An extension of Jensen’s discrete inequality to partially convex functions

Abstract

This paper deals with a new extension of Jensen’s discrete inequality to a partially convex function f, which is defined on a real interval $\mathbb{I}$, convex on a subinterval $[a,b]\subset \mathbb{I}$, decreasing for $u\le c$ and increasing for $u\ge c$, where $c\in [a,b]$. Several relevant applications are given to show the effectiveness of the proposed partially convex function theorem.

MSC:26D07, 26D10, 41A44.

1 Introduction

Let $\mathbf{x}=\{x_1,x_2,\dots ,x_n\}$ be a sequence of real numbers belonging to a given real interval $\mathbb{I}$, and let $\mathbf{p}=\{p_1,p_2,\dots ,p_n\}$ be a sequence of given positive weights associated to x and satisfying $p_1+p_2+\cdots +p_n=1$. If f is a convex function on $\mathbb{I}$, then

$$\sum _{i=1}^n{p}_{i}f(x_i)\ge f\left(\sum _{i=1}^n{p}_i{x}_i\right)$$

is the classical Jensen discrete inequality (see [1, 2]).

In [3], we extended the weighted Jensen discrete inequality to a half convex function f, defined on a real interval $\mathbb{I}$ and convex for $u\le s$ or $u\ge s$, where $s\in \mathbb{I}$.

WHCF-Theorem Let f be a function defined on a real interval $\mathbb{I}$ and convex for $u\le s$ or $u\ge s$, where $s\in \mathbb{I}$, and let $p_1,p_2,\dots ,p_n$ be positive real numbers such that

$$p=min\{p_1,p_2,\dots ,p_n\},p_1+p_2+\cdots +p_n=1.$$

The inequality

$$p_{1}f(x_1)+p_{2}f(x_2)+\cdots +p_{n}f(x_n)\ge f(s)$$

holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ satisfying $p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s$ if and only if

$$pf(x)+(1-p)f(y)\ge f(s)$$

for all $x,y\in \mathbb{I}$ such that $px+(1-p)y=s$.

For the particular case $p_1=p_2=\cdots =p_n=1/n$, from the weighted half convex function theorem, we get the half convex function theorem (see [4, 5]).

HCF-Theorem Let f be a function defined on a real interval $\mathbb{I}$ and convex for $u\le s$ or $u\ge s$, where $s\in \mathbb{I}$. The inequality

$$f(x_1)+f(x_2)+\cdots +f(x_n)\ge nf(s)$$

holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ satisfying $x_1+x_2+\cdots +x_n=ns$ if and only if

$$f(x)+(n-1)f(y)\ge nf(s)$$

for all $x,y\in \mathbb{I}$ which satisfy $x+(n-1)y=ns$.

Applying HCF-Theorem and WHCF-Theorem to the function f defined by $f(u)=g(e^u)$ and replacing s by lnr, x by lnx, y by lny, and each $x_i$ by $ln{a}_i$ for $i=1,2,\dots ,n$, we get the following corollaries, respectively.

HCF-Corollary Let g be a function defined on a positive interval $\mathbb{I}$ such that the function f defined by $f(u)=g(e^u)$ is convex for $e^u\le r$ or $e^u\ge r$, where $r\in \mathbb{I}$. The inequality

$$g(a_1)+g(a_2)+\cdots +g(a_n)\ge ng(r)$$

holds for all $a_1,a_2,\dots ,a_n\in \mathbb{I}$ satisfying $a_1{a}_2\cdots a_n=r^n$ if and only if

$$g(a)+(n-1)g(b)\ge ng(r)$$

for all $a,b\in \mathbb{I}$ which satisfy $a{b}^{n-1}=r^n$.

WHCF-Corollary Let g be a function defined on a positive interval $\mathbb{I}$ such that the function f defined by $f(u)=g(e^u)$ is convex for $e^u\le r$ or $e^u\ge r$, where $r\in \mathbb{I}$, and let $p_1,p_2,\dots ,p_n$ be positive real numbers such that

$$p=min\{p_1,p_2,\dots ,p_n\},p_1+p_2+\cdots +p_n=1.$$

The inequality

$$p_{1}g(a_1)+p_{2}g(a_2)+\cdots +p_{n}g(a_n)\ge g(r)$$

holds for all $a_1,a_2,\dots ,a_n\in \mathbb{I}$ satisfying $a_1^{p_1}{a}_2^{p_2}\cdots a_n^{p_n}=r$ if and only if

$$pg(a)+(1-p)g(b)\ge g(r)$$

for all $a,b\in \mathbb{I}$ such that $a^p{b}^{1-p}=r$.

In this paper, we will use HCF-Theorem and WHCF-Theorem to extend Jensen’s inequality to partially convex functions, which are defined on a real interval $\mathbb{I}$ and convex only on a subinterval $[a,b]\subset \mathbb{I}$.

Remark 1.1 Clearly, HCF-Theorem is a particular case of WHCF-Theorem. However, we posted here the both theorems because HCF-Theorem is much more useful to prove many inequalities of extended Jensen type.

Remark 1.2

Actually, in HCF-Theorem and WHCF-Theorem, it suffices to consider that

$$x\ge s\ge y$$

when f is convex for $u\le s$, and

$$x\le s\le y$$

when f is convex for $u\ge s$ (see [3]). Also, in HCF-Corollary and WHCF-Corollary, it suffices to consider that

$$a\ge r\ge b$$

when f is convex for $e^u\le r$, and

$$a\le r\le b$$

when f is convex for $e^u\ge r$.

2 Main results

The main results of the paper are given by the following two theorems: partially convex function theorem (PCF-Theorem) and weighted partially convex function theorem (WPCF-Theorem).

PCF-Theorem Let f be a function defined on a real interval $\mathbb{I}$, decreasing for $u\le s_0$ and increasing for $u\ge s_0$, where $s_0\in \mathbb{I}$. In addition, assume that f is convex on $[s_0,s]$ or $[s,s_0]$, where $s\in \mathbb{I}$. The inequality

$$f(x_1)+f(x_2)+\cdots +f(x_n)\ge nf(s)$$

holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ satisfying $x_1+x_2+\cdots +x_n=ns$ if and only if

$$f(x)+(n-1)f(y)\ge nf(s)$$

for all $x,y\in \mathbb{I}$ which satisfy $x+(n-1)y=ns$.

