1
$\begingroup$

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 $m=x_1=x_2=\cdots=x_n=M$

Improve this question
$\endgroup$
1
  • $\begingroup$ Why $n$ twice on the left rather than once in the denominator on the right? $\endgroup$
    – Włodzimierz Holsztyński
    Commented Aug 21, 2016 at 4:44

1 Answer 1

Reset to default
2
$\begingroup$

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 $g(t)\geqslant g(t)+g(2s-t)\geqslant 2g(s)$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Dear Dr Fedor Petrov, Thank to You very much for your proof. I think the inequality is true with weighted version. And I think it is a generalization of Kantorovic inequality. What do you think with my remarks? Thank to You again. $\endgroup$
    – Oai Thanh Đào
    Commented Aug 21, 2016 at 14:36
  • $\begingroup$ 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 $m=x_1=x_2=\cdots=x_n=M$ $\endgroup$
    – Oai Thanh Đào
    Commented Aug 22, 2016 at 1:56
  • $\begingroup$ Yes, the proof is the same. $\endgroup$
    – Fedor Petrov
    Commented Aug 22, 2016 at 7:19
  • $\begingroup$ Dear Dr @FedorPetrov , The inequality above is a generalization of the Kantorovich inequality? $\endgroup$
    – Oai Thanh Đào
    Commented Aug 22, 2016 at 7:24
  • $\begingroup$ 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. $\endgroup$
    – Fedor Petrov
    Commented Aug 22, 2016 at 13:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .