Schur-convexity of dual form of some symmetric functions

Abstract

By the properties of a Schur-convex function, Schur-convexity of the dual form of some symmetric functions is simply proved.

MSC:26D15, 05E05, 26B25.

1 Introduction

Throughout the article, ℝ denotes the set of real numbers, $\mathit{x}=(x_1,x_2,\dots ,x_n)$ denotes n-tuple (n-dimensional real vectors), the set of vectors can be written as

$$\begin{array}{c}{\mathbb{R}}^n=\{\mathit{x}=(x_1,\dots ,x_n):x_i\in \mathbb{R},i=1,\dots ,n\},\hfill \\ {\mathbb{R}}_{+}^n=\{\mathit{x}=(x_1,\dots ,x_n):x_i>0,i=1,\dots ,n\}.\hfill \end{array}$$

In particular, the notations ℝ and ${\mathbb{R}}_{+}$ denote ${\mathbb{R}}^1$ and ${\mathbb{R}}_{+}^1$, respectively.

For convenience, we introduce some definitions as follows.

Definition 1 [1, 2]

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

• (i) $\mathit{x}\ge \mathit{y}$ means $x_i\ge y_i$ for all $i=1,2,\dots ,n$.

• (ii) Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$, $\phi :\mathrm{\Omega }\to \mathbb{R}$ is said to be increasing if $\mathit{x}\ge \mathit{y}$ implies $\phi (\mathit{x})\ge \phi (\mathit{y})$. φ is said to be decreasing if and only if −φ is increasing.

Definition 2 [1, 2]

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

• (i) x is said to be majorized by y (in symbols $\mathit{x}\prec \mathit{y}$) if ${\sum }_{i=1}^k{x}_{[i]}\le {\sum }_{i=1}^k{y}_{[i]}$ for $k=1,2,\dots ,n-1$ and ${\sum }_{i=1}^n{x}_i={\sum }_{i=1}^n{y}_i$, where $x_{[1]}\ge \cdots \ge x_{[n]}$ and $y_{[1]}\ge \cdots \ge y_{[n]}$ are rearrangements of x and y in a descending order.

• (ii) Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$, $\phi :\mathrm{\Omega }\to \mathbb{R}$ is said to be a Schur-convex function on Ω if $\mathit{x}\prec \mathit{y}$ on Ω implies $\phi (\mathit{x})\le \phi (\mathit{y})$. φ is said to be a Schur-concave function on Ω if and only if −φ is Schur-convex function on Ω.

Definition 3 [1, 2]

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

• (i) $\mathrm{\Omega }\subset {\mathbb{R}}^n$ is said to be a convex set if $\mathit{x},\mathit{y}\in \mathrm{\Omega }$, $0\le \alpha \le 1$ implies $\alpha \mathit{x}+(1-\alpha )\mathit{y}=(\alpha x_1+(1-\alpha )y_1,\dots ,\alpha x_n+(1-\alpha )y_n)\in \mathrm{\Omega }$.

• (ii) Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$ be a convex set. A function $\phi :\mathrm{\Omega }\to \mathbb{R}$ is said to be a convex function on Ω if

  $$\phi (\alpha \mathit{x}+(1-\alpha )\mathit{y})\le \alpha \phi (\mathit{x})+(1-\alpha )\phi (\mathit{y})$$

for all $\mathit{x},\mathit{y}\in \mathrm{\Omega }$ and all $\alpha \in [0,1]$. φ is said to be a concave function on Ω if and only if −φ is a convex function on Ω.

• (iii) Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$. A function $\phi :\mathrm{\Omega }\to \mathbb{R}$ is said to be a log-convex function on Ω if the function lnφ is convex.

Definition 4 [1]

• (i) $\mathrm{\Omega }\subset {\mathbb{R}}^n$ is called a symmetric set, if $x\in \mathrm{\Omega }$ implies $Px\in \mathrm{\Omega }$ for every $n\times n$ permutation matrix P.

• (ii) The function $\phi :\mathrm{\Omega }\to \mathbb{R}$ is called symmetric if for every permutation matrix P, $\phi (Px)=\phi (x)$ for all $x\in \mathrm{\Omega }$.