WPCF-Theorem Let f be a function defined on a real interval $\mathbb{I}$, decreasing for $u\le s_0$ and increasing for $u\ge s_0$, where $s_0\in \mathbb{I}$, and let $p_1,p_2,\dots ,p_n$ be positive real numbers such that

$$p=min\{p_1,p_2,\dots ,p_n\},p_1+p_2+\cdots +p_n=1.$$

In addition, assume that f is convex on $[s_0,s]$ or $[s,s_0]$, where $s\in \mathbb{I}$. The inequality

$$p_{1}f(x_1)+p_{2}f(x_2)+\cdots +p_{n}f(x_n)\ge f(s)$$

holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ satisfying $p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s$ if and only if

$$pf(x)+(1-p)f(y)\ge f(s)$$

for all $x,y\in \mathbb{I}$ such that $px+(1-p)y=s$.

Applying PCF-Theorem and WPCF-Theorem to the function f defined by $f(u)=g(e^u)$ and replacing $s_0$ by $ln{r}_0$, s by lnr, x by lnx, y by lny, and each $x_i$ by $ln{a}_i$ for $i=1,2,\dots ,n$, we get the following corollaries, respectively.

PCF-Corollary Let g be a function defined on a positive interval $\mathbb{I}$, decreasing for $t\le r_0$ and increasing for $t\ge r_0$, where $r_0\in \mathbb{I}$. In addition, assume that the function f defined by $f(u)=g(e^u)$ is convex for $r_0\le e^u\le r$ or $r\le e^u\le r_0$, where $r\in \mathbb{I}$. The inequality

$$g(a_1)+g(a_2)+\cdots +g(a_n)\ge ng(r)$$

holds for all $a_1,a_2,\dots ,a_n\in \mathbb{I}$ satisfying $a_1{a}_2\cdots a_n=r^n$ if and only if

$$g(a)+(n-1)g(b)\ge ng(r)$$

for all $a,b\in \mathbb{I}$ which satisfy $a{b}^{n-1}=r^n$.

WPCF-Corollary Let g be a function defined on a positive interval $\mathbb{I}$, decreasing for $t\le r_0$ and increasing for $t\ge r_0$, where $r_0\in \mathbb{I}$, and let $p_1,p_2,\dots ,p_n$ be positive real numbers such that

$$p=min\{p_1,p_2,\dots ,p_n\},p_1+p_2+\cdots +p_n=1.$$

In addition, assume that the function f defined by $f(u)=g(e^u)$ is convex for $r_0\le e^u\le r$ or $r\le e^u\le r_0$, where $r\in \mathbb{I}$. The inequality

$$p_{1}g(a_1)+p_{2}g(a_2)+\cdots +p_{n}g(a_n)\ge g(r)$$

holds for all $a_1,a_2,\dots ,a_n\in \mathbb{I}$ satisfying $a_1^{p_1}{a}_2^{p_2}\cdots a_n^{p_n}=r$ if and only if

$$pg(a)+(1-p)g(b)\ge g(r)$$

for all $a,b\in \mathbb{I}$ such that $a^p{b}^{1-p}=r$.

In order to prove WPCF-Theorem, we need the following lemmas.

Lemma 2.1 Let f be a function defined on a real interval $\mathbb{I}$, decreasing for $u\le s_0$ and increasing for $u\ge s_0$, where $s_0\in \mathbb{I}$, and let $p_1,p_2,\dots ,p_n$ be positive real numbers such that

$$p_1+p_2+\cdots +p_n=1.$$

For $s\in \mathbb{I}$, $s\ge s_0$, if the inequality

$$p_{1}f(x_1)+p_{2}f(x_2)+\cdots +p_{n}f(x_n)\ge f(s)$$

holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ such that

$$x_1,x_2,\dots ,x_n\ge s_0,p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s,$$

then it holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ such that

$$p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s.$$

Lemma 2.2 Let f be a function defined on a real interval $\mathbb{I}$, decreasing for $u\le s_0$ and increasing for $u\ge s_0$, where $s_0\in \mathbb{I}$, and let $p_1,p_2,\dots ,p_n$ be positive real numbers such that

$$p_1+p_2+\cdots +p_n=1.$$

For $s\in \mathbb{I}$, $s\le s_0$, if the inequality

$$p_{1}f(x_1)+p_{2}f(x_2)+\cdots +p_{n}f(x_n)\ge f(s)$$

holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ such that

$$x_1,x_2,\dots ,x_n\le s_0,p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s,$$

then it holds for all $x_1,x_2,\dots ,x_n\in \mathbb{I}$ such that

$$p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s.$$

Notice that in the case $s\ge s_0$, WPCF-Theorem follows immediately from Lemma 2.1 and WHCF-Theorem applied to the interval

$${\mathbb{I}}_0=\{u\in \mathbb{I}\mid u\ge s_0\},$$

because f is convex for $u\in {\mathbb{I}}_0$, $u\le s$. Also, in the case $s\le s_0$, WPCF-Theorem follows immediately from Lemma 2.2 and WHCF-Theorem applied to the interval

$${\mathbb{I}}_0=\{u\in \mathbb{I}\mid u\le s_0\},$$

because f is convex for $u\in {\mathbb{I}}_0$, $u\ge s$.

Remark 2.3

According to Remark 1.2, it suffices to consider in PCF-Theorem and WPCF-Theorem that

$$x\ge s\ge y$$

when f is convex on $[s_0,s]$, and

$$x\le s\le y$$

when f is convex on $[s,s_0]$. Also, it suffices to consider in PCF-Corollary and WPCF-Corollary that

$$a\ge r\ge b$$

when f is convex for $r_0\le e^u\le r$, and

$$a\le r\le b$$

when f is convex for $r\le e^u\le r_0$.

