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Schur convexity for the ratios of the Hamy and generalized Hamy symmetric functions

Journal of Inequalities and Applications20112011:131

https://doi.org/10.1186/1029-242X-2011-131

• Accepted: 5 December 2011
• Published:

Abstract

In this paper, we present the Schur convexity and monotonicity properties for the ratios of the Hamy and generalized Hamy symmetric functions and establish some analytic inequalities. The achieved results is inspired by the paper of Hara et al. [J. Inequal. Appl. 2, 387-395, (1998)], and the methods from Guan [Math. Inequal. Appl. 9, 797-805, (2006)]. The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

2010 Mathematics Subject Classification: Primary 05E05; Secondary 26D20.

Keywords

• Hamy symmetric function
• generalized Hamy symmetric function
• Schur convex
• Schur concave

1 Introduction

Throughout this paper, we denote ${ℝ}_{+}^{n}=\left\{x=\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)|{x}_{i}>0,i=1,2,\dots ,n\right\}.$. For $x\in {ℝ}_{+}^{n}$, the Hamy symmetric function  is defined as
${F}_{n}\left(x,r\right)={F}_{n}\left({x}_{1},{x}_{2},\dots ,{x}_{n};r\right)=\sum _{1\le {i}_{1}<{i}_{2}<\cdots <{i}_{r}\le n}{\left(\prod _{j=1}^{r}{x}_{{i}_{j}}\right)}^{\frac{1}{r}},$
(1.1)

where r is an integer and 1 ≤ rn.

The generalized Hamy symmetric function was introduced by Guan  as follows
${F}_{n}^{*}\left(x,r\right)={F}_{n}^{*}\left({x}_{1},{x}_{2},\dots ,{x}_{n};r\right)=\sum _{{i}_{1}+{i}_{2}+\cdots +{i}_{n}=r}{\left({x}_{1}^{{i}_{1}}{x}_{2}^{{i}_{2}}\dots {x}_{n}^{{i}_{n}}\right)}^{\frac{1}{r}},$
(1.2)

where r is a positive integer.

In , Guan proved that both F n (x,r) and ${F}_{n}^{*}\left(x,r\right)$ are Schur concave in ${ℝ}_{+}^{n}$. The main of this paper is to investigate the Schur convexity for the functions $\frac{{F}_{n}\left(x,r\right)}{{F}_{n}\left(x,r-1\right)}$ and $\frac{{F}_{n}^{*}\left(x,r\right)}{{F}_{n}^{*}\left(x,r-1\right)}$ and establish some analytic inequalities by use of the theory of majorization.

For convenience of readers, we recall some definitions as follows, which can be found in many references, such as .

Definition 1.1. The n-tuple x is said to be majorized by the n-tuple y (in symbols x y), if
$\sum _{i=1}^{k}{x}_{\left[i\right]}\le \sum _{i=1}^{k}{y}_{\left[i\right]},\phantom{\rule{1em}{0ex}}\sum _{i=1}^{n}{x}_{\left[i\right]}=\sum _{i=1}^{n}{y}_{\left[i\right]},$

where 1 ≤ kn - 1, and x[i]denotes the i th largest component of x.

Definition 1.2. Let E n be a set. A real-valued function F : E is said to be Schur convex on E if F(x) ≤ F(y) for each pair of n-tuples x = (x1,..., x n ) and y = (y1,..., y n ) in E, such that x y. F is said to be Schur concave if -F is Schur convex.

The theory of Schur convexity is one of the most important theories in the fields of inequalities. It can be used in combinatorial optimization , isoperimetric problems for polytopes , theory of statistical experiments , graphs and matrices , gamma functions , reliability and availability , optimal designs  and other related fields.

Our aim in what follows is to prove the following results.

Theorem 1.1. Let $x\in {ℝ}_{+}^{n},2\le r\le n$ is an integer, then the function ${\varphi }_{r}\left(x\right)=\frac{{F}_{n}\left(x,r\right)}{{F}_{n}\left(x,r-1\right)}$ is Schur concave in ${ℝ}_{+}^{n}$ and increasing with respect to x i (i= 1,2, ...,n).

