Popoviciu's inequality

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

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

If a function f is continuous, then it is convex if and only if the above inequality holds for all x, y, z from ${\displaystyle I}$. When f is strictly convex, the inequality is strict except for x = y = z.[3]

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:[4]

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

${\displaystyle {\frac {1}{k}}{\binom {n-2}{k-2}}\left({\frac {n-k}{k-1}}\sum _{i=1}^{n}f(x_{i})+nf\left({\frac {1}{n}}\sum _{i=1}^{n}x_{i}\right)\right)\geq \sum _{1\leq i_{1}<\dots <i_{k}\leq n}f\left({\frac {1}{k}}\sum _{j=1}^{k}x_{i_{j}}\right)}$

Popoviciu's inequality can also be generalized to a weighted inequality.[5][6][7]