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.

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.