Theorem 1.2. Let $x\in {ℝ}_{+}^{n},2\le r\le n$ is an integer, then the function ${\varphi }_{r}^{*}\left(x\right)=\frac{{F}_{n}^{*}\left(x,r\right)}{{F}_{n}^{*}\left(x,r-1\right)}$ is Schur concave in ${ℝ}_{+}^{n}$ and increasing with respect to x i (i= 1,2,... n).

Corollary 1.1. If ${x}_{i}>0,i=1,2,\dots ,n,\phantom{\rule{2.77695pt}{0ex}}{\sum }_{i=1}^{n}{x}_{i}=s$ and that cs, then
$\frac{{G}_{n}\left(x\right)}{{G}_{n}\left(c-x\right)}=\frac{{F}_{n}\left(x,n\right)}{{F}_{n}\left(c-x,n\right)}\le \frac{{F}_{n}\left(x,n-1\right)}{{F}_{n}\left(c-x,n-1\right)}\le \cdots \le \frac{{F}_{n}\left(x,1\right)}{{F}_{n}\left(c-x,1\right)}=\frac{{A}_{n}\left(x\right)}{{A}_{n}\left(c-x\right)}$
and
$\frac{{G}_{n}\left(x\right)}{{G}_{n}\left(c+x\right)}=\frac{{F}_{n}\left(x,n\right)}{{F}_{n}\left(c+x,n\right)}\le \frac{{F}_{n}\left(x,n-1\right)}{{F}_{n}\left(c+x,n-1\right)}\le \cdots \le \frac{{F}_{n}\left(x,1\right)}{{F}_{n}\left(c+x,1\right)}=\frac{{A}_{n}\left(x\right)}{{A}_{n}\left(c+x\right)},$

where ${A}_{n}\left(x\right)=\frac{1}{n}{\sum }_{i=1}^{n}{x}_{i},\phantom{\rule{2.77695pt}{0ex}}{G}_{n}\left(x\right)={\left({\prod }_{i=1}^{n}{x}_{i}\right)}^{\frac{1}{n}}$ are the arithmetic and geo-metric means of x, respectively.

Corollary 1.2. If ${x}_{i}>0,i=1,2,\dots ,n,\phantom{\rule{2.77695pt}{0ex}}{\sum }_{i=1}^{n}{x}_{i}=s$ and that cs, then
$\frac{{F}_{n}^{*}\left(x,r\right)}{{F}_{n}^{*}\left(c-x,r\right)}\le \frac{{F}_{n}^{*}\left(x,r-1\right)}{{F}_{n}^{*}\left(c-x,r-1\right)}\le \cdots \le \frac{{F}_{n}^{*}\left(x,2\right)}{{F}_{n}^{*}\left(c-x,2\right)}\le \frac{{F}_{n}^{*}\left(x,1\right)}{{F}_{n}^{*}\left(c-x,1\right)}=\frac{{A}_{n}\left(x\right)}{{A}_{n}\left(c-x\right)}$
and
$\frac{{F}_{n}^{*}\left(x,r\right)}{{F}_{n}^{*}\left(c+x,r\right)}\le \frac{{F}_{n}^{*}\left(x,r-1\right)}{{F}_{n}^{*}\left(c+x,r-1\right)}\le \cdots \le \frac{{F}_{n}^{*}\left(x,2\right)}{{F}_{n}^{*}\left(c+x,2\right)}\le \frac{{F}_{n}^{*}\left(x,1\right)}{{F}_{n}^{*}\left(c+x,1\right)}=\frac{{A}_{n}\left(x\right)}{{A}_{n}\left(c+x\right)}.$

2 Lemmas

In order to establish our main results, we need several lemmas, which we present in this section.

Lemma 2.1 (see ). Let E n be a symmetric convex set with nonempty interior intE and φ : E be a continuous symmetric function. If φ is differentiable on intE, then φ is Schur convex (or Schur concave, respectively) on E if and only if
$\left({x}_{i}-{x}_{j}\right)\left(\frac{\partial \phi }{\partial {x}_{i}}-\frac{\partial \phi }{\partial {x}_{j}}\right)\ge 0\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}\left(\mathsf{\text{or}}\le 0,\mathsf{\text{respectively}}\right)$

for all i,j = 1,2,...,n and x = (x1,...,x n ) intE.