Remark 2.4

Let us denote

$$g(u)=\frac{f(u)-f(s)}{u-s},h(x,y)=\frac{g(x)-g(y)}{x-y}.$$

In many applications, it is useful to replace the hypothesis

$$pf(x)+(1-p)f(y)\ge f(s)$$

in WHCF-Theorem and WPCF-Theorem by the equivalent condition:

$$h(x,y)\ge 0\mathrm{\forall }x,y\in \mathbb{I},px+(1-p)y=s.$$

This equivalence is true since

$$\begin{array}{rl}pf(x)+(1-p)f(y)-f(s)& =p[f(x)-f(s)]+(1-p)[f(y)-f(s)]\\ =p(x-s)g(x)+(1-p)(y-s)g(y)\\ =p(1-p)(x-y)[g(x)-g(y)]\\ =p(1-p){(x-y)}^{2}h(x,y).\end{array}$$

In the particular case $p_1=p_2=\cdots =p_n=1/n$, this condition becomes

$$h(x,y)\ge 0\mathrm{\forall }x,y\in \mathbb{I},x+(n-1)y=ns.$$

Remark 2.5 The required inequalities in WHCF-Theorem and WPCF-Theorem turn into equalities for $x_1=x_2=\cdots =x_n=s$. In addition, on the assumption that

$$p=min\{p_1,p_2,\dots ,p_n\},$$

the equality also holds for $x_1=x$ and $x_2=\cdots =x_n=y$ if there exist $x,y\in \mathbb{I}$, $x\ne y$, such that

$$px+(1-p)y=s,pf(x)+(1-p)f(y)=f(s).$$

3 Proof of lemmas

Proof of Lemma 2.1 For $i=1,2,\dots ,n$, define the numbers $y_i\in \mathbb{I}$ as

$$y_i=\{\begin{array}{ll}{s}_0,& x_i\le s_0,\\ x_i,& x_i>s_0.\end{array}$$

Since $y_i\ge x_i$ for $i=1,2,\dots ,n$, we have

$$p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n\ge p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s.$$

In addition, since $f(y_i)\le f(x_i)$ for $x_i\le s_0$ and $f(y_i)=f(x_i)$ for $x_i>s_0$, we have $f(y_i)\le f(x_i)$ for $i=1,2,\dots ,n$, and hence

$$p_{1}f(y_1)+p_{2}f(y_2)+\cdots +p_{n}f(y_n)\le p_{1}f(x_1)+p_{2}f(x_2)+\cdots +p_{n}f(x_n).$$

Thus, it suffices to show that

$$p_{1}f(y_1)+p_{2}f(y_2)+\cdots +p_{n}f(y_n)\ge f(s)$$

for all $y_1,y_2,\dots ,y_n\ge s_0$ such that $p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n\ge s$. By hypothesis, this inequality is true for $y_1,y_2,\dots ,y_n\ge s_0$ and $p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n=s$. Since f is increasing for $u\in \mathbb{I}$, $u\ge s_0$, the more we have $p_{1}f(y_1)+p_{2}f(y_2)+\cdots +p_{n}f(y_n)\ge f(s)$ for $y_1,y_2,\dots ,y_n\ge s_0$ and $p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n\ge s$. □

Proof of Lemma 2.2 For $i=1,2,\dots ,n$, define the numbers $y_i\in \mathbb{I}$ as follows:

$$y_i=\{\begin{array}{ll}{x}_i,& x_i\le s_0,\\ s_0,& x_i>s_0.\end{array}$$

We have $y_i\le s_0$, $y_i\le x_i$ and $f(y_i)\le f(x_i)$ for $i=1,2,\dots ,n$. Therefore,

$$p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n\le p_1{x}_1+p_2{x}_2+\cdots +p_n{x}_n=s$$

and

$$p_{1}f(y_1)+p_{2}f(y_2)+\cdots +p_{n}f(y_n)\le p_{1}f(x_1)+p_{2}f(x_2)+\cdots +p_{n}f(x_n).$$

Thus, it suffices to show that

$$p_{1}f(y_1)+p_{2}f(y_2)+\cdots +p_{n}f(y_n)\ge f(s)$$

for all $y_1,y_2,\dots ,y_n\le s_0$ such that $p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n\le s$. By hypothesis, this inequality is true for $y_1,y_2,\dots ,y_n\le s_0$ and $p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n=s$. Since f is decreasing for $u\in \mathbb{I}$, $u\le s_0$, we have also $p_{1}f(y_1)+p_{2}f(y_2)+\cdots +p_{n}f(y_n)\ge f(s)$ for $y_1,y_2,\dots ,y_n\le s_0$ and $p_1{y}_1+p_2{y}_2+\cdots +p_n{y}_n\le s$. □

4 Applications

Application 4.1 Let $x_1,x_2,\dots ,x_n\ge \frac{-n}{n-2}$ ($n\ge 3$) such that

$$x_1+x_2+\cdots +x_n=n.$$

If $k\ge \frac{n(3n-4)}{{(n-2)}^2}$, then

$$\frac{1-x_1}{k+x_1^2}+\frac{1-x_2}{k+x_2^2}+\cdots +\frac{1-x_n}{k+x_n^2}\ge 0,$$

with equality for $x_1=x_2=\cdots =x_n=1$, and also for $x_1=\frac{-n}{n-2}$ and $x_2=\cdots =x_n=\frac{n}{n-2}$ (or any cyclic permutation).

Proof

Rewrite the desired inequality as

$$f(x_1)+f(x_2)+\cdots +f(x_n)\ge nf(s),$$

where

$$\begin{array}{c}s=1,\hfill \\ f(u)=\frac{1-u}{k+u^2},u\in \mathbb{I}=[\frac{-n}{n-2},\frac{n(2n-3)}{n-2}].\hfill \end{array}$$

We have

$$\begin{array}{c}{f}^{\prime }(u)=\frac{u^2-2u-k}{{(u^2+k)}^2},\hfill \\ f^{″}(u)=\frac{2f_1(u)}{{(u^2+k)}^3},\hfill \end{array}$$

where

$$f_1(u)=-u^3+3u^2+3ku-k=(k+1)(3u-1)-{(u-1)}^3.$$

There are two cases to consider.

