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.