---
abstract: 'The present paper is devoted to the study of Jensen-Mercer-type inequalities. Our results generalize and improve some earlier results in the literature.'
author:
- Hamid Reza Moradi and Shigeru Furuichi
title: 'Improvement and generalization of some Jensen-Mercer-type inequalities'
---

Introduction
============

The well-known Jensen inequality for the convex functions states that if $f$ is a convex function on the interval $\left[ m,M \right]$, then $$\label{3}
f\left( \sum\limits_{i=1}^{n}{{{w}_{i}}{{a}_{i}}} \right)\le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}$$ for all ${{a}_{i}}\in \left[ m,M \right]$ and ${{w}_{i}}\in \left[ 0,1 \right]$ $(i=1,\ldots ,n)$ with $\sum\limits_{i=1}^{n}{{{w}_{i}}}=1$. Various inequalities improving and extending have been studied in [@3; @5; @1].

Mercer [@4] proved that if $f$ is a convex function on $\left[ m,M \right]$, then $$\label{9}
f\left( M+m-\sum\limits_{i=1}^{n}{{{w}_{i}}{{a}_{i}}} \right)\le f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}$$ for all ${{a}_{i}}\in \left[ m,M \right]$ and ${{w}_{i}}\in \left[ 0,1 \right]$ $(i=1,\ldots ,n)$ with $\sum\limits_{i=1}^{n}{{{w}_{i}}}=1$. Several refinements and generalizations of the inequality have been given in [@6; @2].

In [@kian Theorem 2.1] it has been shown that if $f$ is a convex function on the interval $\left[ m,M \right]$, then $$\label{17}
f\left( M+m-\frac{a+b}{2} \right)\le \frac{1}{b-a}\int_{a}^{b}{f\left( M+m-u \right)du}\le f(M)+f(m)-\frac{f\left( a \right)+f\left( b \right)}{2}$$ for all $a,b\in \left[ m,M \right]$.

In this paper we prove the following general result: $$\label{13}
f\left( M+m-\overline{a} \right)\le \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}}\le f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}$$ where $\bar{a} :=\sum\limits_{i=1}^{n}{{{w}_{i}}{{a}_{i}}} $. After that, we show a refinement of inequality in the following form $$\begin{aligned}
f\left( M+m-\overline{a} \right)&\le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\frac{\overline{a}+{{a}_{i}}}{2} \right)} \\ 
& \le \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}} \\ 
& \le f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}.  
\end{aligned}$$ Though we confine our discussion to scalars, by changing the convex function assumption with the operator convex, the inequalities we obtain in this paper can be extended in a natural way to Hilbert space operators.

Main Results
============

The following lemma is well-known in [@4 Lemma 1.3], but we prove it for the reader convenience.

\[1\] Let $f$ be a convex function on $\left[ m,M \right]$, then $$f\left( M+m-{{a}_{i}} \right)\le f\left( M \right)+f\left( m \right)-f\left( {{a}_{i}} \right),\text{ }\left( m\le {{a}_{i}}\le M,~i=1,\ldots ,n \right).$$

If $f:\left[ m,M \right]\to \mathbb{R}$ is a convex function, then for any $x,y\in \left[ m,M \right]$ and $t\in \left[ 0,1 \right]$, we have $$\label{7}
f\left( tx+\left( 1-t \right)y \right)\le tf\left( x \right)+\left( 1-t \right)f\left( y \right).$$ It can be verified that if $m\le {{a}_{i}}\le M\left( i=1,\ldots ,n \right)$, then $\frac{M-{{a}_{i}}}{M-m},\frac{{{a}_{i}}-m}{M-m}\le 1$ and $\frac{M-{{a}_{i}}}{M-m}+\frac{{{a}_{i}}-m}{M-m}=1$. Thanks to , we have $$\label{8}
f\left( {{a}_{i}} \right)\le \frac{M-{{a}_{i}}}{M-m}f\left( m \right)+\frac{{{a}_{i}}-m}{M-m}f\left( M \right).$$ One the other hand, $m\le {{a}_{i}}\le M\left( i=1,\ldots ,n \right)$ implies $m\le M+m-{{a}_{i}}\le M\left( i=1,\ldots ,n \right)$. Thus, from we infer $$\label{10}
f\left( M+m-{{a}_{i}} \right)\le \frac{{{a}_{i}}-m}{M-m}f\left( m \right)+\frac{M-{{a}_{i}}}{M-m}f\left( M \right).$$ Summing up and , we get the desired result.

