Popoviciu's inequality

Question

If $f(x): \mathbb R \to \mathbb R$ is a convex function, prove that $$f(x) + f(y) + f(z) + 3 f(\frac{x + y + z}{3}) \geq 2 f(\frac{x + y}{2}) + 2 f(\frac{x + z}{2}) + 2 f(\frac{y + z}{2})$$

I proved the first case where $\displaystyle y ≤ \frac{x+y+z}{3}$ and I don't know how to prove for the second case where $\displaystyle y ≥ \frac{x+y+z}{3}$

Answer 1

I assume you already proved the following:

Suppose that $f: \Bbb R \to \Bbb R$ is a convex function, $x\le y \le z$ are real numbers. If $$y \le \frac{x+y+z}{3}$$ then $$f(x) + f(y) + f(z) + 3 f(\frac{x + y + z}{3}) \geq 2 f(\frac{x + y}{2}) + 2 f(\frac{x + z}{2}) + 2 f(\frac{y + z}{2})$$

We want to prove the inequality in the case when $y \ge \frac{x+y+z}{3}$.

Let's introduce three new variables $$x' = -z; \quad y'=-y; \quad z'=-x$$ and a new function $f': \Bbb R \to \Bbb R$ defined by $$f'(t)= f(-t)$$ Since $f$ is convex, $f'$ is convex too (I cannot prove it here, since I don't know which definition of convex function you use). Moreover, under the assumptions $x \le y \le z$ and $y \ge \frac{x+y+z}{3}$ we have $$x' \le y' \le z'; \qquad y' \le \frac{x'+y'+z'}{3}$$ Hence we can use the theorem above with $f', x', y', z'$.

This concludes the proof.