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

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

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,

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

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.