Based on this, our first result can be stated as follows:

\[6\] Let $f$ be a convex function on $\left[ m,M \right]$ and $t\in [0,1]$. Then $$\label{5}
\begin{aligned}
 f\left( M+m-\overline{a} \right)&\le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)} \\ 
& \le f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}  
\end{aligned}$$ for all ${{a}_{i}}\in \left[ m,M \right]$ and ${{w}_{i}}\in \left[ 0,1 \right]$ $(i=1,\ldots ,n)$ with $\sum\limits_{i=1}^{n}{{{w}_{i}}}=1$, where $\bar{a} :=\sum\limits_{i=1}^{n}{{{w}_{i}}{{a}_{i}}}$. Moreover, the function $F:\left[ 0,1 \right]\to \mathbb{R}$ defined by $$F\left( t \right)=\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)}.$$ is monotonically nondecreasing and convex on $\left[ 0,1 \right]$.

Firstly, we have $$\label{16}
\begin{aligned}
\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)}&\ge f\left( \sum\limits_{i=1}^{n}{{{w}_{i}}\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)} \right) \\ 
& =f\left( M+m-\overline{a} \right).  
\end{aligned}$$ On the other hand, $$\begin{aligned}
& \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)} \\ 
& =\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( \left( 1-t \right)\left( M+m-\overline{a} \right)+t\left( M+m-{{a}_{i}} \right) \right)} \\ 
& \le \sum\limits_{i=1}^{n}{{{w}_{i}}\left( \left( 1-t \right)f\left( M+m-\overline{a} \right)+tf\left( M+m-{{a}_{i}} \right) \right)} \\ 
& \le \sum\limits_{i=1}^{n}{{{w}_{i}}\left( \left( 1-t \right)\left( f\left( M \right)+f\left( m \right)-\sum\limits_{j=1}^{n}{{{w}_{j}}f\left( {{a}_{j}} \right)} \right)+t\left( f\left( M \right)+f\left( m \right)-f\left( {{a}_{i}} \right) \right) \right)} \\ 
& =f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right).} \\ 
\end{aligned}$$ For the convexity of $F$, we have $$\begin{aligned}
& F\left( \frac{t+s}{2} \right) \\ 
& =\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-\frac{t+s}{2} \right)\overline{a}+\frac{t+s}{2}{{a}_{i}} \right) \right)} \\ 
& =\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \frac{\left( 1-t \right)\overline{a}+t{{a}_{i}}+\left( 1-s \right)\overline{a}+s{{a}_{i}}}{2} \right) \right)} \\ 
& =\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( \frac{M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right)+M+m-\left( \left( 1-s \right)\overline{a}+s{{a}_{i}} \right)}{2} \right)} \\ 
& \le \frac{1}{2}\left[ \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)}+\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\left( \left( 1-s \right)\overline{a}+s{{a}_{i}} \right) \right)} \right] \\ 
& =\frac{F\left( t \right)+F\left( s \right)}{2}. \\ 
\end{aligned}$$ Now, if $0<s<t<1$, then $s=\frac{t-s}{t}\cdot0+\frac{s}{t}\cdot t$ and hence the convexity of $F$ implies $$\begin{aligned}
    F\left( s \right)&=F\left( \frac{t-s}{t}\cdot0+\frac{s}{t}\cdot t \right) \\ 
    & \le \frac{t-s}{t}F\left( 0 \right)+\frac{s}{t}F\left( t \right) \\ 
    & \le \frac{t-s}{t}F\left( t \right)+\frac{s}{t}F\left( t \right) \\ 
    & =F\left( t \right).  
    \end{aligned}$$ We remark that the second inequality in the above follows from and the fact $$F\left( 0 \right)=\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\overline{a} \right)}=f\left( M+m-\overline{a} \right).$$ Therefore $F$ is monotonically nondecreasing on $\left[ 0,1 \right]$.

\[14\] Let all the assumptions of Theorem \[6\] hold, then $$f\left( M+m-\overline{a} \right)\le \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}}\le f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}.$$

