On Satnoianu-Wu’s Inequality

Abstract

By applying techniques in the theory of convex functions and Schur-geometrically convex functions, the author investigates a conjecture of Satnoianu on an algebraic inequality and generalizes some known results in recent years.

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1. Introduction

For $n\in \mathbb{N}=\{\mathrm{1,2},\mathrm{3},\dots \}$ and $x_i\in {ℝ}_{+}=(\mathrm{0},\infty )$ with $i=\mathrm{1,2},\dots ,n$ , let $$\begin{array}{c}\text{(1)}& A_n\left(x\right)=\frac{\mathrm{1}}{n}\sum _{i=\mathrm{1}}^n‍x_i,\hspace{1em}\hspace{1em}{G}_n\left(x\right)={\left(\prod _{i=\mathrm{1}}^n‍x_i\right)}^{\mathrm{1}/n},H_n\left(x\right)=\frac{n}{{\sum }_{i=\mathrm{1}}^n\left(\mathrm{1}/x_i\right)}.\end{array}$$ These quantities are, respectively, called the arithmetic, geometric, and harmonic means of a positive sequence $x=(x_{\mathrm{1}},x_{\mathrm{2}},\dots ,x_n)$ . For more information on the theory of means, see the monograph [1] or the papers [2–5] and plenty of references therein.

For convenience, in what follows, we use the notation $x^q=(x_{\mathrm{1}}^q,x_{\mathrm{2}}^q,\dots ,x_n^q)$ for $$\begin{array}{}\text{(2)}& \mathrm{0}<m\le x_{\mathrm{1}}^q,x_{\mathrm{2}}^q,\dots ,x_n^q\le M,\end{array}$$ and $q\in ℝ=(-\infty ,\infty )$ .

In [6], Satnoianu posed the following conjecture.

Conjecture 1.

For $n\ge \mathrm{2}$ , $x_i>\mathrm{0}$ , and $\lambda \ge n^{n-\mathrm{1}}-\mathrm{1}$ , it is valid that $$\begin{array}{}\text{(3)}& \sum _{i=\mathrm{1}}^n‍{\left(\frac{x_i^{n-\mathrm{1}}}{x_i^{n-\mathrm{1}}+\lambda {\prod }_{k=\mathrm{1},k\ne i}^n{x}_k}\right)}^{\mathrm{1}/(n-\mathrm{1})}\ge \frac{n}{{\left(\mathrm{1}+\lambda \right)}^{\mathrm{1}/(n-\mathrm{1})}}.\end{array}$$

This conjecture has been solved and researched in [7–10]. Among them, Wu obtained in [10] the following result.

Theorem 2 (see [10, Theorem  1]).

Let $n$ be a positive integer and let $\alpha$ ,  $\beta$ , and $x_i$ for $i=\mathrm{1,2},\dots ,n$ be positive numbers. If $p\ne \mathrm{0}$ and $$\begin{array}{}\text{(4)}& \beta \ge (n^{\max \{p,\mathrm{1}\}}-\mathrm{1})\alpha ,\end{array}$$ then $$\begin{array}{}\text{(5)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G\left(x^q\right)}\right]}^{\mathrm{1}/\text{p}}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$

It is clear that inequality (5) generalizes (3). Therefore, we would like to call inequality (5) Satnoianu-Wu’s inequality.

Now we naturally pose the following questions:

(1)

Can one improve the condition (4)?

(2)

Does the reversed inequality of (5) exist?

(3)

Are there other types of inequalities of Satnoianu-Wu type?

The goal of this paper is to answer these questions.

2. Definitions and Lemmas

We need the following definitions and lemmas.

Definition 3 (see [11, page 8]).

Let $x=(x_{\mathrm{1}},x_{\mathrm{2}},\dots ,x_n)\in {ℝ}^n$ and $y=(y_{\mathrm{1}},y_{\mathrm{2}},\dots ,y_n)\in {ℝ}^n$ . We say that $x$ is majorized by $y$ (in symbols $x\prec y$ ) if $$\begin{array}{}\text{(6)}& \sum _{i=\mathrm{1}}^k‍x_{\left[i\right]}\le \sum _{i=\mathrm{1}}^k‍y_{\left[i\right]},\hspace{1em}\hspace{1em}\sum _{i=\mathrm{1}}^n‍x_i=\sum _{i=\mathrm{1}}^n‍y_i,\end{array}$$ for $k=\mathrm{1,2},\dots ,n-\mathrm{1}$ , where $x_{[\mathrm{1}]}\ge \cdots \ge x_{[n]}$ and $y_{[\mathrm{1}]}\ge \cdots \ge y_{[n]}$ are rearrangements of $x$ and $y$ in a descending order.