Case 1: $\sqrt{k+1}\ge \frac{2{(n-1)}^2}{n-2}$. For $u\in \mathbb{I}$, $u\ge 1$, we have

$$f_1(u)>(k+1)(u-1)-{(u-1)}^3=(u-1)[k+1-{(u-1)}^2]\ge 0,$$

since

$$u-1\le \frac{n(2n-3)}{n-2}-1=\frac{2{(n-1)}^2}{n-2}\le \sqrt{k+1}.$$

Therefore, f is convex for $u\in \mathbb{I}$, $u\ge 1$. By HCF-Theorem, we only need to show that $f(x)+(n-1)f(y)\ge nf(1)$ for all $x,y\in \mathbb{I}$ which satisfy $x+(n-1)y=n$. According to Remark 2.4, this is true if $h(x,y)\ge 0$ for $x,y\in \mathbb{I}$ and $x+(n-1)y=n$, where

$$h(x,y)=\frac{g(x)-g(y)}{x-y},g(u)=\frac{f(u)-f(1)}{u-1}.$$

Indeed, we have

$$g(u)=\frac{-1}{u^2+k}$$

and

$$h(x,y)=\frac{x+y}{(x^2+k)(y^2+k)}=\frac{n+(n-2)x}{(n-1)(x^2+k)(y^2+k)}\ge 0.$$

Case 2: $\frac{2(n-1)}{n-2}\le \sqrt{k+1}<\frac{2{(n-1)}^2}{n-2}$. Since

$$1-\sqrt{1+k}\le \frac{-n}{n-2}$$

and

$$1+\sqrt{1+k}<1+\frac{2{(n-1)}^2}{n-2}=\frac{n(2n-3)}{n-2},$$

from the expression of $f^{\prime }$ it follows that f is decreasing on $[\frac{-n}{n-2},s_0]$ and increasing on $[s_0,\frac{n(2n-3)}{n-2}]$, where

$$s_0=1+\sqrt{k+1}>1.$$

On the other hand, for $u\in [1,s_0]$, we have

$$f_1(u)>(k+1)(u-1)-{(u-1)}^3=(u-1)[k+1-{(u-1)}^2]\ge 0,$$

since

$$u-1\le s_0-1=\sqrt{k+1}.$$

Thus, f is convex on $[1,s_0]$. By PCF-Theorem, we only need to show that $f(x)+(n-1)f(y)\ge nf(1)$ for all $x,y\in \mathbb{I}$ such that $x+(n-1)y=n$. We have proved this before (at Case 1). □

Application 4.2 If $x_1,x_2,\dots ,x_n$ ($n\ge 3$) are real numbers such that

$$x_1+x_2+\cdots +x_n=n,$$

then [6]

$$\sum _{i=1}^n\frac{n(n+1)-2x_i}{n^2+(n-2)x_i^2}\le n,$$

with equality for $x_1=x_2=\cdots =x_n=1$, and also for $x_1=n$ and $x_2=\cdots =x_n=0$ (or any cyclic permutation).

Proof The desired inequality is true for $x_1>\frac{n(n+1)}{2}$ since

$$\frac{n(n+1)-2x_1}{n^2+(n-2)x_1^2}<0$$

and

$$\frac{n(n+1)-2x_i}{n^2+(n-2)x_i^2}<\frac{n}{n-1},i=2,3,\dots ,n.$$

Consider further that $x_1,x_2,\dots ,x_n\le \frac{n(n+1)}{2}$ and rewrite the desired inequality as

$$f(x_1)+f(x_2)+\cdots +f(x_n)\ge nf(s),$$

where

$$\begin{array}{c}s=1,\hfill \\ f(u)=\frac{2u-n(n+1)}{(n-2)u^2+n^2},u\in \mathbb{I}=[\frac{n(3-n^2)}{2},\frac{n(n+1)}{2}].\hfill \end{array}$$

We have

$$\frac{f^{\prime }(u)}{2(n-2)}=\frac{n^2+n(n+1)u-u^2}{{[(n-2)u^2+n^2]}^2}$$

and

$$\frac{f^{″}(u)}{2(n-2)}=\frac{f_1(u)}{{[(n-2)u^2+n^2]}^3},$$

where

$$f_1(u)=2(n-2)u^3-3n(n+1)(n-2)u^2-2n^2(2n-3)u+n^3(n+1).$$

From the expression of $f^{\prime }$, it follows that f is decreasing on $[\frac{n(3-n^2)}{2},s_0]$ and increasing on $[s_0,\frac{n(n+1)}{2}]$, where

$$s_0=\frac{n}{2}(n+1-\sqrt{n^2+2n+5})\in (-1,0).$$

On the other hand, for $-1\le u\le 1$, we have

$$\begin{array}{rl}{f}_1(u)& >-2(n-2)-3n(n+1)(n-2)-2n^2(2n-3)+n^3(n+1)\\ =n^2{(n-3)}^2+4(n+1)>0,\end{array}$$

and hence $f^{″}(u)>0$. Since $[s_0,s]\subset [-1,1]$, f is convex on $[s_0,s]$. By PCF-Theorem, we only need to show that $f(x)+(n-1)f(y)\ge nf(1)$ for all $x,y\in \mathbb{I}$ which satisfy $x+(n-1)y=n$. According to Remark 2.4, this is true if $h(x,y)\ge 0$ for $x,y\in \mathbb{I}$ and $x+(n-1)y=n$, where

$$h(x,y)=\frac{g(x)-g(y)}{x-y},g(u)=\frac{f(u)-f(1)}{u-1}.$$

Indeed, we have

$$g(u)=\frac{(n-2)u+n}{(n-2)u^2+n^2}$$

and

$$\begin{array}{rl}\frac{h(x,y)}{n-2}& =\frac{n^2-n(x+y)-(n-2)xy}{[(n-2)x^2+n^2][(n-2)y^2+n^2]}\\ =\frac{(n-1)(n-2)y^2}{[(n-2)x^2+n^2][(n-2)y^2+n^2]}\ge 0.\end{array}$$

 □

Application 4.3 Let $x_1,x_2,\dots ,x_n$ ($n\ge 2$) be positive real numbers such that

$$x_1+x_2+\cdots +x_n\ge n.$$

If $k>1$, then [6]

$$\frac{x_1}{x_1^k+x_2+\cdots +x_n}+\frac{x_2}{x_1+x_2^k+\cdots +x_n}+\cdots +\frac{x_n}{x_1+x_2+\cdots +x_n^k}\le 1,$$

with equality for $x_1=x_2=\cdots =x_n=1$.

