# 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 $\left[a,b\right]\subset \mathbb{I}$, decreasing for $u\le c$ and increasing for $u\ge c$, where $c\in \left[a,b\right]$. Several relevant applications are given to show the effectiveness of the proposed partially convex function theorem.

MSC:26D07, 26D10, 41A44.

## 1 Introduction

Let $\mathbf{x}=\left\{{x}_{1},{x}_{2},\dots ,{x}_{n}\right\}$ be a sequence of real numbers belonging to a given real interval $\mathbb{I}$, and let $\mathbf{p}=\left\{{p}_{1},{p}_{2},\dots ,{p}_{n}\right\}$ 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

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

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

$p=min\left\{{p}_{1},{p}_{2},\dots ,{p}_{n}\right\},\phantom{\rule{1em}{0ex}}{p}_{1}+{p}_{2}+\cdots +{p}_{n}=1.$

The inequality

${p}_{1}f\left({x}_{1}\right)+{p}_{2}f\left({x}_{2}\right)+\cdots +{p}_{n}f\left({x}_{n}\right)\ge f\left(s\right)$

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

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

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

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

$f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right)\ge nf\left(s\right)$

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

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

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

Applying HCF-Theorem and WHCF-Theorem to the function f defined by $f\left(u\right)=g\left({e}^{u}\right)$ 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 $f\left(u\right)=g\left({e}^{u}\right)$ is convex for ${e}^{u}\le r$ or ${e}^{u}\ge r$, where $r\in \mathbb{I}$. The inequality

$g\left({a}_{1}\right)+g\left({a}_{2}\right)+\cdots +g\left({a}_{n}\right)\ge ng\left(r\right)$

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

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

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 $f\left(u\right)=g\left({e}^{u}\right)$ 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

$p=min\left\{{p}_{1},{p}_{2},\dots ,{p}_{n}\right\},\phantom{\rule{1em}{0ex}}{p}_{1}+{p}_{2}+\cdots +{p}_{n}=1.$

The inequality

${p}_{1}g\left({a}_{1}\right)+{p}_{2}g\left({a}_{2}\right)+\cdots +{p}_{n}g\left({a}_{n}\right)\ge g\left(r\right)$

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

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

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 $\left[a,b\right]\subset \mathbb{I}$.

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 $\left[{s}_{0},s\right]$ or $\left[s,{s}_{0}\right]$, where $s\in \mathbb{I}$. The inequality

$f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right)\ge nf\left(s\right)$

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

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

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

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

$p=min\left\{{p}_{1},{p}_{2},\dots ,{p}_{n}\right\},\phantom{\rule{1em}{0ex}}{p}_{1}+{p}_{2}+\cdots +{p}_{n}=1.$

In addition, assume that f is convex on $\left[{s}_{0},s\right]$ or $\left[s,{s}_{0}\right]$, where $s\in \mathbb{I}$. The inequality

${p}_{1}f\left({x}_{1}\right)+{p}_{2}f\left({x}_{2}\right)+\cdots +{p}_{n}f\left({x}_{n}\right)\ge f\left(s\right)$

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

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

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

Applying PCF-Theorem and WPCF-Theorem to the function f defined by $f\left(u\right)=g\left({e}^{u}\right)$ 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 $f\left(u\right)=g\left({e}^{u}\right)$ 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

$g\left({a}_{1}\right)+g\left({a}_{2}\right)+\cdots +g\left({a}_{n}\right)\ge ng\left(r\right)$

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

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

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

$p=min\left\{{p}_{1},{p}_{2},\dots ,{p}_{n}\right\},\phantom{\rule{1em}{0ex}}{p}_{1}+{p}_{2}+\cdots +{p}_{n}=1.$

In addition, assume that the function f defined by $f\left(u\right)=g\left({e}^{u}\right)$ 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

${p}_{1}g\left({a}_{1}\right)+{p}_{2}g\left({a}_{2}\right)+\cdots +{p}_{n}g\left({a}_{n}\right)\ge g\left(r\right)$

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

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

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

${p}_{1}f\left({x}_{1}\right)+{p}_{2}f\left({x}_{2}\right)+\cdots +{p}_{n}f\left({x}_{n}\right)\ge f\left(s\right)$

holds for all ${x}_{1},{x}_{2},\dots ,{x}_{n}\in \mathbb{I}$ such that

${x}_{1},{x}_{2},\dots ,{x}_{n}\ge {s}_{0},\phantom{\rule{1em}{0ex}}{p}_{1}{x}_{1}+{p}_{2}{x}_{2}+\cdots +{p}_{n}{x}_{n}=s,$