Definition 4 (see [12]).

Let $\Omega \subseteq {ℝ}_{+}^n$ .

(1)

The set $\Omega$ is said to be geometrically convex if $(x_{\mathrm{1}}^{\lambda }{y}_{\mathrm{1}}^{\mathrm{1}-\lambda },\dots ,x_n^{\lambda }{y}_n^{\mathrm{1}-\lambda })\in \Omega$ for every $x,y\in \Omega$ and $\lambda \in [\mathrm{0,1}]$ .

(2)

A function $\phi :\Omega \to {ℝ}_{+}$ is said to be Schur-geometrically convex on $\Omega$ if $\ln x=(\ln x_{\mathrm{1}},\dots ,\ln x_n)\prec \ln y=(\ln y_{\mathrm{1}},\dots ,\ln y_n)$ implies $\phi (x)\le \phi (y)$ for every $x,y\in \Omega$ .

(3)

A function $\phi :\Omega \to {ℝ}_{+}$ is said to be Schur-geometrically concave on $\Omega$ if $\ln x=(\ln x_{\mathrm{1}},\dots ,\ln x_n)\prec \ln y=(\ln y_{\mathrm{1}},\dots ,\ln y_n)$ implies $\phi (x)\ge \phi (y)$ for every $x,y\in \Omega$ .

Lemma 5 (see [12]).

Let $\Omega \subseteq {ℝ}_{+}^n$ be a symmetric and geometrically convex set with inner points and $\phi :\Omega \to {ℝ}_{+}$ a symmetric and differentiable function in ${\Omega }^{\circ }$ . Then $\phi$ is a Schur-geometrically convex (or Schur-geometrically concave, resp.) function on $\Omega$ if and only if $$\begin{array}{}\text{(7)}& \left(\ln x_{\mathrm{1}}-\ln x_{\mathrm{2}}\right)\left[x_{\mathrm{1}}\frac{\partial \phi \left(x\right)}{\partial x_{\mathrm{1}}}-x_{\mathrm{2}}\frac{\partial \phi \left(x\right)}{\partial x_{\mathrm{2}}}\right]\ge \mathrm{0}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(or\le \mathrm{0},\hspace{0.17em}resp.\right),\hspace{1em}x\in {\Omega }^{\circ }.\end{array}$$

Lemma 6 (see [1, page 4, Bernoulli's inequality]).

The inequality $$\begin{array}{}\text{(8)}& {\left(\mathrm{1}+t\right)}^r\ge \mathrm{1}+rt\end{array}$$ holds for $r\ge \mathrm{1}$ and $t\ge -\mathrm{1}$ or for $r\le \mathrm{0}$ and $t>-\mathrm{1}$ . If   $\mathrm{0}<r<\mathrm{1}$ and $t\ge -\mathrm{1}$ , inequality (8) is reversed.

3. Main Results

Now we start off to demonstrate our main results.

Theorem 7.

Let $m,M$ be defined as in (2) and $n\ge \mathrm{2}$ , let $\alpha ,\beta ,x_i\in {ℝ}_{+}$ for $i=\mathrm{1,2},\dots ,n$ , and let $p,q\in ℝ$ with $p\ne \mathrm{0}$ . If $(\mathrm{1}-p)m\beta \ge \mathrm{2}pM\alpha$ , then $$\begin{array}{}\text{(9)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$ If $(\mathrm{1}-p)M\beta \le \mathrm{2}pm\alpha$ , inequality (9) is reversed.

Proof.

