Some converse of Jensen’s Inequality and Applications

Revue d’analyse numerique et de theorie de l’approximation

Tome 23, No.1, 1994, pp.71-78

1 Introduction

In Theory of Inequalities, the famous Jensen’s inequality

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

valid for every convex function define on an interval $I$ of real numbers and for every $x_i\in I$ and $p_i\ge 0\left(i=\overline{1,n}\right)$ with $P_n:=\sum _{i=1}^n{p}_i>0$, plays such an important role, that many mathematicians have tried not only to establish (1) in a variety of ways but also to find different extensions, refinements and counterparts; see [2] and [6] where further references are given.

In this paper, we will give some inequalities for differentiable convex functions defined on an interval in connection with this main result.

2 The Main Results

We will start with the following converse of Jensenţs inequality:

Theorem 1

Let $f:I\subseteq ℝ\to ℝ$ be a differentiable convex function on $I$ ($\mathring{I}$ is the interior of $I$), $x_i\in \mathring{I}$ and $p_i\ge 0\left(i=\overline{1,n}\right)$ with $P_n>0$. Then one has the inequality

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-f\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)$ (2)

$\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{f}^{\prime }\left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i{f}^{\prime }\left(x_i\right).$

Proof. Since $f$ is convex on $I$, it follows that

$$f\left(x\right)-f\left(y\right)\ge f^{\prime }\left(y\right)\left(x-y\right)$$

for all $x,y\in \mathring{I}$.

Now, choosing

$$x=\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i,y=x_j\left(j=\overline{1,n}\right)$$

we derive the inequality

$$f\left(\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i\right)-f\left(x_j\right)\ge \left(\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i-x_j\right)f^{\prime }\left(x_j\right)$$

for all $j=1,\mathrm{\dots },n$. If we multiply this inequality with $p_j\ge 0$ and if we add these inequalities, we deduce that

$f\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{j=1}^n}{p}_{j}f\left(x_j\right)\ge$

$\ge {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{j=1}^n}{p}_j{f}^{\prime }\left(x_j\right)-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{j=1}^n}{x}_j{f}^{\prime }\left(x_j\right)p_j$

which is clearly equivalent with (2).

In paper [8], J.E. Pečarić has proved the following interesting converse of Cebyšev’s ineqaulity

$\left|{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{a}_i{b}_i-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{a}_i{\displaystyle \sum _{i=1}^n}{p}_i{b}_i\right|\le$

$\le \left|a_n-a_1\right|\left|b_n-b_1\right|\underset{1\le k\le n-1}{\max }\{P_k{\overline{P}}_{k+1}/P_n^2\}$

where $a=(a_1,\mathrm{\dots },a_n),b=(b_1,\mathrm{\dots },b_n)$ are monotonous $n$-tuples, $p=(p_1,\mathrm{\dots },p_n)$ is positive, i.e., $p_i\ge 0\left(i-\overline{1,n}\right)$ and $P_k:={\sum }_{i=1}^n{p}_i,{\overline{P}}_{k+1}:=P_n-P_k\left(1\le k\le n-1\right)$.

By the use of this result and by Theorem 1, we can state the following corollary.  

Corollary 2

Suppose that $f,p_i,x_i$ are as above and, in addition, $m\le x_i\le M$, for all $i=1,\mathrm{\dots },n$. Then one has the inequality

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-f\left({\displaystyle \frac{1}{P}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)\le$

$\le \left(M-m\right)\left(f\left(M\right)-f^{\prime }\left(m\right)\right)\max \{P_k{\overline{P}}_{k+1}/P_n^2\}.$

In paper [1], D. Andrica and C. Badea have obtained the following inverse of Cebyšev’s inequality

$\left|P_n{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{y}_i-{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i{y}_i\right|\le$

$\le \left(b-a\right)\left(d-c\right)\left({\displaystyle \sum _{i\in N}}{p}_i\right)\left(P_n-{\displaystyle \sum _{i\in N}}{p}_i\right)$

where $a\le xli$ $\le b,c\le y_i\le d\left(i=\overline{1,2}\right)$ and $N$ is the subset of $\{1,\mathrm{\dots },n\}$ which minimize the expresion

$$\left|\sum _{i\in N}{p}_i-P_n/2\right|.$$

By the use of this result, we also have the following corollary:

Corollary 3

Suppose that $f,p_i,x_i$ are as above and, in addition, $m\le x_i\le M$, for all $i=1,\mathrm{\dots },n$. Then one has the inequality

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-f\left({\displaystyle \frac{1}{P_n}}\right)p_i{x}_i\le$

$\le \left(M-m\right)\left(f^{\prime \prime }\left(M\right)-f^{\prime }\left(m\right)\right)\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i\in N}}{p}_i\right)\left(1-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i\in N}}{p}_i\right)$

The second result is embodied in the following theorem.