Theorem A (Schur-convex function decision theorem [[1], p.84])

Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$ be symmetric and have a nonempty interior convex set. ${\mathrm{\Omega }}^0$ is the interior of Ω. $\phi :\mathrm{\Omega }\to \mathbb{R}$ is continuous on Ω and differentiable in ${\mathrm{\Omega }}^0$. Then φ is the Schur-convex (Schur-concave) function if and only if φ is symmetric on Ω and

$$(x_1-x_2)(\frac{\partial \phi }{\partial x_1}-\frac{\partial \phi }{\partial x_2})\ge 0(\le 0)$$

(1)

holds for any $\mathit{x}\in {\mathrm{\Omega }}^0$.

The Schur-convex functions were introduced by Schur in 1923 and have important applications in analytic inequalities, elementary quantum mechanics and quantum information theory. See [1].

In recent years, many scholars use the Schur-convex function decision theorem to determine the Schur-convexity of many symmetric functions.

Xia et al. [3] proved that the symmetric function

$$E_k\left(\frac{\mathit{x}}{1+\mathit{x}}\right)=\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k\frac{x_{i_j}}{1+x_{i_j}},k=1,\dots ,n,$$

(2)

is Schur-convex on ${\mathbb{R}}_{+}^n$.

Chu et al. [4] proved that the symmetric function

$$E_k\left(\frac{\mathit{x}}{1-\mathit{x}}\right)=\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k\frac{x_{i_j}}{1-x_{i_j}},k=1,\dots ,n,$$

(3)

is Schur-convex on ${[\frac{k-1}{2(n-1)},1)}^n$ and Schur-concave on ${[0,\frac{k-1}{2(n-1)}]}^n$.

Xia and Chu [5] proved that the symmetric function

$$E_k\left(\frac{1-\mathit{x}}{\mathit{x}}\right)=\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k\frac{1-x_{i_j}}{x_{i_j}},k=1,\dots ,n,$$

(4)

is Schur-convex on ${(0,\frac{2n-k-1}{2(n-1)}]}^n$ and Schur-concave on ${[\frac{2n-k-1}{2(n-1)},1]}^n$.

Xia and Chu [6] also proved that the symmetric function

$$E_k\left(\frac{1+\mathit{x}}{1-\mathit{x}}\right)=\sum _{1\le i_1<\cdots <i_k\le n}\prod _{j=1}^k\frac{1+x_{i_j}}{1-x_{i_j}},k=1,\dots ,n,$$

(5)

is Schur-convex on ${(0,1)}^n$.

Mei et al. [7] proved that the symmetric function

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

(6)

is Schur-convex on ${(0,1)}^n$. More results for Schur convexity of the symmetric functions, we refer the reader to [8].

In this paper, by the properties of a Schur-convex function, we study Schur-convexity of the dual form of the above symmetric functions, and we obtained the following results.

Theorem 1 The symmetric function

$$E_k^{\ast }\left(\frac{\mathit{x}}{1+\mathit{x}}\right)=\prod _{1\le i_1<\cdots <i_k\le n}\sum _{j=1}^k\frac{x_{i_j}}{1+x_{i_j}},k=1,\dots ,n,$$

(7)

is a Schur-concave function on ${\mathbb{R}}_{+}^n$.

Theorem 2 The symmetric function

$$E_k^{\ast }\left(\frac{\mathit{x}}{1-\mathit{x}}\right)=\prod _{1\le i_1<\cdots <i_k\le n}\sum _{j=1}^k\frac{x_{i_j}}{1-x_{i_j}},k=1,\dots ,n,$$

(8)

is a Schur-convex function on ${[\frac{1}{2},1)}^n$.

Theorem 3 The symmetric function

$$E_k^{\ast }\left(\frac{1-\mathit{x}}{\mathit{x}}\right)=\prod _{1\le i_1<\cdots <i_k\le n}\sum _{j=1}^k\frac{1-x_{i_j}}{x_{i_j}},k=1,\dots ,n,$$

(9)

is a Schur-convex function on ${(0,\frac{1}{2}]}^n$.

Theorem 4 The symmetric function

$$E_k^{\ast }\left(\frac{1+\mathit{x}}{1-\mathit{x}}\right)=\prod _{1\le i_1<\cdots <i_k\le n}\sum _{j=1}^k\frac{1+x_{i_j}}{1-x_{i_j}},k=1,\dots ,n,$$

(10)

is a Schur-convex function on ${(0,1)}^n$.

Theorem 5 The symmetric function

$$E_k^{\ast }(\frac{1}{\mathit{x}}-\mathit{x})=\prod _{1\le i_1<\cdots <i_k\le n}\sum _{j=1}^k(\frac{1}{x_{i_j}}-x_{i_j}),k=1,\dots ,n,$$

(11)

is a Schur-convex function on ${(0,\sqrt{\sqrt{5}-2})}^n$.