For $M^{\prime }>m^{\prime }>\mathrm{0}$ and $r\in ℝ$ with $r\ne \mathrm{0}$ , since $$\begin{array}{}\text{(10)}& {\left[{\left(\frac{t}{\mathrm{1}+t}\right)}^r\right]}^{\mathrm{\prime \prime }}=\frac{r\left(r-\mathrm{1}-\mathrm{2}t\right)t^{r-\mathrm{1}}}{{\left(\mathrm{1}+t\right)}^{r+\mathrm{2}}},\hspace{1em}t\in \left[m^{\prime },M^{\prime }\right],\end{array}$$ then

(1)

for $r<\mathrm{0}$ , or for $r>\mathrm{1}$ and $r-\mathrm{1}-\mathrm{2}{M}^{\prime }>\mathrm{0}$ , the function $(t/(\mathrm{1}+t){)}^r$ is convex on $[m^{\prime },M^{\prime }]$ ;

(2)

for $\mathrm{0}<r\le \mathrm{1}$ , or for $r>\mathrm{1}$ and $r-\mathrm{1}-\mathrm{2}{m}^{\prime }<\mathrm{0}$ , the function $(t/(\mathrm{1}+t){)}^r$ is concave on $[m^{\prime },M^{\prime }]$ .

By Jensen’s inequality, when $r<\mathrm{0}$ , or when $r>\mathrm{1}$ and $r-\mathrm{1}-\mathrm{2}{M}^{\prime }>\mathrm{0}$ , we have $$\begin{array}{}\text{(11)}& \sum _{i=\mathrm{1}}^n‍{\left(\frac{u_i}{\mathrm{1}+u_i}\right)}^r\ge n{\left[\frac{A_n\left(u\right)}{\mathrm{1}+A_n\left(u\right)}\right]}^r,\end{array}$$ where $u=(u_{\mathrm{1}},u_{\mathrm{2}},\dots ,u_n)\in [m^{\prime },M^{\prime }{]}^n$ . When $\mathrm{0}<r\le \mathrm{1}$ , or when $r>\mathrm{1}$ and $r-\mathrm{1}-\mathrm{2}{m}^{\prime }<\mathrm{0}$ , inequality (11) is reversed.

Let $u_i=\alpha x_i^q/\beta A_n(x^q)$ for $i=\mathrm{1,2},\dots ,n$ , $r=1/p$ , $m^{\prime }=\alpha m/\beta M$ , and $M^{\prime }=\alpha M/\beta m$ . Then it is easy to show that $A_n(u)=\alpha /\beta$ . Making use of (11), we obtain inequality (9). The proof of Theorem 7 is complete.

Corollary 8.

Under the conditions of Theorem 7 and when $M=(n-\mathrm{1})m$ , if $(\mathrm{1}-p)\beta >\mathrm{2}p(n-\mathrm{1})\alpha$ , then $$\begin{array}{}\text{(12)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$ if $(\mathrm{1}-p)(n-\mathrm{1})\beta <\mathrm{2}p\alpha$ , inequality (12) is reversed.

Corollary 9.

Let $n\ge \mathrm{2}$ , $\alpha ,\beta ,x_i\in {ℝ}_{+}$ for $i=\mathrm{1,2},\dots ,n$ , and $p,q\in ℝ$ . If $p<\mathrm{0}$ , then $$\begin{array}{}\text{(13)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$ if $p\ge \mathrm{1}$ , inequality (13) is reversed.

Proof.

This follows from Theorem 7 by considering that if $p<\mathrm{0}$ , we have $(\mathrm{1}-p)m\beta >\mathrm{2}pM\alpha$ and that if $p\ge \mathrm{1}$ , we have $(\mathrm{1}-p)M\beta <\mathrm{2}pm\alpha$ .

Theorem 10.

Let $n\ge \mathrm{2}$ , $\alpha ,\beta ,x_i\in {ℝ}_{+}$ for   $i=\mathrm{1,2},\dots ,n$ , $p,q\in ℝ$ with $p\ne \mathrm{0}$ , and $m,M$ defined by (2). If   $m\beta \ge pM\alpha$ , then $$\begin{array}{}\text{(14)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$ if $M\beta \le mp\alpha$ , inequality (14) is reversed.

Proof.

