Popoviciu's inequality

In convex analysis, Popoviciu's inequality is an inequality about convex functions. It is similar to Jensen's inequality and was found in 1965 by Tiberiu Popoviciu, a Romanian mathematician.

Formulation

Let f be a function from an interval $I\subseteq ℝ$ to $ℝ$. If f is convex, then for any three points x, y, z in I,

$\frac{f(x)+f(y)+f(z)}{3}+f\left(\frac{x+y+z}{3}\right)\ge \frac{2}{3}[f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right)].$

If a function f is continuous, then it is convex if and only if the above inequality holds for all x, y, z from $I$. When f is strictly convex, the inequality is strict except for x = y = z.

Generalizations

It can be generalized to any finite number n of points instead of 3, taken on the right-hand side k at a time instead of 2 at a time:

Let f be a continuous function from an interval $I\subseteq ℝ$ to $ℝ$. Then f is convex if and only if, for any integers n and k where n ≥ 3 and $2\le k\le n-1$, and any n points $x_1,\dots ,x_n$ from I,

$\frac{1}{k}(\genfrac{}{}{0ex}{}{n-2}{k-2})(\frac{n-k}{k-1}\sum _{i=1}^{n}f(x_i)+nf\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right))\ge \sum _{1\le i_1<\dots <i_k\le n}f\left(\frac{1}{k}\sum _{j=1}^k{x}_{i_j}\right)$

Weighted inequality

Popoviciu's inequality can also be generalized to a weighted inequality.

Let f be a continuous function from an interval $I\subseteq ℝ$ to $ℝ$. Let $x_1,x_2,x_3$ be three points from $I$, and let $w_1,w_2,w_3$ be three nonnegative reals such that $w_2+w_3\ne 0,w_3+w_1\ne 0$ and $w_1+w_2\ne 0$. Then,

$\begin{array}{cc}& w_{1}f\left(x_1\right)+w_{2}f\left(x_2\right)+w_{3}f\left(x_3\right)+(w_1+w_2+w_3)f\left(\frac{w_1{x}_1+w_2{x}_2+w_3{x}_3}{w_1+w_2+w_3}\right)\\ & \ge (w_2+w_3)f\left(\frac{w_2{x}_2+w_3{x}_3}{w_2+w_3}\right)+(w_3+w_1)f\left(\frac{w_3{x}_3+w_1{x}_1}{w_3+w_1}\right)+(w_1+w_2)f\left(\frac{w_1{x}_1+w_2{x}_2}{w_1+w_2}\right)\end{array}$