The r th elementary symmetric function (see ) is defined as
${E}_{n}\left(x,r\right)={E}_{n}\left({x}_{1},{x}_{2},\dots ,{x}_{n};r\right)=\sum _{1\le {i}_{1}<{i}_{2}<\cdots <{i}_{r}\le n}\left(\prod _{j=1}^{r}{x}_{{i}_{j}}\right),$
(2.1)

where 1 ≤ rn is a positive integer, and E n (x, 0) = 1.

By (2.1) and simple computations, we have the following lemma.

Lemma 2.2. Let $x\in {ℝ}_{+}^{n},\phantom{\rule{2.77695pt}{0ex}}1\le i\le n$, if
$\overline{{x}_{i}}=\left({x}_{1},{x}_{2},\dots ,{x}_{i-1},{x}_{i+1},\dots ,{x}_{n}\right).$
Then,
${E}_{n}\left({x}_{1},{x}_{2},\dots ,{x}_{n};r\right)={x}_{i}{E}_{n-1}\left(\overline{{x}_{i}},r-1\right)+{E}_{n-1}\left(\overline{{x}_{i}},r\right).$
(2.2)

Lemma 2.3 (see ). Let $x\in {ℝ}_{+}^{n},\phantom{\rule{0.3em}{0ex}}r$ is an integer and 1 ≤ rn - 1.

Then,
${\left({E}_{n}\left(x,r\right)\right)}^{2}>{E}_{n}\left(x,r-1\right){E}_{n}\left(x,r+1\right).$
(2.3)
Another important symmetric function is the complete symmetric function (see ), which is defined by
${C}_{r}\left(x\right)={C}_{r}\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)=\sum _{{i}_{1}+{i}_{2}+\cdots +{i}_{n}=r}{x}_{1}^{{i}_{1}}{x}_{2}^{{i}_{2}}\dots {x}_{n}^{{i}_{n}},$

where i1, i2,..., i n are non-negative integer, r {1, 2,...} and C0(x) = 1.

Lemma 2.4 (see ). Let x i > 0, i = 1, 2,..., n, and $\overline{{x}_{i}}=\left({x}_{1},{x}_{2},\dots ,{x}_{i-1},{x}_{i+1},\dots ,{x}_{n}\right)$.

Then,
${C}_{r}\left(x\right)={x}_{i}{C}_{r-1}\left(x\right)+{C}_{r}\left(\overline{{x}_{i}}\right).$
Lemma 2.5 (see ). If $0, then
${C}_{r}\left(x\right){C}_{s-1}\left(x\right)\phantom{\rule{2.77695pt}{0ex}}>{C}_{r-1}{C}_{s}\left(x\right).$
Lemma 2.6 (see ). If ${x}_{i}>0,i=1,2,\dots ,n,{\sum }_{i=1}^{n}{x}_{i}=s$ and cs, then
1. (1)

$\frac{c-x}{\frac{nc}{s}-1}=\left(\frac{c-{x}_{1}}{\frac{nc}{s}-1},\frac{c-{x}_{2}}{\frac{nc}{s}-1},\dots ,\frac{c-{x}_{n}}{\frac{nc}{s}-1}\right)\prec \left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)=x,$,

(2)$\frac{c+x}{s+nc}=\left(\frac{c+{x}_{1}}{s+nc},\frac{c+{x}_{2}}{s+nc},\dots ,\frac{c+{x}_{n}}{s+nc}\right)\prec \left(\frac{{x}_{1}}{s},\frac{{x}_{2}}{s},\dots ,\frac{{x}_{n}}{s}\right)=\frac{x}{s}$.