Proof

Using the substitutions

$$s=\frac{x_1+x_2+\cdots +x_n}{n},$$

and

$$y_1=\frac{x_1}{s},y_2=\frac{x_2}{s},\dots ,y_n=\frac{x_n}{s},$$

the desired inequality becomes

$$\frac{y_1}{s^{k-1}{y}_1^k+y_2+\cdots +y_n}+\cdots +\frac{y_n}{y_1+y_2+\cdots +s^{k-1}{y}_n^k}\le 1,$$

where $s\ge 1$ and $y_1+y_2+\cdots +y_n=n$. Clearly, if this inequality holds for $s=1$, then it holds for any $s\ge 1$. Therefore, we need only to consider the case $s=1$, when $x_1+x_2+\cdots +x_n=n$, and the desired inequality is equivalent to

$$\frac{x_1}{x_1^k-x_1+n}+\frac{x_2}{x_2^k-x_2+n}+\cdots +\frac{x_n}{x_n^k-x_n+n}\le 1.$$

There are two cases to consider: $1<k\le n+1$ and $k>n+1$.

Case 1: $1<k\le n+1$. By Bernoulli’s inequality, we have

$$x_1^k\ge 1+k(x_1-1),$$

and hence

$$x_1^k-x_1+n\ge n-k+1+(k-1)x_1\ge 0.$$

Consequently, it suffices to prove that

$$\sum _{i=1}^n\frac{x_i}{n-k+1+(k-1)x_i}\le 1.$$

For $k=n+1$, this inequality is an equality. Otherwise, for $1<k<n+1$, we rewrite the inequality as

$$\sum _{i=1}^n\frac{1}{n-k+1+(k-1)x_i}\ge 1,$$

which follows from the AM-HM inequality as follows:

$$\sum _{i=1}^n\frac{1}{n-k+1+(k-1)x_i}\ge \frac{n^2}{{\sum }_{i=1}^n[n-k+1+(k-1)x_i]}=1.$$

Case 2: $k>n+1$. Write the desired inequality as

$$f(x_1)+f(x_2)+\cdots +f(x_n)\ge nf(s),$$

where

$$\begin{array}{c}s=1,\hfill \\ f(u)=\frac{-u}{u^k-u+n},u\in \mathbb{I}=(0,n).\hfill \end{array}$$

We have

$$f^{\prime }(u)=\frac{(k-1)u^k-n}{{(u^k-u+n)}^2}$$

and

$$f^{″}(u)=\frac{f_1(u)}{{(u^k-u+n)}^3},$$

where

$$f_1(u)=k(k-1)u^{k-1}(u^k-u+n)-2(k{u}^{k-1}-1)[(k-1)u^k-n].$$

From the expression of $f^{\prime }$, it follows that f is decreasing on $(0,s_0]$ and increasing on $[s_0,n)$, where

$$s_0={\left(\frac{n}{k-1}\right)}^{1/k}<1.$$

On the other hand, for $u\in [s_0,s]$, we have

$$(k-1)u^k-n\ge (k-1)s_0^k-n=0,$$

and hence

$$\begin{array}{rl}{f}_1(u)& \ge k(k-1)u^{k-1}(u^k-u+n)-2k{u}^{k-1}[(k-1)u^k-n]\\ =k{u}^{k-1}[-(k-1)(u^k+u)+n(k+1)]\\ \ge k{u}^{k-1}[-2(k-1)+2(k+1)]=4k{u}^{k-1}>0.\end{array}$$

Thus, $f^{″}(u)>0$, and hence f is convex on $[s_0,s]$. By PCF-Theorem and Remark 2.3, we need to show that $f(x)+(n-1)f(y)\ge nf(1)$ for all positive x, y which satisfy $x\ge 1\ge y>0$ and $x+(n-1)y=n$. Consider the nontrivial case where $x>1>y>0$ and write the inequality $f(x)+(n-1)f(y)\ge nf(1)$ as follows:

$$\begin{array}{c}\frac{x}{x^k-x-n}+\frac{(n-1)y}{y^k-y+n}\le 1,\hfill \\ x^k-x+n\ge \frac{x(y^k-y+n)}{y^k-ny+n},\hfill \\ x^k-x\ge \frac{(n-1)y(y-y^k)}{y^k-ny+n}.\hfill \end{array}$$

Since $y<1$ and $y^k-ny+n>y^k-y+1>0$, it suffices to show that

$$x^k-x\ge \frac{(n-1)(y-y^k)}{y^k-y+1},$$

which is equivalent to

$$h(x)\ge \frac{y-y^k}{(1-y)(y^k-y+1)},$$

where

$$h(x)=\frac{x^k-x}{x-1}.$$

By the weighted AM-GM inequality, we have

$$h^{\prime }(x)=\frac{(k-1)x^k+1-k{x}^{k-1}}{{(x-1)}^2}>0,$$

and hence h is strictly increasing. Since $x-1=(n-1)(1-y)\ge 1-y$, we get

$$h(x)\ge h(2-y)=\frac{{(2-y)}^k-2+y}{1-y}.$$

Thus, it suffices to show that

$${(2-y)}^k-2+y\ge \frac{y-y^k}{y^k-y+1}.$$

Putting $1-y=t$, $0<t<1$, we write this inequality as

$$\begin{array}{c}{(2-y)}^k-1+y\ge \frac{1}{y^k-y+1},\hfill \\ {(1+t)}^k-t\ge \frac{1}{{(1-t)}^k+t},\hfill \\ {(1-t^2)}^k+t{(1+t)}^k\ge 1+t^2+t{(1-t)}^k.\hfill \end{array}$$

By Bernoulli’s inequality,

$${(1-t^2)}^k+t{(1+t)}^k>1-k{t}^2+t(1+kt)=1+t.$$

So, we only need to show that $t(1-t)\ge t{(1-t)}^k$, which is clearly true. □

Application 4.4 Let $x_1,x_2,\dots ,x_n$ ($n\ge 2$) be positive real numbers such that

$$x_1+x_2+\cdots +x_n=n.$$

If $0<k\le \frac{n}{n-1}$, then [6]

$$x_1^{k/x_1}+x_2^{k/x_2}+\cdots +x_n^{k/x_n}\le n$$

with equality for $x_1=x_2=\cdots =x_n=1$.

