# Inspire from Kantorovich Inequality

Inspire from Kantorovich Inequality and my previous question. 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 $m=x_1=x_2=\cdots=x_n=M$

• Why $n$ twice on the left rather than once in the denominator on the right? – Włodzimierz Holsztyński Aug 21 '16 at 4:44

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 $g(t)\geqslant g(t)+g(2s-t)\geqslant 2g(s)$.
• Let $f(x)$ is a real continuous function that is strictly convex on $[m, M]$, let $m \le x_i \le M$, for $i=1,2,\ldots,n$, $\lambda_i > 0$ and $\sum_1^{n} \lambda_i = 1$ then show that: $$f\left(\lambda_1x_1+\cdots+\lambda_nx_n \right)+\left(f(M)+f(m)-2f\left(\frac{M+m}{2}\right)\right) \ge \lambda_1f(x_1)+\cdots+\lambda_nf(x_n)$$ Equality holds if only if $m=x_1=x_2=\cdots=x_n=M$ – Oai Thanh Đào Aug 22 '16 at 1:56
• I do not think so. It would be if you replace coefficient of $f(M)+f(m)-2f((m+M)/2)$ from $n$ to $n/2$. But such inequality is not true in general. – Fedor Petrov Aug 22 '16 at 13:19