Theorem 4

Let $f:I\subseteq ℝ\to ℝ$ be a differentiable convex funciton on $\mathring{I}$, $xli$ $\in I,$ $p_i\ge 0\left(i=1,\mathrm{\dots },n\right)$ with $P_n>0$. Then one has the following inequalities

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_{j}f\left({\displaystyle \frac{x_i+x_j}{2}}\right)\le$ (3)

$\le {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{1}{P_n}}s{s}_{i=1}^n{p}_i{x}_i{f}^{\prime }\left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i{f}^{\prime }\left(x_i\right)\right].$

Proof. The inequality

$$\frac{1}{P_n^2}\sum _{i,j=1}^n{p}_i{p}_{j}f\left(\frac{x_i+x_j}{2}\right)\le \frac{1}{P_n}\sum _{i=1}^n{p}_{i}f\left(x_i\right)$$

was proved by the first author in the paper 3 (see also [7]).

To prove the second part of (3), we observe, by the convexity of $f$, that

$$f\left(\frac{x_i+x_j}{2}\right)-f\left(x_i\right)\ge \left(\frac{x_i+x_j}{2}-x_i\right)f^{\prime }\left(x_i\right)$$

for all $i,j=1,\mathrm{\dots },n$.

If we multiply this inequality with $p_i{p}_j\ge 0$ and if we sum these inequalities, we deduce

${\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_{j}f\left({\displaystyle \frac{x_1+x_j}{2}}\right)-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)\ge$

$\ge {\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j\left({\displaystyle \frac{x_j-x_i}{2}}\right)f^{\prime }\left(x_i\right).$

Since a simple computation shows that

${\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j\left({\displaystyle \frac{x_j-x_i}{2}}\right)f^{\prime }\left(x_i\right)=$

$={\displaystyle \frac{1}{2P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-{\displaystyle \frac{1}{2P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{f}^{\prime }\left(x_i\right)$

the proof of the inequality (3) is finished.

By the use of Pečarič’s result, we have  

Corollary 5

If $f,p_i,x_i$ are as above and, in addition, $m\le x_i\le M,$ for all $i=1,\mathrm{\dots },n$. Then one has the following inequality:

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j,=1}^n}{p}_i{p}_{j}f\left({\displaystyle \frac{x_i+x_j}{2}}\right)\le$

$\le {\displaystyle \frac{1}{2}}\left(M-m\right)\left(f^{\prime \prime }\left(M\right)-f^{\prime }\left(m\right)\right)\underset{1\le k\le n-1}{\max }\{P_k{\overline{P}}_{k+1}/P_n^2\}.$

We also have

Corollary 6

In the above assumptions, one has

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_{j}f\left({\displaystyle \frac{x_i+x_j}{2}}\right)\le$

$\le {\displaystyle \frac{1}{2}}\left(M-m\right)\left(f^{\prime }\left(M\right)-f^{\prime }\left(m\right)\right)\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i\in N}}{p}_i\right)\left(1-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i\in N}}{p}_i\right).$

The proof of tghis fact follows by the result of Andrica and Badea.

The last result is ebodied in the following theorem.

Theorem 7

Let $f:I\subseteq ℝ\to ℝ$ be a differentiable convex mapping on the interval $I$ and $p_i\ge 0,x_i\in \mathring{I}\left(i=\overline{1,n}\right)$ with $P_n>0$. Then

$0$ $\le {\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_{j}f\left({\displaystyle \frac{x_i+x_j}{2}}\right)-f\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)\le$ (4)

$\le {\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{f}^{\prime }\left({\displaystyle \frac{x_i+x_j}{2}}\right)x_i-{\displaystyle \frac{1}{P_n^3}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{f}^{\prime }\left({\displaystyle \frac{x_i+x_j}{2}}\right).$

Proof. The first inequality was proved in [3] (see also [7] for $k=1$).