Proof

According to the power mean inequality, we have

$${\left(\frac{x_1^{p/x_1}+x_2^{p/x_2}+\cdots +x_n^{p/x_n}}{n}\right)}^{1/p}\ge {\left(\frac{x_1^{q/x_1}+x_2^{q/x_2}+\cdots +x_n^{q/x_n}}{n}\right)}^{1/q}$$

for all $p\ge q>0$. Thus, it suffices to prove the desired inequality for

$$k=\frac{n}{n-1},1<k\le 2.$$

Rewrite the desired inequality as

$$f(x_1)+f(x_2)+\cdots +f(x_n)\ge nf(s),$$

where

$$\begin{array}{c}s=1,\hfill \\ f(u)=-u^{k/u},u\in \mathbb{I}=(0,n).\hfill \end{array}$$

We have

$$\begin{array}{c}{f}^{\prime }(u)=k{u}^{\frac{k}{u}-2}(lnu-1),\hfill \\ f^{″}(u)=k{u}^{\frac{k}{u}-4}[u+(1-lnu)(2u-k+klnu)].\hfill \end{array}$$

From the expression of $f^{\prime }$, it follows that f is decreasing on $(0,s_0]$ and increasing on $[s_0,n)$, where

$$s_0=e.$$

In addition, we claim that f is convex on $[s,s_0]$. Indeed, since $1-lnu\ge 0$ and

$$2u-k+klnu\ge 2-k\ge 0,$$

we have $f^{″}>0$ for $u\in [s,s_0]$. Therefore, by PCF-Theorem and Remark 2.3, we only need to show that

$$x^{k/x}+(n-1)y^{k/y}\le n$$

for $0<x\le 1\le y$ and $x+(n-1)y=n$. We have

$$\frac{k}{x}\ge k>1.$$

Also, from

$$\frac{k}{y}=\frac{n}{(n-1)y}>\frac{n}{x+(n-1)y}=1,\frac{k}{y}\le \frac{2}{y}\le 2,$$

we get

$$0<\frac{k}{y}-1\le 1.$$

Therefore, by Bernoulli’s inequality, we have

$$\begin{array}{r}{x}^{k/x}+(n-1)y^{k/y}-n\\ =\frac{1}{{(\frac{1}{x})}^{k/x}}+(n-1)y\cdot y^{k/y-1}-n\\ \le \frac{1}{1+\frac{k}{x}(\frac{1}{x}-1)}+(n-1)y[1+(\frac{k}{y}-1)(y-1)]-n\\ =\frac{x^2}{x^2-kx+k}-(k-1)x^2-(2-k)x\\ =\frac{-{(x-1)}^2[(k-1)x+k(2-k)]}{x^2-kx+k}\le 0.\end{array}$$

 □

Application 4.5 If a, b, c are positive real numbers such that $abc=1$, then

$$\frac{1-a}{17+4a+6a^2}+\frac{1-b}{17+4b+6b^2}+\frac{1-c}{17+4c+6c^2}\ge 0,$$

with equality for $a=b=c=1$ and also for $8a=b=c=2$ (or any cyclic permutation).

Proof

Write the desired inequality as

$$g(a)+g(b)+g(c)\ge 3g(r),$$

where $r=1$ and

$$g(t)=\frac{1-t}{1+pt+q{t}^2},t\in \mathbb{I}=(0,\mathrm{\infty }),$$

where $p=\frac{4}{17}$, $q=\frac{6}{17}$. From

$$g^{\prime }(t)=\frac{q{t}^2-2qt-p-1}{{(1+pt+q{t}^2)}^2},$$

it follows that g is decreasing on $(0,r_0]$ and increasing on $[r_0,\mathrm{\infty })$, where

$$r_0=1+\sqrt{1+\frac{p+1}{q}}>1.$$

We have

$$\begin{array}{c}f(u)=g\left(e^u\right)=\frac{1-e^u}{1+p{e}^u+q{e}^{2u}},\hfill \\ f^{″}(u)=\frac{t\cdot h(t)}{{(1+pt+q{t}^2)}^3},\hfill \end{array}$$

where

$$t=e^u,h(t)=-q^2{t}^4+q(p+4q)t^3+3q(p+2)t^2+(p-4q+p^2)t-p-1.$$

We will show that $h(t)>0$ for $t\in [r,r_0]$, and hence f is convex for

$$e^u\in [r,r_0]=[1,1+\sqrt{1+\frac{p+1}{q}}].$$

We have

$$\begin{array}{c}{h}^{\prime }(t)=-4q^2{t}^3+3q(p+4q)t^2+6q(p+2)t+p-4q+p^2,\hfill \\ h^{″}(t)=6q[-2q{t}^2+(p+4q)t+p+2].\hfill \end{array}$$

Since

$$h^{″}(t)=6q[2(-q{t}^2+2qt+p+1)+p(t-1)]\ge 12q(-q{t}^2+2qt+p+1)\ge 0,$$

$h^{\prime }(t)$ is increasing,

$$h^{\prime }(t)\ge h^{\prime }(1)=p^2+9pq+8q^2+p+8q>0,$$

h is increasing, and hence

$$\begin{array}{rl}h(t)\ge h(1)& =p^2+4pq+3q^2+2q-1={(p+2q)}^2-{(q-1)}^2\\ =(p+q+1)(p+3q-1)>0.\end{array}$$

By PCF-Corollary, we only need to prove that $g(a)+2g(b)\ge 3g(1)$ for $a{b}^2=1$; that is,

$$\frac{1-a}{1+pa+q{a}^2}+\frac{2(1-b)}{1+pb+q{b}^2}\ge 0,$$

$$\frac{b^2(b^2-1)}{b^4+p{b}^2+q}+\frac{2(1-b)}{1+pb+q{b}^2}\ge 0,$$

$$pA+qB\ge C,$$

where

$$\begin{array}{c}A=b^2{(b-1)}^2(b+2),\hfill \\ B={(b-1)}^2(b^4+2b^3+2b^2+2b+2),\hfill \\ C=b^2{(b-1)}^2(2b+1).\hfill \end{array}$$

Indeed, we have

$$17(pA+qB-C)=3{(b-1)}^2{(b-2)}^2(2b^2+2b+1)\ge 0.$$

 □

Application 4.6 If a, b, c are positive real numbers such that $abc=1$, then [6]

$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\ge 1,$$

with equality for $a=b=c=1$ and also for $8a=b=c=2$ (or any cyclic permutation).