Integrating the inequality over $t\in \left[ 0,1 \right]$, we get . Here we used the fact $$\begin{aligned}
 \int_{0}^{1}{f\left( M+m-\left( \left( 1-t \right)\overline{a}+t{{a}_{i}} \right) \right)dt}&=\int_{0}^{1}{f\left( \left( 1-t \right)\left( M+m-\overline{a} \right)+t\left( M+m-{{a}_{i}} \right) \right)dt} \\ 
& =\int_{0}^{1}{f\left( t\left( M+m-\overline{a} \right)+\left( 1-t \right)\left( M+m-{{a}_{i}} \right) \right)dt} \\ 
& =\frac{1}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}.  
\end{aligned}$$

Put $n=2$, ${{w}_{1}}={{w}_{2}}={1}/{2}\;$, ${{a}_{1}}=a$, and ${{a}_{2}}=b$ in Corollary \[14\], then $$\begin{aligned}
    f\left( M+m-\frac{a+b}{2} \right)& \le \frac{1}{b-a}\int_{a}^{b}{f\left( M+m-u \right)du} \\ 
    & \le f\left( M \right)+f\left( m \right)-\frac{f\left( a \right)+f\left( b \right)}{2}  
    \end{aligned}$$ which shows that our inequality generalizes inequality .

We give a more precise estimate in the next theorem.

\[15\] Let $f$ be a convex function on $\left[ m,M \right]$. Then $$\label{4}
\begin{aligned}
 f\left( M+m-\overline{a} \right)&\le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\frac{\overline{a}+{{a}_{i}}}{2} \right)} \\ 
& \le \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}} \\ 
& \le f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}  
\end{aligned}$$ for all ${{a}_{i}}\in \left[ m,M \right]$ and ${{w}_{i}}\in \left[ 0,1 \right]$ $(i=1,\ldots ,n)$ with $\sum\limits_{i=1}^{n}{{{w}_{i}}}=1$.

If $f:\left[ m,M \right]\to \mathbb{R}$ is a convex function, then we have for any $a,b\in \left[ m,M \right]$ $$\begin{aligned}
 f\left( \frac{a+b}{2} \right)&=f\left( \frac{ta+\left( 1-t \right)b+tb+\left( 1-t \right)a}{2} \right) \\ 
& \le \frac{f\left( ta+\left( 1-t \right)b \right)+f\left( tb+\left( 1-t \right)a \right)}{2} \\ 
& \le \frac{f\left( a \right)+f\left( b \right)}{2}.  
\end{aligned}$$ Replacing $a$ and $b$ by $M+m-a$ and $M+m-b$, respectively, we get $$\begin{aligned}
 f\left( M+m-\frac{a+b}{2} \right)&\le \frac{f\left( M+m-\left( ta+\left( 1-t \right)b \right) \right)+f\left( M+m-\left( tb+\left( 1-t \right)a \right) \right)}{2} \\ 
& \le \frac{f\left( M+m-a \right)+f\left( M+m-b \right)}{2}.  
\end{aligned}$$ Integrating the inequality over $t\in \left[ 0,1 \right]$, and using the fact $$\int_{0}^{1}{f\left( tx+\left( 1-t \right)y \right)dt}=\int_{0}^{1}{f\left( ty+\left( 1-t \right)x \right)dt},$$ we infer that $$\begin{aligned}
 f\left( M+m-\frac{a+b}{2} \right)&\le \int_{0}^{1}{f\left( M+m-\left( ta+\left( 1-t \right)b \right) \right)dt} \\ 
& \le \frac{f\left( M+m-a \right)+f\left( M+m-b \right)}{2}.  
\end{aligned}$$ Since ${{a}_{i}}, \bar{a} \in \left[ m,M \right]$, we can write $$\begin{aligned}
 f\left( M+m-\frac{\overline{a}+{{a}_{i}}}{2} \right)&\le \frac{1}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt} \\ 
& \le \frac{f\left( M+m-\overline{a} \right)+f\left( M+m-{{a}_{i}} \right)}{2},  
\end{aligned}$$ due to $$\int_{0}^{1}{f\left( M+m-\left( t\overline{a}+\left( 1-t \right){{a}_{i}} \right) \right)dt}=\frac{1}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}.$$ Multiplying by ${{w}_{i}}>0$ $\left( i=1,\ldots ,n \right)$ and summing over $i$ from $1$ to $n$ we may deduce $$\label{2}
\begin{aligned}
 \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\frac{\overline{a}+{{a}_{i}}}{2} \right)}&\le \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{f\left( t \right)dt}} \\ 
& \le \frac{f\left( M+m-\overline{a} \right)+\sum\nolimits_{i=1}^{n}{{{w}_{i}}f\left( M+m-{{a}_{i}} \right)}}{2}.  
\end{aligned}$$ On the other hand, by $$\label{11}
\begin{aligned}
 f\left( M+m-\bar{a} \right)&=f\left( \sum\limits_{i=1}^{n}{{{w}_{i}}\left( M+m-\frac{\bar{a}+{{a}_{i}}}{2} \right)} \right) \\ 
& \le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( M+m-\frac{\bar{a}+{{a}_{i}}}{2} \right)}  
\end{aligned}$$ and by Lemma \[1\] $$\label{12}
\begin{aligned}
& \frac{f\left( M+m-\bar{a} \right)+{\sum\nolimits_{i=1}^{n}{{{w}_{i}}f\left( M+m-{{a}_{i}} \right)}}}{2} \\ 
& \le \frac{f\left( M \right)+f\left( m \right)-\sum\nolimits_{j=1}^{n}{{{w}_{j}}f\left( {{a}_{j}} \right)}+{f\left( M \right)+f\left( m \right)-\sum\nolimits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}}}{2} \\ 
& =f\left( M \right)+f\left( m \right)-\sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}.
\end{aligned}$$ Combining , , and , we get .