To prove the second part of (4), we observe, by the convexity of $f$, that

$f\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)-f\left({\displaystyle \frac{x_{i+x_j}}{2}}\right)\ge$

$\ge \left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i-{\displaystyle \frac{x_i+x_j}{2}}\right)f^{\prime }\left({\displaystyle \frac{x_i+x_j}{2}}\right)$

for all $i,j=1,\mathrm{\dots },n$.  

Proof. If we multiply with $p_i{p}_j\ge 0\left(i,j=\overline{1,n}\right)$ and if we sum these inequalities, we derive

$f\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_{j}f\left({\displaystyle \frac{x_i+x_j}{2}}\right)\ge$

$\ge {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{f}^{\prime }\left({\displaystyle \frac{x_i+x_j}{2}}\right)-$

$-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j\left({\displaystyle \frac{x_i+x_j}{2}}\right)f^{\prime }\left({\displaystyle \frac{x_i+x_j}{2}}\right).$

Since a simple computation shows that

$$\sum _{i,j=1}^n{p}_i{p}_j\left(\frac{x_i+x_j}{2}\right)f^{\prime }\left(\frac{x_i+x_j}{2}\right)=\sum _{i,j=1}^n{p}_i{p}_j{x}_i{f}^{\prime }\left(\frac{x_i+x_j}{2}\right)$$

the proof is thus finished.  

3 Applications

a.

Let $x_i>0,p_i\ge 0\left(i=1,\mathrm{\dots },n\right)$ with $P_n>0$. Then one has

$$1\le \frac{1/P_n{\sum }_{i=1}^n{p}_i{x}_i}{{\left(\prod _{i=1}^n{x}_i^{p_i}\right)}^{1/P_n}}\le \exp \left(\frac{1}{P_n^2}\sum _{i=1}^n{p}_i{x}_i\sum _{i=1}^n{p}_i/x_i-1\right)$$

and

$$1\le \frac{{\left[\prod _{i,j=1}^n{\left(\frac{x_i+x_j}{2}\right)}^{p_i{p}_j}\right]}^{1/P_n^2}}{{\left(\prod _{i\in 1}^n{x}_i^{p_i}\right)}_{1/P_n}}\le \exp \left[\frac{1}{2}\cdot \left(\frac{1}{P_n^2}\sum _{i=1}^n{p}_i{x}_i\sum _{i=1}^n{p}_i/x_i-1\right)\right]$$

and

$1$ $\le {\displaystyle \frac{1/P_n{\sum }_{i=1}^n{p}_i{x}_i}{{\left[\prod _{i,j=1}^n{\left(\frac{x_i+x_j}{2}\right)}^{p_i{p}_j}\right]}^{1/P_n}}}\le$

$\le \exp \left[{\displaystyle \frac{1}{P_n^3}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i,j=1}^n}{\displaystyle \frac{2p_i{p}_j}{\left(x_i+x_j\right)}}-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{\displaystyle \frac{2p_i{p}_j}{\left(x_i+x_j\right)}}\right]$

respectively.

The proofs follow by Theorems 1, 4 and 7 for the convex mapping $f:(0,\mathrm{\infty })\to ℝ$,$f\left(x\right)=-\ln x.$

b.

Let $x_i\ge 0,p_i\ge 0(-=\overline{1,n})$ with $P_n>0$ and $p\ge 1$. Then

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i^p-{\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)}^p$

$\le p\left[{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i^p-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i{x}_i^{p-1}\right]$

and

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i^p-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{\left({\displaystyle \frac{x_i+x_j}{2}}\right)}^p\le$

$\le {\displaystyle \frac{p}{2}}\left[{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i^p-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i{x}_i^{p-1}\right]$

and

$0$ $\le {\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{\left({\displaystyle \frac{x_i+x_j}{2}}\right)}^p-{\left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)}^p\le$

$\le p\left[{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{x}_i{\left({\displaystyle \frac{x_i+x_j}{2}}\right)}^{p-1}-{\displaystyle \frac{1}{P_n^3}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i,j=1}^n}{p}_i{p}_j{\left({\displaystyle \frac{x_i+x_j}{2}}\right)}^{p-1}\right]$

respectively.

The proofs follow by Theorems 1, 4 and 7 for the convex mapping $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty }),f\left(x\right):=x^p\left(p\ge 1\right)$.

c.