For $M^{\prime }>m^{\prime }>\mathrm{0}$ , let $$\begin{array}{}\text{(15)}& f\left(u\right)=\sum _{i=\mathrm{1}}^n‍{\left(\mathrm{1}+u_i\right)}^r,\hspace{1em}u\in {\left[m^{\prime },M^{\prime }\right]}^n.\end{array}$$ Then $$\begin{array}{c}\text{(16)}& \frac{\partial f\left(u\right)}{\partial u_i}=r{\left(\mathrm{1}+u_i\right)}^{r-\mathrm{1}},\frac{\partial }{\partial u_i}\left[u_i\frac{\partial f\left(u\right)}{\partial u_i}\right]=r{\left(\mathrm{1}+u_i\right)}^{r-\mathrm{2}}\left(\mathrm{1}+r{u}_i\right),\end{array}$$ for $\mathrm{1}\le i\le n$ . When $r>\mathrm{0}$ , or when $r<\mathrm{0}$ and $\mathrm{1}+r{m}^{\prime }<\mathrm{0}$ , we have $$\begin{array}{}\text{(17)}& \left(\ln u_{\mathrm{1}}-\ln u_{\mathrm{2}}\right)\left[u_{\mathrm{1}}\frac{\partial f\left(u\right)}{\partial u_{\mathrm{1}}}-u_{\mathrm{2}}\frac{\partial f\left(u\right)}{\partial u_{\mathrm{2}}}\right]\ge \mathrm{0},\hspace{1em}u\in {\left[m^{\prime },M^{\prime }\right]}^n;\end{array}$$ when $r<\mathrm{0}$ and $\mathrm{1}+r{M}^{\prime }>\mathrm{0}$ , inequality (17) is reversed.

Using Lemma 5, we have the following conclusions:

(1)

if $r>\mathrm{0}$ , or if $r<\mathrm{0}$ and $\mathrm{1}+r{m}^{\prime }<\mathrm{0}$ , the function $f(u)$ is Schur-geometrically convex on $[m^{\prime },M^{\prime }{]}^n$ ;

(2)

if $r<\mathrm{0}$ and $\mathrm{1}+r{M}^{\prime }>\mathrm{0}$ , the function $f(u)$ is Schur-geometrically concave on $[m^{\prime },M^{\prime }{]}^n$ .

By the fact that $$\begin{array}{}\text{(18)}& \left(\ln G_n\left(u\right),\ln G_n\left(u\right),\dots ,\ln G_n\left(u\right)\right)\hspace{1em}\hspace{1em}\prec \left(\ln u_{\mathrm{1}},\ln u_{\mathrm{2}},\dots ,\ln u_n\right),\hspace{1em}u\in {ℝ}_{+}^n,\end{array}$$ and by Definition 4, if $r>\mathrm{0}$ , or if $r<\mathrm{0}$ and $\mathrm{1}+r{m}^{\prime }<\mathrm{0}$ , we have $$\begin{array}{}\text{(19)}& \sum _{i=\mathrm{1}}^n‍{\left(\mathrm{1}+u_i\right)}^r\ge n{\left(\mathrm{1}+G_n\left(u\right)\right)}^r,\hspace{1em}u\in {\left[m^{\prime },M^{\prime }\right]}^n;\end{array}$$ if $r<\mathrm{0}$ and $\mathrm{1}+r{M}^{\prime }>\mathrm{0}$ , inequality (19) is reversed.

Letting $u_i=\beta G_n(x^q)/\alpha x_i^q$ for $i=\mathrm{1,2},\dots ,n$ leads to $G_n(u)=\beta /\alpha$ . Putting $r=-\mathrm{1}/p$ , $m^{\prime }=\beta m/\alpha M$ and $M^{\prime }=\beta M/\alpha m$ in inequality (19) results in inequality (14). The proof of Theorem 10 is complete.

Corollary 11.

Under the conditions of Theorem 10 and when $M=(n-\mathrm{1})m$ , if $\beta \ge p(n-\mathrm{1})\alpha$ , then $$\begin{array}{}\text{(20)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$ if $(n-\mathrm{1})\beta \le p\alpha$ , inequality (20) is reversed.

Remark 12.

When $n\ge \mathrm{2}$ and $p>\mathrm{1}$ , from Lemma 6 and (4), it follows that $$\begin{array}{}\text{(21)}& \beta \ge \left(n^{\max \left\{p,\mathrm{1}\right\}}-\mathrm{1}\right)\alpha >p\left(n-\mathrm{1}\right)\alpha .\end{array}$$

Theorem 13.