2 Lemmas

To prove the above three theorems, we need the following lemmas.

Lemma 1 ([[1], p.97], [2])

If φ is symmetric and convex (concave) on a symmetric convex set Ω, then φ is Schur-convex (Schur-concave) on Ω.

Lemma 2 [[2], p.64]

Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$, $\phi :\mathrm{\Omega }\to {\mathbb{R}}_{+}$. Then logφ is Schur-convex (Schur-concave) if and only if φ is Schur-convex (Schur-concave).

Lemma 3 ([[1], p.642], [2])

Let $\mathrm{\Omega }\subset {\mathbb{R}}^n$ be an open convex set, $\phi :\mathrm{\Omega }\to \mathbb{R}$. For $\mathit{x},\mathit{y}\in \mathrm{\Omega }$, define one variable function $g(t)=\phi (t\mathit{x}+(1-t)\mathit{y})$ on the interval $(0,1)$. Then φ is convex (concave) on Ω if and only if g is convex (concave) on $[0,1]$ for all $\mathit{x},\mathit{y}\in \mathrm{\Omega }$.

Lemma 4 Let $\mathit{x}=(x_1,\dots ,x_m)$ and $\mathit{y}=(y_1,\dots ,y_m)\in {\mathbb{R}}_{+}^m$. Then the function $p(t)=logg(t)$ is concave on $[0,1]$, where

$$g(t)=\sum _{j=1}^m\frac{t{x}_j+(1-t)y_j}{1+t{x}_j+(1-t)y_j}.$$

Proof

$$p^{\mathrm{\prime }}(t)=\frac{g^{\mathrm{\prime }}(t)}{g(t)},$$

where

$$\begin{array}{c}{g}^{\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{x_j-y_j}{{(1+t{x}_j+(1-t)y_j)}^2},\hfill \\ p^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{g^{\mathrm{\prime }\mathrm{\prime }}(t)g(t)-{(g^{\mathrm{\prime }}(t))}^2}{g^2(t)},\hfill \end{array}$$

where

$$g^{\mathrm{\prime }\mathrm{\prime }}(t)=-\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(1+t{x}_j+(1-t)y_j)}^3}.$$

Thus,

$$\begin{array}{c}{g}^{\mathrm{\prime }\mathrm{\prime }}(t)g(t)-{(g^{\mathrm{\prime }}(t))}^2\hfill \\ =(-\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(1+t{x}_j+(1-t)y_j)}^3})\left(\sum _{j=1}^m\frac{t{x}_j+(1-t)y_j}{1+t{x}_j+(1-t)y_j}\right)\hfill \\ -{\left(\sum _{j=1}^m\frac{x_j-y_j}{{(1+t{x}_j+(1-t)y_j)}^2}\right)}^2\hfill \\ \le 0,\hfill \end{array}$$

and then $p^{\mathrm{\prime }\mathrm{\prime }}(t)\le 0$, that is, $p(t)$ is concave on $[0,1]$.

The proof of Lemma 4 is completed. □

Lemma 5 Let $\mathit{x}=(x_1,\dots ,x_m)$ and $\mathit{y}=(y_1,\dots ,y_m)\in {[\frac{1}{2},1)}^m$. Then the function $q(t)=log\psi (t)$ is convex on $[0,1]$, where

$$\psi (t)=\sum _{j=1}^m\frac{t{x}_j+(1-t)y_j}{1-t{x}_j-(1-t)y_j}.$$

Proof

$$q^{\mathrm{\prime }}(t)=\frac{{\psi }^{\mathrm{\prime }}(t)}{\psi (t)},$$

where

$$\begin{array}{c}{\psi }^{\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{x_j-y_j}{{(1-t{x}_j-(1-t)y_j)}^2},\hfill \\ q^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{{\psi }^{\mathrm{\prime }\mathrm{\prime }}(t)\psi (t)-{({\psi }^{\mathrm{\prime }}(t))}^2}{{\psi }^2(t)},\hfill \end{array}$$

where

$${\psi }^{\mathrm{\prime }\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(1-t{x}_j-(1-t)y_j)}^3}.$$

