I proposed my conjecture as follows:


Let $f(x)$ is a [real continuous function](http://mathworld.wolfram.com/ContinuousFunction.html) on $[m, M]$ and $f'>0, f''>0$ on $[m, M]$, let $m \le x\_i \le M$, for $i=1, 2,..., n$. Then


$$\frac{f(x\_1)+f(x\_2)+.....+f(x\_n)}{n}-f(\frac{x\_1+x\_2+....x\_n}{n}) \le \frac{f(M)+ f(m)}{2}-f(\frac{M+m}{2}) $$


Equality holds if only if $m=x\_1=x\_2=....=x\_n=M$



An inequality of real continuous function with f'>0 and f''>0
This conjecture is false. E.g., let $n=3$, $m=x\_1=0$, $M=x\_2=x\_3=3$, and (say) $f(x):=0\vee(x-2)$. Then the inequality does not hold. Replacing now $f$ by its convolution with (say) the pdf of the centered normal distribution with a small enough variance, one can satisfy the conditions $f'>0, f''>0$ on $[m, M]$, whereas, by continuity, the inequality will still be false. 


For instance, if the variance of the centered normal distribution is $1/4$, then the resulting convolution, given by the formula 
\begin{equation}
 f(x)=\frac{1}{2} \left((x-2) \text{erf}\left(\sqrt{2} (x-2)\right)+x+\frac{e^{-2
 (x-2)^2}}{\sqrt{2 \pi }}-2\right),
\end{equation}
will satisfy the conditions $f'>0, f''>0$ on $\mathbb R$ but will fail to satisfy the inequality in question. 


