3.9) \\ = p(x_1-x_2)^2(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \left[ \frac{p-1}{(\zeta+1)^p} \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p)}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p)} - 1 \right]. \end{aligned} $$

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

$$ \frac{1}{(\zeta + 1)^p} \ge \frac{k-1}{(n-k)(p-1)}. \quad (3.10) $$

Consequently, inequalities from (3.9) to (3.10) imply $\Delta(E_k((\mathbf{x}+\mathbf{1})^p)) \ge 0$ for $\mathbf{x} \in (0, a^{1/p} - 1]^n$. If $a < 1$ and $\mathbf{x} \in (0, a^{-1/p} - 1]^n$, we obtain

$$ \begin{aligned} \Delta_A(E_k((\mathbf{x}+\mathbf{1})^p)) &= p(x_1-x_2)^2(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \\ &\quad \times \left[ -1 + \frac{p-1}{(\zeta+1)^p} \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p)}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p)} \right] \\ &\le p(x_1-x_2)^2(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \left[ -1 + \frac{(n-k)(p-1)}{(k-1)(n-2)(\zeta+1)^p} \sum_{i=3}^{n} (x_i+1)^p \right] \\ &\le p(x_1-x_2)^2(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \\ &\quad \times \left\{ -1 + \frac{(n-k)(p-1)}{(k-1)(\zeta+1)^p} \left[ \left( \left( \frac{k-1}{(n-k)(p-1)} \right)^{1/p} - 1 \right) + 1 \right] \right\} \\ &= p(x_1-x_2)^2(x_1+1)^{p-1}(x_2+1)^{p-1}E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p) \left[ \frac{1}{(\zeta+1)^p} - 1 \right] < 0. \end{aligned} $$

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

Using (3.3) and by Definitions 2 and 3, inequalities in (3.1) to (3.2) hold. The proof of Theorem 1 is complete. $\square$
**Theorem 2.** Let $2 \le k \le n$ and $p \in \mathbb{R} \setminus \{0\}, i_1, \dots, i_k \in \mathbb{N}$, and

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

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

$$\binom{n}{k} [(G(\mathbf{x}) + 1)^p + 1]^k \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^k [(x_{i_j} + 1)^p + 1] \le \binom{n}{k} [(A(\mathbf{x}) + 1)^p + 1]^k. \quad (3.11)$$

If $2 \le k \le n, p < 0$, and $x \in (0, \frac{1}{|p|}]^n$, then the double inequality (3.11) is reversed.

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

$$\binom{n}{k} [(A(\mathbf{x}) + 1)^p + 1]^k \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^k [(x_{i_j} + 1)^p + 1]. \quad (3.12)$$

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

$$\binom{n}{k} [(G(\mathbf{x}) + 1)^p + 1]^k \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^k [(x_{i_j} + 1)^p + 1]. \quad (3.13)$$

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

$$[(G(\mathbf{x}) + 1)^p + 1]^n \le \prod_{i=1}^{n} [(x_i + 1)^p + 1]. \quad (3.14)$$

**Proof.** When $k = n$, a direct differentiation yields

$$\frac{\partial E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})}{\partial x_1} = \frac{p(x_1 + 1)^{p-1}}{(x_1 + 1)^p + 1} E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1}). \quad (3.15)$$

From Lemma 4, it follows that

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

and

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

Therefore, from Lemmas 1 and 2, we acquire

1. if $0 < p \le 2$, then $E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-concave on $\mathbb{R}_+^n$;

2. if $p > 2$, then $E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-convex on $(0, (p-1)^{1/p} - 1]^n$;

3. if $p > 2$, then $E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-concave on $((p-1)^{1/p} - 1, \infty)^n$;

4. if $p < 0$, then $E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-convex on $\mathbb{R}_+^n$;

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

6. if $p < 0$, then $E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-geometrically concave on $(0, \frac{1}{|p|}]^n$;

7. if $p < 0$, then $E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-geometrically convex on $[\frac{2}{|p|}, \infty)^n$.

From the relations (3.3) and (3.4), employing Definitions 2 and 3, we conclude inequalities in (3.11), (3.12), (3.13), and (3.14) for $k = n$.

