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 ⊆ R {\displaystyle I\subseteq \mathbb {R} } to R {\displaystyle \mathbb {R} } . If f is convex, then for any three points x, y, z in I,
f ( x ) + f ( y ) + f ( z ) 3 + f ( x + y + z 3 ) ≥ 2 3 [ f ( x + y 2 ) + f ( y + z 2 ) + f ( z + x 2 ) ] . {\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 I {\displaystyle 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 ⊆ R {\displaystyle I\subseteq \mathbb {R} } to R {\displaystyle \mathbb {R} } . Then f is convex if and only if, for any integers n and k where n ≥ 3 and 2 ≤ k ≤ n − 1 {\displaystyle 2\leq k\leq n-1} , and any n points x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}} from I,
1 k ( n − 2 k − 2 ) ( n − k k − 1 ∑ i = 1 n f ( x i ) + n f ( 1 n ∑ i = 1 n x i ) ) ≥ ∑ 1 ≤ i 1 < ⋯ < i k ≤ n f ( 1 k ∑ j = 1 k x i j ) {\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)}
Weighted inequality Popoviciu's inequality can also be generalized to a weighted inequality. Let f be a continuous function from an interval I ⊆ R {\displaystyle I\subseteq \mathbb {R} } to R {\displaystyle \mathbb {R} } . Let x 1 , x 2 , x 3 {\displaystyle x_{1},x_{2},x_{3}} be three points from I {\displaystyle I} , and let w 1 , w 2 , w 3 {\displaystyle w_{1},w_{2},w_{3}} be three nonnegative reals such that w 2 + w 3 ≠ 0 , w 3 + w 1 ≠ 0 {\displaystyle w_{2}+w_{3}\neq 0,w_{3}+w_{1}\neq 0} and w 1 + w 2 ≠ 0 {\displaystyle w_{1}+w_{2}\neq 0} . Then,
w 1 f ( x 1 ) + w 2 f ( x 2 ) + w 3 f ( x 3 ) + ( w 1 + w 2 + w 3 ) f ( w 1 x 1 + w 2 x 2 + w 3 x 3 w 1 + w 2 + w 3 ) ≥ ( w 2 + w 3 ) f ( w 2 x 2 + w 3 x 3 w 2 + w 3 ) + ( w 3 + w 1 ) f ( w 3 x 3 + w 1 x 1 w 3 + w 1 ) + ( w 1 + w 2 ) f ( w 1 x 1 + w 2 x 2 w 1 + w 2 ) {\displaystyle {\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}}}
Notes