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

$\endgroup$
  • $\begingroup$ Why $n$ twice on the left rather than once in the denominator on the right? $\endgroup$ – Włodzimierz Holsztyński Aug 21 '16 at 4:44
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Aug 21 '16 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 Aug 22 '16 at 1:56
  • $\begingroup$ Yes, the proof is the same. $\endgroup$ – Fedor Petrov Aug 22 '16 at 7:19
  • $\begingroup$ Dear Dr @FedorPetrov , The inequality above is a generalization of the Kantorovich inequality? $\endgroup$ – Oai Thanh Đào Aug 22 '16 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 Aug 22 '16 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.