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

${p}_{1}f\left({x}_{1}\right)+{p}_{2}f\left({x}_{2}\right)+\cdots +{p}_{n}f\left({x}_{n}\right)\ge f\left(s\right)$

holds for all ${x}_{1},{x}_{2},\dots ,{x}_{n}\in \mathbb{I}$ such that

${x}_{1},{x}_{2},\dots ,{x}_{n}\le {s}_{0},\phantom{\rule{1em}{0ex}}{p}_{1}{x}_{1}+{p}_{2}{x}_{2}+\cdots +{p}_{n}{x}_{n}=s,$

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

${\mathbb{I}}_{0}=\left\{u\in \mathbb{I}\mid u\ge {s}_{0}\right\},$

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

${\mathbb{I}}_{0}=\left\{u\in \mathbb{I}\mid u\le {s}_{0}\right\},$

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 $\left[{s}_{0},s\right]$, and

$x\le s\le y$

when f is convex on $\left[s,{s}_{0}\right]$. 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

$g\left(u\right)=\frac{f\left(u\right)-f\left(s\right)}{u-s},\phantom{\rule{2em}{0ex}}h\left(x,y\right)=\frac{g\left(x\right)-g\left(y\right)}{x-y}.$

In many applications, it is useful to replace the hypothesis

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

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

$h\left(x,y\right)\ge 0\phantom{\rule{0.25em}{0ex}}\mathrm{\forall }x,y\in \mathbb{I},\phantom{\rule{1em}{0ex}}px+\left(1-p\right)y=s.$

This equivalence is true since

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

In the particular case ${p}_{1}={p}_{2}=\cdots ={p}_{n}=1/n$, this condition becomes

$h\left(x,y\right)\ge 0\phantom{\rule{0.25em}{0ex}}\mathrm{\forall }x,y\in \mathbb{I},\phantom{\rule{1em}{0ex}}x+\left(n-1\right)y=ns.$

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

$p=min\left\{{p}_{1},{p}_{2},\dots ,{p}_{n}\right\},$

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

$px+\left(1-p\right)y=s,\phantom{\rule{2em}{0ex}}pf\left(x\right)+\left(1-p\right)f\left(y\right)=f\left(s\right).$

## 3 Proof of lemmas

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

${y}_{i}=\left\{\begin{array}{ll}{s}_{0},& {x}_{i}\le {s}_{0},\\ {x}_{i},& {x}_{i}>{s}_{0}.\end{array}$

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 $f\left({y}_{i}\right)\le f\left({x}_{i}\right)$ for ${x}_{i}\le {s}_{0}$ and $f\left({y}_{i}\right)=f\left({x}_{i}\right)$ for ${x}_{i}>{s}_{0}$, we have $f\left({y}_{i}\right)\le f\left({x}_{i}\right)$ for $i=1,2,\dots ,n$, and hence

${p}_{1}f\left({y}_{1}\right)+{p}_{2}f\left({y}_{2}\right)+\cdots +{p}_{n}f\left({y}_{n}\right)\le {p}_{1}f\left({x}_{1}\right)+{p}_{2}f\left({x}_{2}\right)+\cdots +{p}_{n}f\left({x}_{n}\right).$

Thus, it suffices to show that

${p}_{1}f\left({y}_{1}\right)+{p}_{2}f\left({y}_{2}\right)+\cdots +{p}_{n}f\left({y}_{n}\right)\ge f\left(s\right)$

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 ${p}_{1}f\left({y}_{1}\right)+{p}_{2}f\left({y}_{2}\right)+\cdots +{p}_{n}f\left({y}_{n}\right)\ge f\left(s\right)$ 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:

${y}_{i}=\left\{\begin{array}{ll}{x}_{i},& {x}_{i}\le {s}_{0},\\ {s}_{0},& {x}_{i}>{s}_{0}.\end{array}$

We have ${y}_{i}\le {s}_{0}$, ${y}_{i}\le {x}_{i}$ and $f\left({y}_{i}\right)\le f\left({x}_{i}\right)$ 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

${p}_{1}f\left({y}_{1}\right)+{p}_{2}f\left({y}_{2}\right)+\cdots +{p}_{n}f\left({y}_{n}\right)\le {p}_{1}f\left({x}_{1}\right)+{p}_{2}f\left({x}_{2}\right)+\cdots +{p}_{n}f\left({x}_{n}\right).$

Thus, it suffices to show that

${p}_{1}f\left({y}_{1}\right)+{p}_{2}f\left({y}_{2}\right)+\cdots +{p}_{n}f\left({y}_{n}\right)\ge f\left(s\right)$

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 ${p}_{1}f\left({y}_{1}\right)+{p}_{2}f\left({y}_{2}\right)+\cdots +{p}_{n}f\left({y}_{n}\right)\ge f\left(s\right)$ 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 $k\ge \frac{n\left(3n-4\right)}{{\left(n-2\right)}^{2}}$, 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

$f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right)\ge nf\left(s\right),$

where

$\begin{array}{c}s=1,\hfill \\ f\left(u\right)=\frac{1-u}{k+{u}^{2}},\phantom{\rule{1em}{0ex}}u\in \mathbb{I}=\left[\frac{-n}{n-2},\frac{n\left(2n-3\right)}{n-2}\right].\hfill \end{array}$

We have

$\begin{array}{c}{f}^{\prime }\left(u\right)=\frac{{u}^{2}-2u-k}{{\left({u}^{2}+k\right)}^{2}},\hfill \\ {f}^{″}\left(u\right)=\frac{2{f}_{1}\left(u\right)}{{\left({u}^{2}+k\right)}^{3}},\hfill \end{array}$

where

${f}_{1}\left(u\right)=-{u}^{3}+3{u}^{2}+3ku-k=\left(k+1\right)\left(3u-1\right)-{\left(u-1\right)}^{3}.$

There are two cases to consider.

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

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

since

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

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

$h\left(x,y\right)=\frac{g\left(x\right)-g\left(y\right)}{x-y},\phantom{\rule{2em}{0ex}}g\left(u\right)=\frac{f\left(u\right)-f\left(1\right)}{u-1}.$

Indeed, we have

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

and

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

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

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

and

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

from the expression of ${f}^{\prime }$ it follows that f is decreasing on $\left[\frac{-n}{n-2},{s}_{0}\right]$ and increasing on $\left[{s}_{0},\frac{n\left(2n-3\right)}{n-2}\right]$, where

${s}_{0}=1+\sqrt{k+1}>1.$

On the other hand, for $u\in \left[1,{s}_{0}\right]$, we have

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

since

$u-1\le {s}_{0}-1=\sqrt{k+1}.$

Thus, f is convex on $\left[1,{s}_{0}\right]$. By PCF-Theorem, we only need to show that $f\left(x\right)+\left(n-1\right)f\left(y\right)\ge nf\left(1\right)$ for all $x,y\in \mathbb{I}$ such that $x+\left(n-1\right)y=n$. 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]

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

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 ${x}_{1}>\frac{n\left(n+1\right)}{2}$ since

$\frac{n\left(n+1\right)-2{x}_{1}}{{n}^{2}+\left(n-2\right){x}_{1}^{2}}<0$

and

$\frac{n\left(n+1\right)-2{x}_{i}}{{n}^{2}+\left(n-2\right){x}_{i}^{2}}<\frac{n}{n-1},\phantom{\rule{1em}{0ex}}i=2,3,\dots ,n.$

Consider further that ${x}_{1},{x}_{2},\dots ,{x}_{n}\le \frac{n\left(n+1\right)}{2}$ and rewrite the desired inequality as

$f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right)\ge nf\left(s\right),$

where

$\begin{array}{c}s=1,\hfill \\ f\left(u\right)=\frac{2u-n\left(n+1\right)}{\left(n-2\right){u}^{2}+{n}^{2}},\phantom{\rule{1em}{0ex}}u\in \mathbb{I}=\left[\frac{n\left(3-{n}^{2}\right)}{2},\frac{n\left(n+1\right)}{2}\right].\hfill \end{array}$

We have

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

and

$\frac{{f}^{″}\left(u\right)}{2\left(n-2\right)}=\frac{{f}_{1}\left(u\right)}{{\left[\left(n-2\right){u}^{2}+{n}^{2}\right]}^{3}},$

where

${f}_{1}\left(u\right)=2\left(n-2\right){u}^{3}-3n\left(n+1\right)\left(n-2\right){u}^{2}-2{n}^{2}\left(2n-3\right)u+{n}^{3}\left(n+1\right).$

From the expression of ${f}^{\prime }$, it follows that f is decreasing on $\left[\frac{n\left(3-{n}^{2}\right)}{2},{s}_{0}\right]$ and increasing on $\left[{s}_{0},\frac{n\left(n+1\right)}{2}\right]$, where

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

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

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

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

$h\left(x,y\right)=\frac{g\left(x\right)-g\left(y\right)}{x-y},\phantom{\rule{2em}{0ex}}g\left(u\right)=\frac{f\left(u\right)-f\left(1\right)}{u-1}.$

Indeed, we have

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

and

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

□

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

${y}_{1}=\frac{{x}_{1}}{s},\phantom{\rule{2em}{0ex}}{y}_{2}=\frac{{x}_{2}}{s},\phantom{\rule{2em}{0ex}}\dots ,\phantom{\rule{2em}{0ex}}{y}_{n}=\frac{{x}_{n}}{s},$

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 and $k>n+1$.

Case 1: $1. By Bernoulli’s inequality, we have

${x}_{1}^{k}\ge 1+k\left({x}_{1}-1\right),$

and hence

${x}_{1}^{k}-{x}_{1}+n\ge n-k+1+\left(k-1\right){x}_{1}\ge 0.$

Consequently, it suffices to prove that

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

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

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

which follows from the AM-HM inequality as follows:

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

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

$f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right)\ge nf\left(s\right),$

where

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

We have

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

and

${f}^{″}\left(u\right)=\frac{{f}_{1}\left(u\right)}{{\left({u}^{k}-u+n\right)}^{3}},$

where

${f}_{1}\left(u\right)=k\left(k-1\right){u}^{k-1}\left({u}^{k}-u+n\right)-2\left(k{u}^{k-1}-1\right)\left[\left(k-1\right){u}^{k}-n\right].$

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

${s}_{0}={\left(\frac{n}{k-1}\right)}^{1/k}<1.$

On the other hand, for $u\in \left[{s}_{0},s\right]$, we have

$\left(k-1\right){u}^{k}-n\ge \left(k-1\right){s}_{0}^{k}-n=0,$

and hence

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

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

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

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

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

which is equivalent to

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

where

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

By the weighted AM-GM inequality, we have

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

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

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

Thus, it suffices to show that

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

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

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

By Bernoulli’s inequality,

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

So, we only need to show that $t\left(1-t\right)\ge t{\left(1-t\right)}^{k}$, 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, 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},\phantom{\rule{1em}{0ex}}1

Rewrite the desired inequality as

$f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right)\ge nf\left(s\right),$

where

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

We have

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

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

${s}_{0}=e.$

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

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

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

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

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

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

Also, from

$\frac{k}{y}=\frac{n}{\left(n-1\right)y}>\frac{n}{x+\left(n-1\right)y}=1,\phantom{\rule{1em}{0ex}}\frac{k}{y}\le \frac{2}{y}\le 2,$

we get

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

Therefore, by Bernoulli’s inequality, we have

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

□

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

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

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

Proof

Write the desired inequality as

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

where $r=1$ and

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

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

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

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

${r}_{0}=1+\sqrt{1+\frac{p+1}{q}}>1.$

We have

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

where

$t={e}^{u},\phantom{\rule{1em}{0ex}}h\left(t\right)=-{q}^{2}{t}^{4}+q\left(p+4q\right){t}^{3}+3q\left(p+2\right){t}^{2}+\left(p-4q+{p}^{2}\right)t-p-1.$

We will show that $h\left(t\right)>0$ for $t\in \left[r,{r}_{0}\right]$, and hence f is convex for

${e}^{u}\in \left[r,{r}_{0}\right]=\left[1,1+\sqrt{1+\frac{p+1}{q}}\right].$

We have

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

Since

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

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

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

h is increasing, and hence

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

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

$\frac{1-a}{1+pa+q{a}^{2}}+\frac{2\left(1-b\right)}{1+pb+q{b}^{2}}\ge 0,$
$\frac{{b}^{2}\left({b}^{2}-1\right)}{{b}^{4}+p{b}^{2}+q}+\frac{2\left(1-b\right)}{1+pb+q{b}^{2}}\ge 0,$
$pA+qB\ge C,$

where

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

Indeed, we have

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

□

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

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

where $r=1$ and

$g\left(t\right)=\frac{7-6t}{2+{t}^{2}},\phantom{\rule{1em}{0ex}}t\in \mathbb{I}=\left(0,\mathrm{\infty }\right).$

From

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

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

${r}_{0}=3.$

We have

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

where

$t={e}^{u},\phantom{\rule{1em}{0ex}}h\left(t\right)=-3{t}^{4}+14{t}^{3}+36{t}^{2}-28t-12.$

We will show that $h\left(t\right)>0$ for $t\in \left[r,{r}_{0}\right]$, and hence f is convex for

${e}^{u}\in \left[r,{r}_{0}\right]=\left[1,3\right].$

We have

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

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

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

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

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

then [6]

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

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

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

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

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

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

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

where $r=1$ and

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

From

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

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

${r}_{0}=\frac{\sqrt{3}}{9}.$

We have

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

where

$t={e}^{u},\phantom{\rule{1em}{0ex}}h\left(t\right)=-{t}^{4}-4k{t}^{3}+6{t}^{2}+4kt-1.$

We will show that $h\left(t\right)>0$ for $t\in \left[{r}_{0},r\right]$, and hence f is convex for

${e}^{u}\in \left[{r}_{0},r\right]=\left[\frac{\sqrt{3}}{9},1\right].$

Indeed, since

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

we have

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

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

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

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

Application 4.8 If a, b, c are positive real numbers and $0, 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,

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

Thus, we can write the inequality as

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

where $r=1$ and

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

From

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

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

${r}_{0}=\sqrt[3]{\frac{-k-2+\sqrt{\left(k+1\right)\left(k+4\right)}}{k}}\approx 0.4149.$

We have

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

where

$t={e}^{u},\phantom{\rule{1em}{0ex}}h\left(t\right)={k}^{2}{t}^{9}-k\left(4k+1\right){t}^{6}+\left(13k+16\right){t}^{3}-1.$

We have $h\left(t\right)\ge 0$ for $t\in \left[{t}_{1},{t}_{2}\right]$, where ${t}_{1}\approx 0.2345$ and ${t}_{2}\approx 1.02$. Since

$\left[{r}_{0},r\right]\subset \left[{t}_{1},{t}_{2}\right],$

f is convex for ${e}^{u}\in \left[{r}_{0},r\right]$. Then, by PCF-Corollary, it suffices to show that $g\left(a\right)+2g\left(b\right)\ge 3g\left(1\right)$ 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

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

□

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

$g\left({a}_{1}\right)+g\left({a}_{2}\right)+g\left({a}_{3}\right)+g\left({a}_{4}\right)+g\left({a}_{5}\right)\ge 3g\left(r\right),$

where $r=1$ and

$g\left(t\right)=\frac{1-t}{1+{t}^{2}},\phantom{\rule{1em}{0ex}}t\in \mathbb{I}=\left(0,\mathrm{\infty }\right).$

From

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

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

${r}_{0}=1+\sqrt{2}.$

We have

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

where

$t={e}^{u},\phantom{\rule{1em}{0ex}}h\left(t\right)=-{t}^{4}+4{t}^{3}+6{t}^{2}-4t-1.$

We will show that $h\left(t\right)>0$ for $t\in \left[r,{r}_{0}\right]$, and hence f is convex for

${e}^{u}\in \left[r,{r}_{0}\right]=\left[1,1+\sqrt{2}\right].$

Indeed,

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

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

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

Since

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

it suffices to show that

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

This inequality is equivalent to ${\left(1-b\right)}^{4}\ge 0$, 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

$\frac{-{a}_{1}\left({a}_{1}+1\right)}{1+{a}_{1}^{2}}\ge 0,$

which is false for

${a}_{1}=\frac{1}{{a}_{2}{a}_{3}{a}_{4}{a}_{5}{a}_{6}}=\frac{1}{32}.$

## References

1. Jensen JLWV: Sur les fonctions convexes et les inegalites entre les valeurs moyennes. Acta Math. 1906, 30(1):175–193. 10.1007/BF02418571

2. Mitrinović DS, Pečarić JE, Fink AM: Classical and New Inequalities in Analysis. Kluwer Academic, Dordrecht; 1993.

3. Cirtoaje V, Baiesu A: An extension of Jensen’s discrete inequality to half convex functions. J. Inequal. Appl. 2011., 2011: Article ID 101

4. Cirtoaje V: A generalization of Jensen’s inequality. Gaz. Mat., Ser. A 2005, 2: 124–138.

5. Cirtoaje V: Algebraic Inequalities-Old and New Methods. GIL Publishing House, Zalau; 2006.

6. Art of Problem Solving. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45516/115797/220970/242881/246265/453476/465869

## Acknowledgements

The author is grateful to the referees for their useful comments.

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Cirtoaje, V. An extension of Jensen’s discrete inequality to partially convex functions. J Inequal Appl 2013, 54 (2013). https://doi.org/10.1186/1029-242X-2013-54