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

$$ \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})}{\partial x_1} = p(x_1 + 1)^{p-1}[(x_2 + 1)^p + 1]E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p + 1) + p(x_1 + 1)^{p-1}E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p + 1). $$

Therefore, we have

$$ \begin{aligned} \Delta_A(E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})) &\triangleq (x_1 - x_2) \left[ \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})}{\partial x_1} - \frac{\partial E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})}{\partial x_2} \right] \\ &= p(x_1 - x_2) \left[ \frac{(x_1 + 1)^{p-1}}{(x_1 + 1)^p + 1} - \frac{(x_2 + 1)^{p-1}}{(x_2 + 1)^p + 1} \right] \prod_{i=1}^{2} [(x_i + 1)^p + 1] E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p + 1) \quad (3.16) \\ &\quad + p(x_1 - x_2) [(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}] E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p + 1). \end{aligned} $$

By the equation (3.16) and Lemma 1, it is easy to obtain that

1. if $p < 0$, then $E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-convex on $\mathbb{R}_+^n$;

2. if $0 < p \le 1$, then $E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-concave on $\mathbb{R}_+^n$;

3. if $p > 1$ and $\mathbf{x} \in \mathbb{R}_+^n$,

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

(b) when $x_1 \ne x_2$, using Cauchy's mean value theorem, we have

$$ \frac{\frac{(x_1+1)^{p-1}}{(x_1+1)^{p+1}} - \frac{(x_2+1)^{p-1}}{(x_2+1)^{p+1}}}{(x_1+1)^{p-1} - (x_2+1)^{p-1}} = \frac{p-1-(\zeta+1)^p}{(p-1)[(\zeta+1)^p+1]^2} $$

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

$$ \begin{aligned} \Delta_A(E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})) &= p(x_1 - x_2) [(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}] E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p + 1) \\ &\quad \times \left\{ \frac{\frac{(x_1+1)^{p-1}}{(x_1+1)^{p+1}} - \frac{(x_2+1)^{p-1}}{(x_2+1)^{p+1}}}{(x_1+1)^{p-1} - (x_2+1)^{p-1}} \prod_{i=1}^{2} [(x_i+1)^p + 1] + \frac{E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p + 1)}{E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p + 1)} \right\} \\ &= p(x_1 - x_2) [(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}] E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p + 1) \\ &\quad \times \left\{ \frac{p-1-(\zeta+1)^p}{(p-1)[(\zeta+1)^p+1]^2} \prod_{i=1}^{2} [(x_i+1)^p + 1] + \frac{E_{k-1}((\hat{\mathbf{x}} + \mathbf{1})^p + 1)}{E_{k-2}((\hat{\mathbf{x}} + \mathbf{1})^p + 1)} \right\}. \end{aligned} $$
If $p > 2$ and $x \in (0, (p-1)^{1/p} - 1]$, then $p - 1 - (x+1)^p \ge 0$, so $\Delta_A(E_k((\mathbf{x}+\mathbf{1})^p + \mathbf{1})) \ge 0$ for $\mathbf{x} \in (0, (p-1)^{1/p} - 1]^n$.

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

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

and Newton's inequality lead to

$$
\begin{aligned}
\Delta_A(E_k((\mathbf{x}+\mathbf{1})^p + \mathbf{1})) &= p(x_1-x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p + 1) \\
&\quad \times \left\{ \frac{p-1-(\zeta+1)^p}{(p-1)[(\zeta+1)^p+1]^2} \prod_{i=1}^2 [(x_i+1)^p+1] + \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p + 1)}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p + 1)} \right\} \\
&\ge p(x_1-x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p + 1) \\
&\quad \times \left\{ \frac{p-1-(\zeta+1)^p}{(p-1)[(\zeta+1)^p+1]^2} \prod_{i=1}^2 [(x_i+1)^p+1] + \frac{n-k}{k-1} \frac{n-2}{\sum_{i=3}^n [(x_i+1)^p+1]^{-1}} \right\} \\
&\ge p(x_1-x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p + 1) \\
&\quad \times \left\{ \frac{p-1-(b+1)^p}{2(p-1)} [(b+1)^p+1] + \frac{2(n-k)}{k-1} \right\} \\
&= \frac{p}{2(p-1)}(x_1-x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p + 1) \\
&\quad \times \left\{ -(b+1)^p [(b+1)^p+1] + \frac{(4n-3k-1)(p-1)}{k-1} \right\} \ge 0.
\end{aligned}
$$

