# Convex function Inequality (3 - point version)

I was reading this article on inequalities (which some of you may find useful) here.

On page 7, I came across this question by Titu Andreescu, which I shall reproduce here:

Question: Let f be a convex function on $\mathbb{R}$. If $x_1$, $x_2$ and $x_3$ lie in it's domain, prove that: $$f(x_1)+f(x_2)+f(x_3)+f\left(\frac{x_1+x_2+x_3}{3}\right) \geq \frac{4}{3}\left[f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+f\left(\frac{x_3+x_1}{2}\right)\right]$$

My Attempt:

I rewrote $f(x_1)+f(x_2)+f(x_3)$ as $\frac{f(x_1) + f(x_2)}{2}+\frac{f(x_2) + f(x_3)}{2}+\frac{f(x_3) + f(x_1)}{2}$. By Jensen's inequality,

$$f(x_1)+f(x_2)+f(x_3) \geq f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+f\left(\frac{x_3+x_1}{2}\right)$$ However since $$f\left(\frac{x_1+x_2+x_3}{3}\right)= f\left(\frac{1}{3}\left[\frac{x_1+x_2}{2}+\frac{x_2+x_3}{2}+\frac{x_3+x_1}{2}\right]\right) \leq \frac{1}{3}\left[f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+f\left(\frac{x_3+x_1}{2}\right)\right]$$

I am stuck since the flipped inequality prevents this approach. I feel that there may be another manipulation that would work but I cannot see it as of now. I appreciate any help and hope you find the embedded article helpful to those who like symmetric inequalities etc.

• Look up Popoviciu's inequality. Your strategy doesn't work since this is the case where you have $A \ge B \ge C$ and you want to prove something like $A+C \ge 2B$. Let me add that this is a non-trivial inequality (which means it doesn't follow directly from Jensen). The proof I have seen uses the concept of majorization and is probably a good thing to know about. – user27126 Jun 8 '13 at 19:59
• But this question was asked in the earlier section when only AMGM and Jensen's was covered... – Gautam Shenoy Jun 8 '13 at 22:33
• I saw the document, and I'm not sure why it is arranged that way. – user27126 Jun 8 '13 at 22:54