Popoviciu's Inequality

Popoviciu’s inequality will be used in the same manner as Jensen’s inequality. But we must note that this inequality is stronger, i.e. in some cases this inequality can be a powerful tool for proving other inequalities where Jensen’s inequality does not work.

Popoviciu’s Inequality

Let $f : [a, b] \to \mathbb{R}$ be a convex function on the interval $[a, b].$ Then for any $x, y, z \in [a, b]$ we have

$f \left(\dfrac{x+y+z}{3}\right) + \dfrac{f(x) + f(y) + f(z)}{3} \ge \dfrac{2}{3}\Bigg(f\left(\dfrac{x+y}{2}\right) + f\left(\dfrac{y+z}{2}\right) + f\left(\dfrac{z+x}{2}\right)\Bigg).$

Without loss of generality, we assume that $x \le y \le z.$

If $x\le y \le \frac{x+y+z}{3},$ then

$\dfrac{x+y+z}{3} \le \dfrac{x+z}{2} \le z\quad \text{ and }\quad \dfrac{x+y+z}{3} \le \dfrac{y+z}{2} \le z.$

Therefore, there exist $s, t \in [0, 1]$ such that

$\begin{aligned} \dfrac{x+z}{2} &= \left(\dfrac{x+y+z}{3}\right)s + z(1-s)\qquad \text{(A)}\\\\ \dfrac{y+z}{2} &= \left(\dfrac{x+y+z}{3}\right)t + z(1-t).\qquad \text{(B)} \end{aligned}$

On adding (A) and (B),

$\dfrac{x+y-2z}{2}=\dfrac{x+y-2z}{3}(s+t)\implies s+t = \dfrac{3}{2}.$

As $f$ is a convex function,

$\begin{aligned} f\left(\dfrac{x+z}{2}\right) \le s\cdot f\left(\dfrac{x+y+z}{3}\right) + (1-s)\cdot f(z)\\\\ f\left(\dfrac{y+z}{2}\right) \le t\cdot f\left(\dfrac{x+y+z}{3}\right) + (1-t)\cdot f(z) \end{aligned}$

and

$f\left(\dfrac{x+y}{2}\right) \le \dfrac{1}{2}\cdot f(x) + \dfrac{1}{2}\cdot f(y).$

After adding together the last three inequalities we obtain the required inequality.

The case when $\frac{x+y+z}{3} \le y$ is considered similarly, bearing in mind that $x \le \frac{x+z}{2} \le \frac{x+y+z}{3}$ and $x \le \frac{y+z}{2} \le \frac{x+y+z}{3}.\ _\square$

Note: When $f$ is a concave function, the inequality gets reversed.