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

When $2 \le k \le n-1$, from (3.15), we obtain

$$ (\ln x_1 - \ln x_2) \left[ x_1 \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p + \mathbf{1})}{\partial x_1} - x_2 \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p + \mathbf{1})}{\partial x_2} \right] = p(\ln x_1 - \ln x_2) \\ \times \left[ \frac{x_1(x_1+1)^{p-1}}{(x_1+1)^p+1} - \frac{x_2(x_2+1)^{p-1}}{(x_2+1)^p+1} \right] \prod_{i=1}^2 [(x_i+1)^p+1] E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p + 1) \tag{3.17} \\ + p(x_1-x_2)[x_1(x_1+1)^{p-1} - x_2(x_2+1)^{p-1}] E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p + 1). $$

Using the equation (3.17) and Lemma 2, we obtain that

1. if $p > 0$, then $E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-geometrically convex on $\mathbb{R}_+^n$;

2. if $p < 0$, then $E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-geometrically concave on $(0, \frac{1}{|p|}]^n$;

3. if $p < 0$, then $E_k((\mathbf{x} + \mathbf{1})^p + \mathbf{1})$ is Schur-geometrically convex on $[\frac{2}{|p|}, \infty)^n$.

By (3.3) and Lemmas 1 and 2, we arrive at inequalities (3.11) to (3.14). The proof of Theorem 2 is complete. $\square$

**Theorem 3.** Let $2 \le k \le n$, $p \in \mathbb{R} \setminus \{0\}$, $i_1, \dots, i_k \in \mathbb{N}$,

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

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

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

$$\binom{n}{k} [(G(\mathbf{x}) + 1)^p - 1]^k \le \sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} [(x_{i_j} + 1)^p - 1] \le \binom{n}{k} [(A(\mathbf{x}) + 1)^p - 1]^k. \quad (3.18)$$

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

$$\sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} [(x_{i_j} - 1)^p + 1] \le \binom{n}{k} [(G(\mathbf{x}) + 1)^p - 1]^k.$$

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

$$\sum_{1 \le i_1 < \dots < i_k \le n} \prod_{j=1}^{k} [(x_{i_j} - 1)^p + 1] \le \binom{n}{k} [(A(\mathbf{x}) + 1)^p - 1]^k. \quad (3.19)$$

If $p < 0, k = n$ is an odd integer, and $x \in \mathbb{R}_+^n$, or if $2 \le k \le n-1, k$ is an odd integer, $p < 0$, and $x \in (0, \frac{1}{|p|}]^n$, then the inequality (3.19) is reversed.

**Proof.** The proof is divided into three cases.

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

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

From Lemma 4, it follows that

$$\begin{align*}
& (x_1 - x_2) \left[ \frac{\partial E_n((\mathbf{x} + \mathbf{1})^p - \mathbf{1})}{\partial x_1} - \frac{\partial E_n((\mathbf{x} + \mathbf{1})^p - \mathbf{1})}{\partial x_2} \right] \\
&= p(x_1 - x_2) \left[ \frac{(x_1 + 1)^{p-1}}{(x_1 + 1)^p - 1} - \frac{(x_2 + 1)^{p-1}}{(x_2 + 1)^p - 1} \right] E_n((\mathbf{x} + \mathbf{1})^p - \mathbf{1}) \\
&\begin{cases}
\le 0, & \text{if } p > 0; \\
\le 0, & \text{if } p < 0 \text{ and } n \text{ is an even integer}; \\
\ge 0, & \text{if } p < 0 \text{ and } n \text{ is an odd integer}
\end{cases}
\end{align*}$$