3 Proof of Theorems

Proof of Theorem 1.1. It is obvious that ϕ r (x) is symmetric and has continuous partial derivatives in ${ℝ}_{+}^{n}$. By Lemma 2.1, we only need to prove that
$\left({x}_{1}-{x}_{2}\right)\left(\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{1}}-\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{2}}\right)\le 0.$
(3.1)
For any fixed 2 ≤ rn, let ${u}_{i}=\sqrt[r]{{x}_{i}},\phantom{\rule{2.77695pt}{0ex}}i=1,2,\dots ,n$ and $u=\left({u}_{1},{u}_{2},\dots ,{u}_{n}\right)\in {ℝ}_{+}^{n}$, we have
${\varphi }_{r}\left(x\right)=\frac{{F}_{n}\left(x,r\right)}{{F}_{n}\left(x,r-1\right)}=\frac{{E}_{n}\left(u,r\right)}{{E}_{n}\left(u,r-1\right)}.$
Differentiating ϕ r (x) with respect to x1 yields
$\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{1}}=\frac{1}{{E}_{n}^{2}\left(u,r-1\right)}\left[{E}_{n}\left(u,r-1\right)\frac{\partial {E}_{n}\left(u,r\right)}{\partial {u}_{1}}\frac{\partial {u}_{1}}{\partial {x}_{1}}-{E}_{n}\left(u,r\right)\frac{\partial {E}_{n}\left(u,r-1\right)}{\partial {u}_{1}}\frac{\partial {u}_{1}}{\partial {x}_{1}}\right].$
(3.2)
Using Lemma 2.2 repeatedly, we get
$\begin{array}{ll}\hfill {E}_{n}\left(u,r\right)& ={u}_{1}{u}_{2}{E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-2\right)+\left({u}_{1}+{u}_{2}\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-1\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r\right).\phantom{\rule{2em}{0ex}}\end{array}$
(3.3)
Equations (3.2) and (3.3) lead to
$\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{1}}=\frac{1}{r{E}_{n}^{2}\left(u,r-1\right)}\left({u}_{1}^{1-r}{u}_{2}A+{u}_{1}^{1-r}B\right),$
(3.4)
where
$A={E}_{n}\left(u,r-1\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-2\right)-{E}_{n}\left(u,r\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-3\right)$
and
$B={E}_{n}\left(u,r-1\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-1\right)-{E}_{n}\left(u,r\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-2\right).$
Similarly, we can deduce that
$\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{2}}=\frac{1}{r{E}_{n}^{2}\left(u,r-1\right)}\left({u}_{1}{u}_{2}^{1-r}A+{u}_{2}^{1-r}B\right).$
(3.5)
From (3.4) and (3.5), one has
$\begin{array}{c}\left({x}_{1}-{x}_{2}\right)\left(\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{1}}-\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{2}}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{{x}_{1}-{x}_{2}}{r{E}_{n}^{2}\left(u,r-1\right)}\left[{x}_{1}^{\frac{1}{r}}{x}_{2}^{\frac{1}{r}}\left({x}_{1}^{-1}-{x}_{2}^{-1}\right)A+\left({x}_{1}^{\frac{1}{r}-1}-{x}_{2}^{\frac{1}{r}-1}\right)B\right].\end{array}$
(3.6)
It follows from (3.3) and Lemma 2.3 that
$\begin{array}{c}A=\left({u}_{1}+{u}_{2}\right)\left[{E}_{n-2}^{2}\left({u}_{3},\dots ,{u}_{n};r-2\right)-{E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-1\right)\\ \phantom{\rule{0.5em}{0ex}}×{E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-3\right)\right]+{E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-1\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-2\right)\\ \phantom{\rule{0.5em}{0ex}}-{E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r\right){E}_{n-2}\left({u}_{3},\dots ,{u}_{n};r-3\right)\\ >0.\end{array}$

Similarly, we can get B > 0.

It follows from the function ${x}^{\frac{k-r}{r}}\left(k=0,1\right)$ is decreasing in (0, +∞) that
$\left({x}_{1}-{x}_{2}\right)\left({x}_{1}^{\frac{k-r}{r}}-{x}_{2}^{\frac{k-r}{r}}\right)\le 0,\left(k=0,1\right).$
(3.7)

Therefore, inequality (3.1) follows from (3.6) and (3.7) together with A > 0 and B > 0.