Proof

Write the desired inequality as

$$g(a)+g(b)+g(c)\ge 3g(r),$$

where $r=1$ and

$$g(t)=\frac{7-6t}{2+t^2},t\in \mathbb{I}=(0,\mathrm{\infty }).$$

From

$$g^{\prime }(t)=\frac{2(3t+2)(t-3)}{{(2+t^2)}^2},$$

it follows that g is decreasing on $(0,r_0]$ and increasing on $[r_0,\mathrm{\infty })$, where

$$r_0=3.$$

We have

$$\begin{array}{c}f(u)=g\left(e^u\right)=\frac{7-6e^u}{2+e^{2u}},\hfill \\ f^{″}(u)=\frac{2t\cdot h(t)}{{(2+t^2)}^3},\hfill \end{array}$$

where

$$t=e^u,h(t)=-3t^4+14t^3+36t^2-28t-12.$$

We will show that $h(t)>0$ for $t\in [r,r_0]$, and hence f is convex for

$$e^u\in [r,r_0]=[1,3].$$

We have

$$\begin{array}{rl}h(t)& =3(t^2-1)(9-t^2)+14t^3+6t^2-28t+15\\ =3(t^2-1)(9-t^2)+14t^2(t-1)+14{(t-1)}^2+6t^2+1>0.\end{array}$$

By PCF-Corollary, we only need to prove that $g(a)+2g(b)\ge 3g(1)$ for $a{b}^2=1$; that is,

$$\begin{array}{c}\frac{7-6a}{2+a^2}+\frac{2(7-6b)}{2+b^2}\ge 1,\hfill \\ \frac{b^2(7b^2-6)}{2b^4+1}+\frac{2(7-6b)}{2+b^2}\ge 1,\hfill \\ {(b-1)}^2{(b-2)}^2(5b^2+6b+3)\ge 0.\hfill \end{array}$$

Since the last inequality is true, the proof is completed. □

Application 4.7 Let a, b, c be positive real numbers such that $abc=1$. If

$$k\in [\frac{-13}{3\sqrt{3}},\frac{13}{3\sqrt{3}}],$$

then [6]

$$\frac{a+k}{a^2+1}+\frac{b+k}{b^2+1}+\frac{c+k}{c^2+1}\le \frac{3(1+k)}{2},$$

with equality for $a=b=c=1$. If $k=\frac{13}{3\sqrt{3}}$, then the equality holds also for $a=7+4\sqrt{3}$ and $b=c=2-\sqrt{3}$ (or any cyclic permutation). If $k=\frac{-13}{3\sqrt{3}}$, then the equality holds also for $a=7-4\sqrt{3}$ and $b=c=2+\sqrt{3}$ (or any cyclic permutation).

Proof

The desired inequality is equivalent to

$$\sum \frac{{(a-1)}^2}{a^2+1}\ge k(\sum \frac{2}{a^2+1}-3).$$

Thus, it suffices to prove this inequality for only $|k|=\frac{13}{3\sqrt{3}}$. On the other hand, replacing a, b, c by $1/a$, $1/b$, $1/c$, the inequality becomes

$$\sum \frac{{(a-1)}^2}{a^2+1}\ge k(3-\sum \frac{2}{a^2+1}).$$

Thus, we only need to prove the desired inequality for $k=\frac{13}{3\sqrt{3}}$. Write this inequality as

$$g(a)+g(b)+g(c)\ge 3g(r),$$

where $r=1$ and

$$g(t)=\frac{-t-k}{t^2+1},t\in \mathbb{I}=(0,\mathrm{\infty }).$$

From

$$g^{\prime }(t)=\frac{t^2+2kt-1}{{(t^2+1)}^2},$$

it follows that g is decreasing on $(0,r_0]$ and increasing on $[r_0,\mathrm{\infty })$, where

$$r_0=\frac{\sqrt{3}}{9}.$$

We have

$$\begin{array}{c}f(u)=g\left(e^u\right)=\frac{-e^u-k}{e^{2u}+1},\hfill \\ f^{″}(u)=\frac{t\cdot h(t)}{{(t^2+1)}^3},\hfill \end{array}$$

where

$$t=e^u,h(t)=-t^4-4k{t}^3+6t^2+4kt-1.$$

We will show that $h(t)>0$ for $t\in [r_0,r]$, and hence f is convex for

$$e^u\in [r_0,r]=[\frac{\sqrt{3}}{9},1].$$

Indeed, since

$$4kt=\frac{52t}{3\sqrt{3}}=\frac{52}{27}>1,$$

we have

$$h(t)=-t^4+6t^2-1+4kt(1-t^2)\ge -t^4+6t^2-1+(1-t^2)=t^2(5-t^2)>0.$$

By PCF-Corollary, we only need to prove that $g(a)+2g(b)\ge 3g(1)$ for $a{b}^2=1$; that is,

$$\begin{array}{c}\frac{a+k}{a^2+1}+\frac{2(b+k)}{b^2+1}\le \frac{3(1+k)}{2},\hfill \\ \frac{b^2(k{b}^2+1)}{b^4+1}+\frac{2(b+k)}{b^2+1}\le \frac{3(1+k)}{2},\hfill \\ 3b^6-4b^5+b^4+b^2-4b+3-k{(1-b^2)}^3\ge 0,\hfill \\ {(b-1)}^2[(3+k)b^4+2(1+k)b^3+2b^2+2(1-k)b+3-k]\ge 0,\hfill \\ {(b-1)}^2{(b-2+\sqrt{3})}^2[(27+13\sqrt{3})b^2+24(2+\sqrt{3})b+33+17\sqrt{3}]\ge 0.\hfill \end{array}$$

The last inequality is clearly true, and the proof is completed. □

Application 4.8 If a, b, c are positive real numbers and $0<k\le 2+2\sqrt{2}$, then [6]

$$\frac{a^3}{k{a}^2+bc}+\frac{b^3}{k{b}^2+ca}+\frac{c^3}{k{c}^2+ab}\ge \frac{a+b+c}{k+1},$$

with equality for $a=b=c=1$. If $k=2+2\sqrt{2}$, then the equality holds also for $\frac{a}{\sqrt{2}}=b=c$ (or any cyclic permutation).