and

$$\begin{align*}
& (\ln x_1 - \ln x_2) \left[ x_1 \frac{\partial E_n((\mathbf{x} + \mathbf{1})^p - \mathbf{1})}{\partial x_1} - x_2 \frac{\partial E_n((\mathbf{x} + \mathbf{1})^p + \mathbf{1})}{\partial x_2} \right] \\
&= p(\ln x_1 - \ln x_2) \left[ \frac{x_1(x_1+1)^{p-1}}{(x_1+1)^p-1} - \frac{x_2(x_2+1)^{p-1}}{(x_2+1)^p-1} \right] E_n((\mathbf{x} + \mathbf{1})^p - \mathbf{1}) \\
&\begin{cases}
\ge 0, & \text{if } p \ge 1; \\
\le 0, & \text{if } 0 < p < 1; \\
\le 0, & \text{if } p < 0 \text{ and } n \text{ is an even integer}; \\
\ge 0, & \text{if } p < 0 \text{ and } n \text{ is an odd integer}.
\end{cases}
\end{align*}$$

Therefore, from Lemmas 1 and 2, we have
1. if $p > 0$, then $E_n((x+1)^p - 1)$ is Schur-concave on $\mathbb{R}_+^n$; if $p < 0$ and $n$ is an even (or odd, respectively) integer, then $E_n((x+1)^p - 1)$ is Schur-concave (or Schur-convex, respectively) on $\mathbb{R}_+^n$;

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

For $k = n$, by the relations (3.3) and (3.4) and by Definitions 2 and 3, the inequalities in (3.18) to (3.19) hold.

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

$$ \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})}{\partial x_1} = p(x_1+1)^{p-1}[(x_2+1)^p - 1]E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ + p(x_1+1)^{p-1}E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}), \quad (3.20) $$

we have

$$ \begin{aligned} \Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) &\triangleq (x_1 - x_2) \left[ \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})}{\partial x_1} - \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})}{\partial x_2} \right] \\ &= p(x_1 - x_2) \{(x_1+1)^{p-1}[(x_2+1)^p - 1] - [(x_1+1)^p - 1](x_2+1)^{p-1}\} \\ &\times E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) + p(x_1 - x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \end{aligned} \quad (3.21) \\ &= p(x_1 - x_2) \left[ \frac{(x_1+1)^{p-1}}{(x_1+1)^p - 1} - \frac{(x_2+1)^{p-1}}{(x_2+1)^p - 1} \right] \prod_{i=1}^{2} [(x_i+1)^p - 1]E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad + p(x_1 - x_2)[(x_1+1)^{p-1} - (x_2+1)^{p-1}]E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}). \end{aligned} $$

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

1. if $0 < p \le 1$, then $\Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) \le 0$ for $\mathbf{x} \in \mathbb{R}_+^n$, so the function $E_n((\mathbf{x}+\mathbf{1})^p - \mathbf{1})$ is Schur-concave function on $\mathbb{R}_+^n$;

2. if $p < 0$ and $k$ is an even (or odd, respectively) integer, then $\Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) \ge 0$ for $\mathbf{x} \in \mathbb{R}_+^n$. This shows from Lemma 1 that, if $p < 0$ and $n$ is an even (or odd, respectively) integer, the function $E_n((\mathbf{x}+\mathbf{1})^p - \mathbf{1})$ is Schur-concave (or Schur-convex, respectively) on $\mathbb{R}_+^n$;

3. if $p > 1$ and $\mathbf{x} \in (0, c]^n$, from (3.21), it follows that

$$ \begin{aligned} \Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) &= p(x_1 - x_2) \{-(x_1 - x_2)(x_1 + 1)^{p-1}(x_2 + 1)^{p-1} \\ &\quad - [(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}]\} E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad + p(x_1 - x_2)[(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}] E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &= p(x_1 - x_2) E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \left\{ -(x_1 - x_2)(x_1 + 1)^{p-1}(x_2 + 1)^{p-1} \right. \\ &\quad \left. + [(x_1 + 1)^{p-1} - (x_2 + 1)^{p-1}] \left[ \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})} - 1 \right] \right\}. \end{aligned} \quad (3.22) $$