Next, we prove that ${\varphi }_{r}\left(x\right)=\frac{{F}_{n}\left(x,r\right)}{{F}_{n}\left(x,r-1\right)}$ is increasing with respect to x i (i= 1,2,...,n).

By the symmetry of ϕ r (x) with respect to x i (i = 1, 2,..., n), we only need to prove that
$\frac{\partial {\varphi }_{r}\left(x\right)}{\partial {x}_{1}}\ge 0,$

which can be derived directly from A > 0 and B > 0 together with Equation (3.4).

Proof of Theorem 1.2. It is obvious that ${\varphi }_{r}^{*}\left(x\right)$ is symmetric and has continuous partial derivatives in ${ℝ}_{+}^{n}$. By Lemma 2.1, we only need to prove that
$\left({x}_{1}-{x}_{2}\right)\left(\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{1}}-\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{2}}\right)\le 0.$
(3.8)
For any fixed 2 ≤ rn, let ${u}_{i}=\sqrt[r]{{x}_{i}},\phantom{\rule{2.77695pt}{0ex}}i=1,2,\dots ,n$ and $u=\left({u}_{1},{u}_{2},\dots ,{u}_{n}\right)\in {ℝ}_{+}^{n}$. Then,
${\varphi }_{r}^{*}\left(x\right)=\frac{{F}_{n}^{*}\left(x,r\right)}{{F}_{n}^{*}\left(x,r-1\right)}=\frac{{C}_{r}\left(u\right)}{{C}_{r-1}\left(u\right)}.$
(3.9)
Differentiating ${\varphi }_{r}^{*}\left(x\right)$ with respect to x1, we have
$\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{1}}=\frac{1}{{C}_{r-1}^{2}\left(u\right)}\left[{C}_{r-1}\left(u\right)\frac{\partial {C}_{r}\left(u\right)}{\partial {u}_{1}}\frac{\partial {u}_{1}}{\partial {x}_{1}}-{C}_{r}\left(u\right)\frac{\partial {C}_{r-1}\left(u\right)}{\partial {u}_{1}}\frac{\partial {u}_{1}}{\partial {x}_{1}}\right].$
(3.10)
It follows from Lemma 2.4 that
$\begin{array}{ll}\hfill \frac{\partial {C}_{r}\left(u\right)}{\partial {u}_{1}}& ={C}_{r-1}\left(u\right)+{u}_{1}\frac{\partial {C}_{r-1}\left(u\right)}{\partial {u}_{1}}\phantom{\rule{2em}{0ex}}\\ ={C}_{r-1}\left(u\right)+{u}_{1}\left[{C}_{r-2}\left(u\right)+{u}_{1}\frac{\partial {C}_{r-2}\left(u\right)}{\partial {u}_{1}}\right]\phantom{\rule{2em}{0ex}}\\ ={C}_{r-1}\left(u\right)+{u}_{1}{C}_{r-2}\left(u\right)+{u}_{1}^{2}\frac{\partial {C}_{r-2}\left(u\right)}{\partial {u}_{1}}\phantom{\rule{2em}{0ex}}\\ =\cdots \cdots \phantom{\rule{2em}{0ex}}\\ ={C}_{r-1}\left(u\right)+{u}_{1}{C}_{r-2}\left(u\right)+{u}_{1}^{2}{C}_{r-3}\left(u\right)+\cdots +{u}_{1}^{r-2}{C}_{1}\left(u\right)+{u}_{1}^{r-1}.\phantom{\rule{2em}{0ex}}\end{array}$
(3.11)
Equations (3.10) and (3.11) lead to