Let $n\ge \mathrm{2}$ , $\alpha ,\beta ,x_i\in {ℝ}_{+}$ for $i=\mathrm{1,2},\dots ,n$ , and $p,q\in ℝ$ with $p\ne \mathrm{0}$ . If $p\ge -\mathrm{1}$ , then $$\begin{array}{}\text{(22)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta H_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$ if $p\le -\mathrm{1}$ , inequality (22) is reversed.

Proof.

Since $(\mathrm{1}+t{)}^r$ is a convex (or concave, resp.) function on ${ℝ}_{+}$ for $r\ge \mathrm{1}$ or $r<\mathrm{0}$ (or for $\mathrm{0}<r\le \mathrm{1}$ , resp.), by Jensen's inequality, if $r\ge \mathrm{1}$ or $r<\mathrm{0}$ , we have $$\begin{array}{}\text{(23)}& \sum _{i=\mathrm{1}}^n‍{\left(\mathrm{1}+u_i\right)}^r\ge n{\left(\mathrm{1}+A_n\left(u\right)\right)}^r,\hspace{1em}u\in {ℝ}_{+}^n;\end{array}$$ if $\mathrm{0}<r\le \mathrm{1}$ , inequality (23) is reversed.

Letting $u_i=\beta H_n(x^q)/\alpha x_i^q$ and $r=-\mathrm{1}/p$ shows $A_n(u)=\beta /\alpha$ . Further from (23), we obtain inequality (22). The proof of Theorem 13 is complete.

Remark 14.

It is clear that inequalities (9) and (22) both generalize inequality (3).

Corollary 15.

Let $n\ge \mathrm{2}$ , $\alpha ,\beta ,x_i\in {ℝ}_{+}$ for $i=\mathrm{1,2},\dots ,n$ , $p,q\in ℝ$ with $p\ne \mathrm{0}$ , and $m,M$ defined as in (2).

(1)

When   $-\mathrm{1}\le p<\mathrm{0}$ , one has $$\begin{array}{}\text{(24)}& \sum _{i=\mathrm{1}}^n{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}‍\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta H_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$

(2)

When $\mathrm{0}<p<\mathrm{1}$ and $(\mathrm{1}-p)m\beta >\mathrm{2}pM\alpha$ , one has $$\begin{array}{}\text{(25)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta H_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$

(3)

When $p>\mathrm{0}$ and $m\beta >pM\alpha$ , one has $$\begin{array}{}\text{(26)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta H_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$

(4)

When $p>\mathrm{0}$ and $M\beta <mp\alpha$ , one has $$\begin{array}{}\text{(27)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\le \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\le \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$

Proof.

This follows from utilizing the well-known harmonic-geometric-arithmetic mean inequality $$\begin{array}{}\text{(28)}& H_n\left(x\right)\le G_n\left(x\right)\le A_n\left(x\right)\end{array}$$ and Theorems 7 to 13.

Corollary 16.

Under the conditions of Corollary 15 and when $M=(n-\mathrm{1})m$ ,

(1)

if   $\mathrm{0}<p<\mathrm{1}$ and $(\mathrm{1}-p)\beta >\mathrm{2}p(n-\mathrm{1})\alpha$ , one has $$\begin{array}{}\text{(29)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta H_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$

(2)

if $p>\mathrm{0}$ and $\beta >p(n-\mathrm{1})\alpha$ , one has $$\begin{array}{}\text{(30)}& \sum _{i=\mathrm{1}}^n{\left[\frac{x_i^q}{\alpha x_i^q+\beta H_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\ge \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}};\end{array}$$

(3)

if $p>\mathrm{0}$ and $(n-\mathrm{1})\beta <p\alpha$ , one has $$\begin{array}{}\text{(31)}& \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta A_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\le \sum _{i=\mathrm{1}}^n‍{\left[\frac{x_i^q}{\alpha x_i^q+\beta G_n\left(x^q\right)}\right]}^{\mathrm{1}/p}\le \frac{n}{{\left(\alpha +\beta \right)}^{\mathrm{1}/p}}.\end{array}$$

Acknowledgments

The author thanks four anonymous referees for their careful corrections to and valuable comments on the original version of this paper. This work was partially supported by the Foundation of the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region under Grant no. NJZY13159, China.

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