By the Cauchy inequality, we have

$$\begin{array}{c}{\psi }^{\mathrm{\prime }\mathrm{\prime }}(t)\psi (t)-{({\psi }^{\mathrm{\prime }}(t))}^2\hfill \\ =\left(\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(1-t{x}_j-(1-t)y_j)}^3}\right)\left(\sum _{j=1}^m\frac{t{x}_j+(1-t)y_j}{1-t{x}_j-(1-t)y_j}\right)-{\left(\sum _{j=1}^m\frac{x_j-y_j}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2\hfill \\ \ge {\left(\sum _{j=1}^m\frac{\sqrt{2}|x_j-y_j|}{{(1-t{x}_j-(1-t)y_j)}^{\frac{3}{2}}}\frac{\sqrt{t{x}_j+(1-t)y_j}}{\sqrt{1-t{x}_j-(1-t)y_j}}\right)}^2-{\left(\sum _{j=1}^m\frac{x_j-y_j}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2\hfill \\ ={\left(\sum _{j=1}^m\frac{\sqrt{2}|x_j-y_j|\sqrt{t{x}_j+(1-t)y_j}}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2-{\left(\sum _{j=1}^m\frac{x_j-y_j}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2.\hfill \end{array}$$

From $x_j,y_j\in [\frac{1}{2},1)$ it follows that $\sqrt{2}\sqrt{t{x}_j+(1-t)y_j}\ge 1$, hence ${\psi }^{\mathrm{\prime }\mathrm{\prime }}(t)\psi (t)-{({\psi }^{\mathrm{\prime }}(t))}^2\ge 0$, and then $q^{\mathrm{\prime }\mathrm{\prime }}(t)\ge 0$, that is, $q(t)$ is convex on $[0,1]$.

The proof of Lemma 5 is completed. □

Lemma 6 Let $\mathit{x}=(x_1,\dots ,x_m)$ and $\mathit{y}=(y_1,\dots ,y_m)\in {(0,\frac{1}{2}]}^m$. Then the function $r(t)=log\phi (t)$ is convex on $[0,1]$, where

$$\phi (t)=\sum _{j=1}^m\frac{1-t{x}_j-(1-t)y_j}{t{x}_j+(1-t)y_j}.$$

Proof

$$r^{\mathrm{\prime }}(t)=\frac{{\phi }^{\mathrm{\prime }}(t)}{\phi (t)},$$

where

$$\begin{array}{c}{\phi }^{\mathrm{\prime }}(t)=-\sum _{j=1}^m\frac{x_j-y_j}{{(t{x}_j+(1-t)y_j)}^2},\hfill \\ r^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{{\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)\phi (t)-{({\phi }^{\mathrm{\prime }}(t))}^2}{{\phi }^2(t)},\hfill \end{array}$$

where

$${\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(t{x}_j+(1-t)y_j)}^3}.$$

By the Cauchy inequality, we have

$$\begin{array}{c}{\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)\phi (t)-{({\phi }^{\mathrm{\prime }}(t))}^2\hfill \\ =\left(\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(t{x}_j+(1-t)y_j)}^3}\right)\left(\sum _{j=1}^m\frac{1-t{x}_j-(1-t)y_j}{t{x}_j+(1-t)y_j}\right)-{(-\sum _{j=1}^m\frac{x_j-y_j}{{(t{x}_j+(1-t)y_j)}^2})}^2\hfill \\ \ge {\left(\sum _{j=1}^m\frac{\sqrt{2}|x_j-y_j|}{{(t{x}_j+(1-t)y_j)}^{\frac{3}{2}}}\frac{\sqrt{1-t{x}_j-(1-t)y_j}}{\sqrt{t{x}_j+(1-t)y_j}}\right)}^2-{\left(\sum _{j=1}^m\frac{x_j-y_j}{{(t{x}_j+(1-t)y_j)}^2}\right)}^2\hfill \\ ={\left(\sum _{j=1}^m\frac{\sqrt{2}|x_j-y_j|\sqrt{1-t{x}_j-(1-t)y_j}}{{(t{x}_j+(1-t)y_j)}^2}\right)}^2-{\left(\sum _{j=1}^m\frac{x_j-y_j}{{(t{x}_j+(1-t)y_j)}^2}\right)}^2.\hfill \end{array}$$

From $x_j,y_j\in (0,\frac{1}{2}]$ it follows that $\sqrt{2}\sqrt{1-t{x}_j-(1-t)y_j}\ge 1$, hence ${\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)\phi (t)-{({\phi }^{\mathrm{\prime }}(t))}^2\ge 0$, and then $r^{\mathrm{\prime }\mathrm{\prime }}(t)\ge 0$, that is, $r(t)$ is convex on $[0,1]$.