$\begin{array}{c}\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{1}}=\frac{1}{{C}_{r-1}^{2}\left(u\right)}\left\{\left[{C}_{r-1}^{2}\left(u\right)-{C}_{r}\left(u\right){C}_{r-2}\left(u\right)\right]+{u}_{1}\left[{C}_{r-1}\left(u\right){C}_{r-2}\left(u\right)\\ -{C}_{r}\left(u\right){C}_{r-3}\left(u\right)\right]+\cdots +{u}_{1}^{r-2}\left[{C}_{r-1}\left(u\right){C}_{1}\left(u\right)-{C}_{r}\left(u\right){C}_{0}\left(u\right)\right]\\ +{C}_{r-1}\left(u\right){u}_{1}^{r-1}\right\}\frac{1}{r}{u}_{1}^{1-r}.\end{array}$
(3.12)
Similarly, we have
$\begin{array}{c}\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{2}}=\frac{1}{{C}_{r-1}^{2}\left(u\right)}\left\{\left[{C}_{r-1}^{2}\left(u\right)-{C}_{r}\left(u\right){C}_{r-2}\left(u\right)\right]+{u}_{2}\left[{C}_{r-1}\left(u\right){C}_{r-2}\left(u\right)\\ -{C}_{r}\left(u\right){C}_{r-3}\left(u\right)\right]+\cdots +{u}_{2}^{r-2}\left[{C}_{r-1}\left(u\right){C}_{1}\left(u\right)-{C}_{r}\left(u\right){C}_{0}\left(u\right)\right]\\ +{C}_{r-1}\left(u\right){u}_{2}^{r-1}\right\}\frac{1}{r}{u}_{2}^{1-r}.\end{array}$
(3.13)
From (3.12) and (3.13), one has
$\begin{array}{l}\left({x}_{1}-{x}_{2}\right)\left(\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{1}}-\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{2}}\right)\\ =\frac{{x}_{1}-{x}_{2}}{r{C}_{r-1}^{2}\left(u\right)}\left\{\left[{C}_{r-1}^{2}\left(u\right)-{C}_{r}\left(u\right){C}_{r-2}\left(u\right)\right]\left({x}_{1}^{\frac{1-r}{r}}-{x}_{2}^{\frac{1-r}{r}}\right)+\left[{C}_{r-1}\left(u\right){C}_{r-2}\left(u\right)\\ -{C}_{r}\left(u\right){C}_{r-3}\left(u\right)\right]\left({x}_{1}^{\frac{2-r}{r}}-{x}_{2}^{\frac{2-r}{r}}\right)+\cdots +\left[{C}_{r-1}\left(u\right){C}_{1}\left(u\right)-{C}_{r}\left(u\right){C}_{0}\left(u\right)\right]\\ ×\left({x}_{1}^{\frac{\left(r-1\right)-r}{r}}-{x}_{2}^{\frac{\left(r-1\right)-r}{r}}\right)\right\}.\end{array}$
(3.14)
By Lemma 2.5, we know that
$\begin{array}{c}{C}_{r-1}^{2}\left(u\right)-{C}_{r}\left(u\right){C}_{r-2}\left(u\right)>0,\\ {C}_{r-1}\left(u\right){C}_{r-2}\left(u\right)-{C}_{r}\left(u\right){C}_{r-3}\left(u\right)>0,\\ \cdots \cdots \phantom{\rule{0.3em}{0ex}},\\ {C}_{r-1}\left(u\right){C}_{1}\left(u\right)-{C}_{r}\left(u\right){C}_{0}\left(u\right)>0.\end{array}$
(3.15)
The monotonicity of the function ${x}^{\frac{j-r}{r}}\left(1\le j\le r-1\right)$ in (0, +∞) leads to the conclusion that
$\left({x}_{1}-{x}_{2}\right)\left({x}_{1}^{\frac{j-r}{r}}-{x}_{2}^{\frac{j-r}{r}}\right)\le 0.$
(3.16)