In paper [9], C.L. Wang has obtained the following inequality

$${\left[\prod _{i=1}^n{\left(x_i/\left(1-x_i\right)\right)}^{p_i}\right]}^{1/P_n}\le \left(\sum _{i=1}^n{p}_i{x}_i\right)/\left(\sum _{i=1}^n{p}_i\left(1-x_i\right)\right)$$

where $p_i>0,x_i$ $\in (0,1/2]\left(i=\overline{1,n}\right),$ which shows that Ky Fan’s inequality [2] also holds for weighted means.

Let sau ler?? $x_i,p_i\left(i=\overline{1,n}\right)$ be as above, then one has the inequalities:

$1\le \left({\displaystyle \frac{\sum _{i=1}^n{p}_i{x}_i}{{\sum }_{i=1}^n{p}_i\left(1-x_i\right)}}\right)|{\left[{\displaystyle \prod _{i=1}^n}{\left({\displaystyle \frac{x_i}{1-x_i}}\right)}^{p_i}\right]}^{1/P_n}\le$

$\le \exp \left[{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{p_i}{x_i\left(1-x_i\right)}}-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{p_i}{1-x_i}}\right]$

and

$1$ $\le {\left[{\displaystyle \prod _{i,j=1}^n}{\left({\displaystyle \frac{x_i+x_j}{2-x_i-x_j}}\right)}^{p_i{p}_j}\right]}^{1/P_n^2}\mathrm{╱}{\left[{\displaystyle \prod _{i=1}^n}{\left({\displaystyle \frac{x_i}{1-x_i}}\right)}^{p_i}\right]}^{1/P_n}$

$\le \exp \left\{{\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{p_i}{x_i\left(1-x_i\right)}}-{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{p_i}{1-x_i}}\right]\right\}$

and

$1$ $\le \left({\displaystyle \frac{{\sum }_{i=1}^n{p}_i{x}_i}{{\sum }_{i=1}^n{p}_i\left(1-x_i\right)}}\right)\mathrm{╱}{\left[{\displaystyle \prod _{i.j=1}^{n^{-}}}{\left({\displaystyle \frac{x_i+x_j}{2-x_i-x_j}}\right)}^{p_i{p}_j}\right]}^{1/P_n^2}\le$

$\le \exp [{\displaystyle \frac{1}{P_n^3}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i,j=1}^n}{\displaystyle \frac{p_i{p}_j}{\left(x_i+x_j\right)\left(2-x_i-x_j\right)}}-$

$-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i,j=1}^n}{\displaystyle \frac{2x_i{p}_i{p}_j}{\left(x_i+x_j\right)\left(2-x_i-x_j\right)}}]$

respectively.

The proofs follow by Theorems 1, 4 and 7 applied for the convex mapping ,$f\left(x\right)=-\ln \left[x/\left(1-x\right)\right]$.

d.

Let $x_i\in ℝ,p_i\ge 0\left(i=\overline{1,n}\right)$ with $P_n>0$. Then one has the inequalities

$0$ $\ge {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i\exp \left(x_i\right)=\exp \left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)\le$

$\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\exp \left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i\exp \left(x_i\right)$

and

$0$ $\le {\displaystyle \frac{1}{P_n}}{\displaystyle \sum i}=1^n{p}_i\exp \left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\left({\displaystyle \sum _{i=1}^n}{p}_i\exp \left(x_i/2\right)\right)}^2\le$

$\le {\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\exp \left(x_i\right)-{\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\displaystyle \sum _{i=1}^n}{p}_i\exp \left(x_i\right)\right]$

and

$0$ $\le {\displaystyle \frac{1}{P_n^2}}({\displaystyle \sum _{i=1}^n}{p}_i\exp {(x_i/2)}^2-\exp \left({\displaystyle \frac{1}{P_n}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)\le$

$\le {\displaystyle \frac{1}{P_n^2}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\exp \left(x_i/2\right){\displaystyle \sum _{i=1}^n}{p}_i\exp \left(x_i/2\right)-{\displaystyle \frac{1}{P_n^3}}{\displaystyle \sum _{i=1}^n}{p}_i{x}_i{\left({\displaystyle \sum _{i=1}^n}{p}_i\exp \left(x_i\right)/2\right)}^2$

respectively.

The proofs follow form Theorems 1, 4 and 7applied for the convex mapping $f:ℝ\to {ℝ}_{+},f\left(x\right)=\exp \left(x\right)$.

Received 10 II 1993        Department of Mathematics

Timişoara University

B-dul V. Pârvan

R-1900 Timişoara, România