The proof of Lemma 6 is completed. □

Lemma 7 Let $\mathit{x}=(x_1,\dots ,x_m)$ and $\mathit{y}=(y_1,\dots ,y_m)\in {(0,1)}^m$. Then the function $h(t)=logf(t)$ is convex on $[0,1]$, where

$$f(t)=\sum _{j=1}^m\frac{1+t{x}_j+(1-t)y_j}{1-t{x}_j-(1-t)y_j}.$$

Proof

$$h^{\mathrm{\prime }}(t)=\frac{f^{\mathrm{\prime }}(t)}{f(t)},$$

where

$$\begin{array}{c}{f}^{\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{2(x_j-y_j)}{{(1-t{x}_j-(1-t)y_j)}^2},\hfill \\ h^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{f^{\mathrm{\prime }\mathrm{\prime }}(t)f(t)-{(f^{\mathrm{\prime }}(t))}^2}{f^2(t)},\hfill \end{array}$$

where

$$f^{\mathrm{\prime }\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{4{(x_j-y_j)}^2}{{(1-t{x}_j-(1-t)y_j)}^3}.$$

By the Cauchy inequality, we have

$$\begin{array}{c}{f}^{\mathrm{\prime }\mathrm{\prime }}(t)f(t)-{(f^{\mathrm{\prime }}(t))}^2\hfill \\ =\left(\sum _{j=1}^m\frac{4{(x_j-y_j)}^2}{{(1-t{x}_j-(1-t)y_j)}^3}\right)\left(\sum _{j=1}^m\frac{1+t{x}_j+(1-t)y_j}{1-t{x}_j-(1-t)y_j}\right)\hfill \\ -{\left(\sum _{j=1}^m\frac{2(x_j-y_j)}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2\hfill \\ \ge {\left(\sum _{j=1}^m\frac{2|x_j-y_j|}{{(1-t{x}_j-(1-t)y_j)}^{\frac{3}{2}}}\frac{\sqrt{1+t{x}_j+(1-t)y_j}}{\sqrt{1-t{x}_j-(1-t)y_j}}\right)}^2-{\left(\sum _{j=1}^m\frac{2(x_j-y_j)}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2\hfill \\ ={\left(\sum _{j=1}^m\frac{2|x_j-y_j|\sqrt{1+t{x}_j+(1-t)y_j}}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2-{\left(\sum _{j=1}^m\frac{2(x_j-y_j)}{{(1-t{x}_j-(1-t)y_j)}^2}\right)}^2.\hfill \end{array}$$

From $x_j,y_j\in (0,1)$ it follows that $\sqrt{2}\sqrt{1+t{x}_j+(1-t)y_j}\ge 1$, hence $f^{\mathrm{\prime }\mathrm{\prime }}(t)f(t)-{(f^{\mathrm{\prime }}(t))}^2\ge 0$, and then $h^{\mathrm{\prime }\mathrm{\prime }}(t)\ge 0$, that is, $h(t)$ is convex on $[0,1]$.

The proof of Lemma 7 is completed. □

Lemma 8 Let $\mathit{x}=(x_1,\dots ,x_m)$ and $\mathit{y}=(y_1,\dots ,y_m)\in {(0,\sqrt{\sqrt{5}-2})}^m$. Then the function $s(t)=logw(t)$ is convex on $[0,1]$, where

$$w(t)=\sum _{j=1}^m(\frac{1}{t{x}_j+(1-t)y_j}-(t{x}_j+(1-t)y_j)).$$

Proof

$$s^{\mathrm{\prime }}(t)=\frac{w^{\mathrm{\prime }}(t)}{w(t)},$$

where

$$\begin{array}{c}{w}^{\mathrm{\prime }}(t)=-\sum _{j=1}^m(x_j-y_j)(\frac{1}{{(t{x}_j+(1-t)y_j)}^2}+1),\hfill \\ s^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{w^{\mathrm{\prime }\mathrm{\prime }}(t)w(t)-{(w^{\mathrm{\prime }}(t))}^2}{w^2(t)},\hfill \end{array}$$

where

$$w^{\mathrm{\prime }\mathrm{\prime }}(t)=\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(t{x}_j+(1-t)y_j)}^3}.$$