Proof For $k<2+2\sqrt{2}$, the proof is similar to the one of the main case $k=2+2\sqrt{2}$. For this reason, we consider further only the case where

$$k=2+2\sqrt{2}.$$

Due to homogeneity, we may assume that $abc=1$. On this hypothesis,

$$\sum \frac{a^3}{k{a}^2+bc}-\frac{1}{k+1}\sum a=\sum (\frac{a^4}{k{a}^3+1}-\frac{a}{k+1})=\frac{1}{k+1}\sum \frac{a^4-a}{k{a}^3+1}.$$

Thus, we can write the inequality as

$$g(a)+g(b)+g(c)\ge 3g(r),$$

where $r=1$ and

$$g(t)=\frac{t^4-t}{k{t}^3+1},t\in \mathbb{I}=(0,\mathrm{\infty }).$$

From

$$g^{\prime }(t)=\frac{k{t}^6+2(k+2)t^3-1}{{(k{t}^3+1)}^2},$$

it follows that g is decreasing on $(0,r_0]$ and increasing on $[r_0,\mathrm{\infty })$, where

$$r_0=\sqrt[3]{\frac{-k-2+\sqrt{(k+1)(k+4)}}{k}}\approx 0.4149.$$

We have

$$\begin{array}{c}f(u)=g\left(e^u\right)=\frac{e^{4u}-e^u}{k{e}^{3u}+1},\hfill \\ f^{″}(u)=\frac{t\cdot h(t)}{{(k{t}^3+1)}^3},\hfill \end{array}$$

where

$$t=e^u,h(t)=k^2{t}^9-k(4k+1)t^6+(13k+16)t^3-1.$$

We have $h(t)\ge 0$ for $t\in [t_1,t_2]$, where $t_1\approx 0.2345$ and $t_2\approx 1.02$. Since

$$[r_0,r]\subset [t_1,t_2],$$

f is convex for $e^u\in [r_0,r]$. Then, by PCF-Corollary, it suffices to show that $g(a)+2g(b)\ge 3g(1)$ for $a{b}^2=1$. This is true if the original inequality holds for $b=c=1$. Thus, we need to show that

$$\frac{a^3}{k{a}^2+1}+\frac{2}{a+k}\ge \frac{a+2}{k+1},$$

which is equivalent to the obvious inequality

$${(a-1)}^2{(a-\sqrt{2})}^2\ge 0.$$

 □

Application 4.9 If $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ are positive real numbers such that

$$a_1{a}_2{a}_3{a}_4{a}_5=1,$$

then [6]

$$\frac{1-a_1}{1+a_1^2}+\frac{1-a_2}{1+a_2^2}+\frac{1-a_3}{1+a_3^2}+\frac{1-a_4}{1+a_4^2}+\frac{1-a_5}{1+a_5^2}\ge 0,$$

with equality for $a_1=a_2=a_3=a_4=a_5=1$.

Proof

Write the inequality as

$$g(a_1)+g(a_2)+g(a_3)+g(a_4)+g(a_5)\ge 3g(r),$$

where $r=1$ and

$$g(t)=\frac{1-t}{1+t^2},t\in \mathbb{I}=(0,\mathrm{\infty }).$$

From

$$g^{\prime }(t)=\frac{t^2-2t-1}{{(t^2+1)}^2},$$

it follows that g is decreasing on $(0,r_0]$ and increasing on $[r_0,\mathrm{\infty })$, where

$$r_0=1+\sqrt{2}.$$

We have

$$\begin{array}{c}f(u)=g\left(e^u\right)=\frac{1-e^u}{1+e^{2u}},\hfill \\ f^{″}(u)=\frac{t\cdot h(t)}{{(t^2+1)}^3},\hfill \end{array}$$

where

$$t=e^u,h(t)=-t^4+4t^3+6t^2-4t-1.$$

We will show that $h(t)>0$ for $t\in [r,r_0]$, and hence f is convex for

$$e^u\in [r,r_0]=[1,1+\sqrt{2}].$$

Indeed,

$$h(t)\ge -t^4+6t^2-1=8-{(3-t^2)}^2\ge 4.$$

By PCF-Corollary, we only need to prove that $g(a)+4g(b)\ge 5g(1)$ for $a{b}^4=1$; that is,

$$\begin{array}{c}\frac{1-a}{1+a^2}+\frac{4(1-b)}{1+b^2}\ge 0,\hfill \\ \frac{b^4(b^4-1)}{1+b^8}+\frac{4(1-b)}{1+b^2}\ge 0,\hfill \\ 1+\frac{4(1-b)}{1+b^2}\ge \frac{1+b^4}{1+b^8}.\hfill \end{array}$$

Since

$$\frac{1+b^4}{1+b^8}\le \frac{2}{1+b^4}\le \frac{4}{{(1+b^2)}^2},$$

it suffices to show that

$$1+\frac{4(1-b)}{1+b^2}\ge \frac{4}{{(1+b^2)}^2}.$$

This inequality is equivalent to ${(1-b)}^4\ge 0$, and the proof is completed. □

Remark 4.1

The inequality

$$\frac{1-a_1}{1+a_1^2}+\frac{1-a_2}{1+a_2^2}+\frac{1-a_3}{1+a_3^2}+\frac{1-a_4}{1+a_4^2}+\frac{1-a_5}{1+a_5^2}+\frac{1-a_6}{1+a_6^2}\ge 0$$

is not true for any positive numbers $a_1$, $a_2$, $a_3$, $a_4$, $a_5$, $a_6$ satisfying $a_1{a}_2{a}_3{a}_4{a}_5{a}_6=1$. Indeed, for $a_2=a_3=a_4=a_5=a_6=2$, the inequality becomes

$$\frac{-a_1(a_1+1)}{1+a_1^2}\ge 0,$$

which is false for

$$a_1=\frac{1}{a_2{a}_3{a}_4{a}_5{a}_6}=\frac{1}{32}.$$