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I proposed my conjecture as follows:

Let $f(x)$ is a real continuous function 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 $m=x_1=x_2=....=x_n=M$

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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.

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    $\begingroup$ Or to skip the convolution, take $n=3$, $m=x_1=0$, and $M=x_2=x_3=1$, and $f(x)=x^k$. Then the inequality reads $$\frac{2}{3} - \left(\frac{2}{3}\right)^k \le \frac{1}{2} - \left(\frac{1}{2}\right)^k$$ which is clearly false for large $k$. (Indeed, for $k \ge 4$.) $\endgroup$
    – Nate Eldredge
    Commented Jul 25, 2016 at 2:13
  • $\begingroup$ Yes, this will also do. $\endgroup$
    – Iosif Pinelis
    Commented Jul 25, 2016 at 2:15

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