By the Cauchy inequality, we have

$$\begin{array}{c}{w}^{\mathrm{\prime }\mathrm{\prime }}(t)w(t)-{(w^{\mathrm{\prime }}(t))}^2\hfill \\ =\left(\sum _{j=1}^m\frac{2{(x_j-y_j)}^2}{{(t{x}_j+(1-t)y_j)}^3}\right)\left(\sum _{j=1}^m(\frac{1}{t{x}_j+(1-t)y_j}-(t{x}_j+(1-t)y_j))\right)\hfill \\ -{(-\sum _{j=1}^m(x_j-y_j)(\frac{1}{{(t{x}_j+(1-t)y_j)}^2}+1))}^2\hfill \\ \ge {\left(\sum _{j=1}^m\frac{\sqrt{2}|x_j-y_j|}{{(t{x}_j+(1-t)y_j)}^{\frac{3}{2}}}\sqrt{\frac{1}{t{x}_j+(1-t)y_j}-(t{x}_j+(1-t)y_j)}\right)}^2\hfill \\ -{(\sum _{j=1}^m(x_j-y_j)(\frac{1}{{(t{x}_j+(1-t)y_j)}^2}+1))}^2\hfill \\ ={\left(\sum _{j=1}^m\frac{\sqrt{2}|x_j-y_j|\sqrt{1-{(t{x}_j+(1-t)y_j)}^2}}{{(t{x}_j+(1-t)y_j)}^2}\right)}^2-{(\sum _{j=1}^m(x_j-y_j)\frac{1+{(t{x}_j+(1-t)y_j)}^2}{{(t{x}_j+(1-t)y_j)}^2})}^2.\hfill \end{array}$$

Let $u_j:=t{x}_j+(1-t)y_j$. From $x_j,y_j\in (0,\sqrt{\sqrt{5}-2})$ it follows that $u_j^2\le \sqrt{5}-2$. Since

$$\begin{array}{c}{u}_j^2\le \sqrt{5}-2\iff {(u_j^2+2)}^2\le 5\iff u_j^4+4u_j^2-1\le 0\hfill \\ \iff 2(1-u_j^2)\ge {(1+u_j^2)}^2\iff \sqrt{2}\sqrt{1-u_j^2}\ge 1+u_j^2,\hfill \end{array}$$

so $w^{\mathrm{\prime }\mathrm{\prime }}(t)w(t)-{(w^{\mathrm{\prime }}(t))}^2\ge 0$, and then $s^{\mathrm{\prime }\mathrm{\prime }}(t)\ge 0$, that is, $s(t)$ is convex on $[0,1]$.

The proof of Lemma 8 is completed. □

3 Proof of main results

Proof of Theorem 4 For any $1\le i_1<\cdots <i_k\le n$, by Lemma 3 and Lemma 7, it follows that $log{\sum }_{j=1}^k\frac{1+x_{i_j}}{1-x_{i_j}}$ is convex on ${(0,1)}^k$. Obviously, $log{\sum }_{j=1}^k\frac{1+x_{i_j}}{1-x_{i_j}}$ is also convex on ${(0,1)}^n$, and then $log{E}_k^{\ast }(\frac{1+\mathit{x}}{1-\mathit{x}})={\sum }_{1\le i_1<\cdots <i_k\le n}log{\sum }_{j=1}^k\frac{1+x_{i_j}}{1-x_{i_j}}$ is convex on ${(0,1)}^n$. Furthermore, it is clear that $log{E}_k^{\ast }(\frac{1+\mathit{x}}{1-\mathit{x}})$ is symmetric on ${(0,1)}^n$. By Lemma 1, it follows that $log{E}_k^{\ast }(\frac{1+\mathit{x}}{1-\mathit{x}})$ is Schur-convex on ${(0,1)}^n$, and then from Lemma 2 we conclude that $E_k^{\ast }(\frac{1+\mathit{x}}{1-\mathit{x}})$ is also Schur-convex on ${(0,1)}^n$.

The proof of Theorem 4 is completed. □

Similar to the proof of Theorem 4, we can use Lemma 4, Lemma 5, Lemma 6 and Lemma 8 respectively to prove Theorem 1, Theorem 2, Theorem 3 and Theorem 5; therefore we omit the details of the proof.

Remark 1 Using the Schur-convex function decision theorem, Liu et al. [9] have proved Theorem 3. Xia and Chu [10] have proved that the symmetric function

$$E_k^{\ast }\left(\frac{1+\mathit{x}}{\mathit{x}}\right)=\prod _{1\le i_1<\cdots <i_k\le n}\sum _{j=1}^k\frac{1+x_{i_j}}{x_{i_j}},k=1,\dots ,n,$$

(12)

is a Schur-convex function on ${\mathbb{R}}_{+}^n$.

The reader may wish to prove the inequality (12) by the properties of a Schur-convex function.