// Workers AI · dad joke modeDoes Popoviciu's inequality have friends? It's unequal to others.

Not to be confused with Popoviciu's inequality on variances.

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 $I\subseteq \mathbb {R}$ to $\mathbb {R}$. If f is convex, then for any three points x, y, z in I,

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

If a function f is continuous, then it is convex if and only if the above inequality holds for all x, y, z from $I$. 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 $I\subseteq \mathbb {R}$ to $\mathbb {R}$. Then f is convex if and only if, for any integers n and k where n ≥ 3 and $2\leq k\leq n-1$, and any n points $x_{1},\dots ,x_{n}$ from I,

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

Weighted inequality

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

Let f be a continuous function from an interval $I\subseteq \mathbb {R}$ to $\mathbb {R}$. 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 $w_{2}+w_{3}\neq 0,w_{3}+w_{1}\neq 0$ and $w_{1}+w_{2}\neq 0$. Then,

${\begin{aligned}&w_{1}f\left(x_{1}\right)+w_{2}f\left(x_{2}\right)+w_{3}f\left(x_{3}\right)+\left(w_{1}+w_{2}+w_{3}\right)f\left({\frac {w_{1}x_{1}+w_{2}x_{2}+w_{3}x_{3}}{w_{1}+w_{2}+w_{3}}}\right)\\&\geq \left(w_{2}+w_{3}\right)f\left({\frac {w_{2}x_{2}+w_{3}x_{3}}{w_{2}+w_{3}}}\right)+\left(w_{3}+w_{1}\right)f\left({\frac {w_{3}x_{3}+w_{1}x_{1}}{w_{3}+w_{1}}}\right)+\left(w_{1}+w_{2}\right)f\left({\frac {w_{1}x_{1}+w_{2}x_{2}}{w_{1}+w_{2}}}\right)\end{aligned}}$