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

$$ (x_1 + 1)^{p-1} - (x_2 + 1)^{p-1} = (p-1)(x_1 - x_2)(\zeta + 1)^{p-2} \quad (3.23) $$

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

$$ \frac{(x_1 + 1)^{p-1}(x_2 + 1)^{p-1}}{(p-1)(\zeta + 1)^{p-2}} \ge \frac{1}{(p-1)(\zeta + 1)^p} \ge \frac{1}{(p-1)(c+1)^p}. $$
For $k = 2, \dots, n$ and $x \in (0, c]^n$, using Lemma 5 yields

$$ \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})} \le \frac{(n-k)\hat{F}_1((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})}{(k-1)\hat{F}_0((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})} \le \frac{n-k}{k-1}(c+1)^p - \frac{n-k}{k-1}. \quad (3.24) $$

For $x \in (0, c]^n$, by the equation (3.22) and the inequalities in (3.23) and (3.24), we obtain

$$ \begin{aligned} \Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) &= p(x_1 - x_2)^2 E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad \times \left\{ -(x_1+1)^{p-1}(x_2+1)^{p-1} + (p-1)(\zeta+1)^{p-2} \left[ \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})} - 1 \right] \right\} \\ &\le p(x_1 - x_2)^2 (p-1)(\zeta+1)^{p-2} E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad \times \left[ -\frac{(x_1+1)^{p-1}(x_2+1)^{p-1}}{(p-1)(\zeta+1)^{p-2}} + \frac{n-k}{(k-1)(n-2)} \sum_{i=3}^{n} (x_i+1)^p - \frac{n-1}{k-1} \right] \\ &\le p(x_1 - x_2)^2 (p-1)(\zeta+1)^{p-2} E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad \times \left[ -\frac{1}{(p-1)(c+1)^p} + \frac{n-k}{k-1}(c+1)^p - \frac{n-1}{k-1} \right] = 0. \end{aligned} $$

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

1. if $0 < p \le 1$, the function $E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})$ is Schur-concave on $\mathbb{R}_+^n$;

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

3. if $p > 1$, the function $E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})$ is Schur-concave on $(0, c]^n$.

Case 3. If $2 \le k \le n-1$, then from (3.20) it follows that

$$ \begin{aligned} \Delta_G(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) &\triangleq (\ln x_1 - \ln x_2) \left[ x_1 \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})}{\partial x_1} - x_2 \frac{\partial E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})}{\partial x_2} \right] \\ &= p(\ln x_1 - \ln x_2) \left[ \frac{x_1(x_1+1)^{p-1}}{(x_1+1)^p - 1} - \frac{x_2(x_2+1)^{p-1}}{(x_2+1)^p - 1} \right] \prod_{i=1}^2 [(x_i+1)^p - 1] E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad + p(\ln x_1 - \ln x_2) [x_1(x_1+1)^{p-1} - x_2(x_2+1)^{p-1}] E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}). \end{aligned} $$

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

1. if $p \ge 1$, then $\Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) \ge 0$ for $x \in \mathbb{R}_+^n$;

2. if $p < 0$ and $k$ is an even (or odd, respectively) integer, then $\Delta(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) \le 0$ for $x \in (0, \frac{1}{|p|}]^n$;

3. if $0 < p < 1$, then

$$ \begin{aligned} \Delta_G(E_k((\mathbf{x}+\mathbf{1})^p - \mathbf{1})) &= p(\ln x_1 - \ln x_2) \{x_1(x_1+1)^{p-1}[(x_2+1)^p - 1] \\ &\quad -[(x_2+1)^p - 1]x_2(x_2+1)^{p-1}\} E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &\quad +p(\ln x_1 - \ln x_2) [x_1(x_1+1)^{p-1} - x_2(x_2+1)^{p-1}] E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \\ &= p(\ln x_1 - \ln x_2) E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1}) \left\{ (x_1 - x_2)(x_1+1)^{p-1}(x_2+1)^{p-1} \right. \\ &\quad \left. + [x_1(x_1+1)^{p-1} - x_2(x_2+1)^{p-1}] \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p - \mathbf{1})} - 1 \right\}. \end{aligned} \quad (3.25) $$