\[18\] Let ${{a}_{i}}\in \left[ m,M \right]$ and ${{w}_{i}}\in \left[ 0,1 \right]$ $(i=1,\ldots ,n)$ with $\sum\limits_{i=1}^{n}{{{w}_{i}}}=1$. Then $$\begin{aligned}
 \frac{Mm}{\prod\limits_{i=1}^{n}{a_{i}^{{{w}_{i}}}}}&\le \exp \left[ \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{\overline{a}-{{a}_{i}}}\int_{M+m-\overline{a}}^{M+m-{{a}_{i}}}{\log tdt}} \right] \\ 
& \le \prod\limits_{i=1}^{n}{{{\left( M+m-\frac{\overline{a}+{{a}_{i}}}{2} \right)}^{{{w}_{i}}}}} \\ 
& \le M+m-\sum\limits_{i=1}^{n}{{{w}_{i}}{{a}_{i}}}.  
\end{aligned}$$

Put $f(t)=-\log t$, $(0<t \le 1)$ in Theorem \[15\].

If we set $n=2$, $a_1=m,a_2=M$ and $w_1=w_2=1/2$ in Corollary \[18\], then we have $$\sqrt{Mm} \leq \frac{M^{\frac{M}{M-m}}}{e m^{\frac{m}{M-m}}} \leq \frac{1}{4}\sqrt{(M+3m)(m+3M)} \leq \frac{1}{2}\left(M+m\right).$$ One can obtain the inequalities for the weighted parameter in means of two variables $m$ and $M$ by elementary calculations. We leave it to the interested readers.

Let all the assumptions of Theorem \[15\] hold, then $$\begin{aligned}
 f\left( \overline{a} \right)&\le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( \frac{\overline{a}+{{a}_{i}}}{2} \right)} \\ 
& \le \sum\limits_{i=1}^{n}{\frac{{{w}_{i}}}{{{a}_{i}}-\overline{a}} \int_{\overline{a}}^{{{a}_{i}}}{f\left( t \right)dt}} \\ 
& \le \frac{f\left( \overline{a} \right)+\sum\nolimits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}}{2} \\ 
& \le \sum\limits_{i=1}^{n}{{{w}_{i}}f\left( {{a}_{i}} \right)}.  
\end{aligned}$$ The proof is in the same spirit as that of Theorem \[15\] (see also [@dragomir Corollary 3]).

Acknowledgements {#acknowledgements .unnumbered}
================

The author (S.F.) was partially supported by JSPS KAKENHI Grant Number 16K05257.

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[0.3 true cm ]{}

[(H.R. Moradi) Department of Mathematics, Payame Noor University (PNU), P.O. Box 19395-4697, Tehran, Iran.]{}

[*E-mail address:* hrmoradi@mshdiau.ac.ir ]{}

[0.3 true cm ]{}

[(S. Furuichi) Department of Information Science, College of Humanities and Sciences, Nihon University, 3-25-40, Sakurajyousui, Setagaya-ku, Tokyo, 156-8550, Japan.]{}

[*E-mail address:* furuichi@chs.nihon-u.ac.jp]{}
