Inspire from [Kantorovich Inequality](http://mathworld.wolfram.com/KantorovichInequality.html) and [my previous question](https://mathoverflow.net/questions/245092/a-conjecture-generalization-of-karamata-inequality). I am looking for a proof of the nice inequality as following:


Let $f(x)$ is a real continuous function that is strictly convex on $[m, M]$, let $m \le x\_i \le M$, for $i=1,2,\ldots,n$ then show that:


$$nf\left(\frac{x\_1+\cdots+x\_n}{n}\right)+n\left(f(M)+f(m)-2f\left(\frac{M+m}{2}\right)\right) \ge f(x\_1)+\cdots+f(x\_n)$$


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



Inspire from Kantorovich Inequality
This is true. Choose a linear function $\ell(x)$ such that $\ell(m)=f(m),\ell(M)=f(M)$. Denoting $g=f-\ell$ we get $g(m)=g(M)=0$ and have to prove $$ng\left(\frac{x\_1+\cdots+x\_n}{n}\right)-2ng\left(\frac{M+m}{2}\right) \geqslant g(x\_1)+\cdots+g(x\_n).$$
But $g$ is convex, thus $g$ is non-positive on $[m,M$]. So, RHS is non-positive, and it suffices to prove that LHS is non-negative. Indeed, if we denote $t=\frac{x\_1+\dots+x\_n}n$, $s=\frac{m+M}2$, then $g(t)\geqslant g(t)+g(2s-t)\geqslant 2g(s)$.