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

$$ x_1(x_1+1)^{p-1} - x_2(x_2+1)^{p-1} = (x_1 - x_2)(\zeta+1)^{p-2}(1+p\zeta) \quad (3.26) $$
for some $\zeta \in (\min\{x_1, x_2\}, \max\{x_1, x_2\})$. Therefore, using Lemma 5 leads to

$$ \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1})}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1})} \le \frac{n-k}{(k-1)(n-2)} \sum_{i=3}^{n} (x_i+1)^p - \frac{n-k}{k-1}, \quad \mathbf{x} \in \mathbb{R}_+^n. \quad (3.27) $$

For $\mathbf{x} \in (0, d]^n$, substituting (3.26) and (3.27) into (3.25) yields

$$ \begin{aligned} & \Delta_G(E_k((\mathbf{x}+\mathbf{1})^p-\mathbf{1})) = p(\ln x_1 - \ln x_2)(x_1-x_2)E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1}) \\ & \times \left\{ (x_1+1)^{p-1}(x_2+1)^{p-1} + (\zeta+1)^{p-2}(1+p\zeta) \left[ \frac{E_{k-1}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1})}{E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1})} - 1 \right] \right\} \\ & \le p(\ln x_1 - \ln x_2)(x_1-x_2)E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1}) \left\{ (x_1+1)^{p-1}(x_2+1)^{p-1} \right. \\ & \qquad \left. + (\zeta+1)^{p-2}(1+p\zeta) \left[ \frac{n-k}{(k-1)(n-2)} \sum_{i=3}^{n} (x_i+1)^p - \frac{n-1}{k-1} \right] \right\} \\ & = p(\ln x_1 - \ln x_2)(x_1-x_2)(\zeta+1)^{p-2}(1+p\zeta)E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1}) \\ & \times \left[ \frac{(x_1+1)^{p-1}(x_2+1)^{p-1}}{(\zeta+1)^{p-2}(1+p\zeta)} + \frac{n-k}{(k-1)(n-2)} \sum_{i=3}^{n} (x_i+1)^p - \frac{n-1}{k-1} \right] \\ & \le p(\ln x_1 - \ln x_2)(x_1-x_2)(\zeta+1)^{p-2}(1+p\zeta)E_{k-2}((\hat{\mathbf{x}}+\mathbf{1})^p-\mathbf{1}) \\ & \times \left[ (d+1)^{2p} + \frac{n-k}{k-1}(d+1)^p - \frac{n-1}{k-1} \right] = 0. \end{aligned} $$

Therefore, using Lemma 2, we obtain that

1. if $p \ge 1$, the function $E_k((\mathbf{x}+\mathbf{1})^p-\mathbf{1})$ is Schur-geometrically convex on $\mathbb{R}_+^n$;

2. if $p < 0$ and $k$ is an even (or odd, respectively) integer, the function $E_k((\mathbf{x}+\mathbf{1})^p-\mathbf{1})$ is Schur-geometrically concave (or convex, respectively) function on $(0, \frac{1}{|p|}]^n$;

3. if $0 < p < 1$, the function $E_k((\mathbf{x}+\mathbf{1})^p-\mathbf{1})$ is Schur-geometrically concave on $(0, d]^n$.

For $k = n$, by the relations (3.3) and (3.4) and by Definitions 2 and 3, the inequalities in (3.18) and (3.19) hold. Theorem 3 is thus proved. $\square$

**Remark 1.** This paper is a corrected and revised version of the preprint [28].

**Acknowledgments:** The authors are thankful to anonymous referees for their careful corrections to and valuable comments on the original version of this paper.

**Author Contributions:** The authors contributed equally to this work. The authors read and approved the final manuscript.

**Funding:** The first two authors and the last author were partially supported by the National Natural Science Foundation of China under Grant No. 11361038, by the Foundation of the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region under Grant No. NJZZ18154, and by the Inner Mongolia Autonomous Region Natural Science Foundation Project under Grant No. 2018LH01002, China.

**Conflicts of Interest:** The authors declare no conflict of interest.

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