Therefore, inequality (3.8) follows from (3.14)-(3.16).

Next, we prove that ${\varphi }_{r}^{*}\left(x\right)=\frac{{F}_{n}^{*}\left(x,r\right)}{{F}_{n}^{*}\left(x,r-1\right)}$ is increasing with respect to x i (i= 1,2,...,n).

From (3.12) and (3.15), we clearly see that
$\frac{\partial {\varphi }_{r}^{*}\left(x\right)}{\partial {x}_{1}}\ge 0.$
(3.17)

Inequality (3.17) implies that ${\varphi }_{r}^{*}\left(x\right)$ is increasing with respect to x1, then from the symmetry of ${\varphi }_{r}^{*}\left(x\right)$ with respect to x i (i = 1, 2,..., n) we know that ${\varphi }_{r}^{*}\left(x\right)$ is increasing with respect to each x i (i = 1, 2,..., n).

Proof of Corollary 1.1. By Theorem 1.1 and Lemma 2.6, we have ${\varphi }_{r}\left(\frac{c-x}{\frac{nc}{s}-1}\right)\ge {\varphi }_{r}\left(x\right)$ and ${\varphi }_{r}\left(\frac{c+x}{s+nc}\right)\ge {\varphi }_{r}\left(\frac{x}{s}\right)$ which imply Corollary 1.1.

Remark 1. Let $0<{x}_{i}\le \frac{1}{2},\phantom{\rule{2.77695pt}{0ex}}i=1,2,\dots ,n$, then
$\frac{{G}_{n}\left(x\right)}{{G}_{n}\left(1-x\right)}\le \frac{{A}_{n}\left(x\right)}{{A}_{n}\left(1-x\right)},$
(3.18)

where (1 - x) = (1 - x1, 1 - x2,... , 1 - x n ), commonly referred to as Ky Fan inequality (see ), which has attracted the attention of a considerable number of mathematicians (see ).

Letting ${\sum }_{i=1}^{n}{x}_{i}\le 1$ and taking c = 1 in Corollary 1.1, we get
$\frac{{G}_{n}\left(x\right)}{{G}_{n}\left(1-x\right)}=\frac{{F}_{n}\left(x,n\right)}{{F}_{n}\left(1-x,n\right)}\le \frac{{F}_{n}\left(x,n-1\right)}{{F}_{n}\left(1-x,n-1\right)}\le \cdots \le \frac{{F}_{n}\left(x,1\right)}{{F}_{n}\left(1-x,1\right)}=\frac{{A}_{n}\left(x\right)}{{A}_{n}\left(1-x\right)}.$
(3.19)

It is obvious that inequality (3.19) can be called Ky Fan-type inequality.

Remark 2. Let x i > 0, i = 1, 2,..., n, the following inequalities
$\prod _{i=1}^{n}\left({x}_{i}^{-1}-1\right)\ge {\left(n-1\right)}^{n}$
and
$\prod _{i=1}^{n}\left({x}_{i}^{-1}+1\right)\ge {\left(n+1\right)}^{n}$

are the well-known Weierstrass inequalities (see ).

Taking c = s = 1 in Corollary 1.1, one has
$\prod _{i=1}^{n}\left({x}_{i}^{-1}-1\right)\ge {\left(\frac{{F}_{n}\left(1-x,n-1\right)}{{F}_{n}\left(x,n-1\right)}\right)}^{n}\ge \cdots \ge {\left(\frac{{F}_{n}\left(1-x,2\right)}{{F}_{n}\left(x,2\right)}\right)}^{n}\ge {\left(n-1\right)}^{n}$
and
$\prod _{i=1}^{n}\left({x}_{i}^{-1}+1\right)\ge {\left(\frac{{F}_{n}\left(1+x,n-1\right)}{{F}_{n}\left(x,n-1\right)}\right)}^{n}\ge \cdots \ge {\left(\frac{{F}_{n}\left(1+x,2\right)}{{F}_{n}\left(x,2\right)}\right)}^{n}\ge {\left(n+1\right)}^{n}.$

It is obvious that our inequalities can be called Weierstrass-type inequalities.

Proof of Corollary 1.2. By Theorem 1.2 and Lemma 2.6, we have ${\varphi }_{r}^{*}\left(\frac{c-x}{\frac{nc}{s}-1}\right)\ge {\varphi }_{r}^{*}\left(x\right)$ and ${\varphi }_{r}^{*}\left(\frac{c+x}{s+nc}\right)\ge {\varphi }_{r}^{*}\left(\frac{x}{s}\right)$, which imply Corollary 1.2.

Declarations

Acknowledgements

This work was supported by NSF of China under grant No. 11071069.

Authors’ Affiliations

(1)
Huzhou Broadcast and TV University, Huzhou, 313000, China

References 