Chebyshev's sum inequality

In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if

${\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}\quad }$ and ${\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}$

then

${\displaystyle {1 \over n}\sum _{k=1}^{n}a_{k}b_{k}\geq \left({1 \over n}\sum _{k=1}^{n}a_{k}\right)\!\!\left({1 \over n}\sum _{k=1}^{n}b_{k}\right)\!.}$

In words, if we are given two sequences that are both non-increasing or non-decreasing, then the product of their averages is less than the average of their (termwise) product.

Similarly, if

${\displaystyle a_{1}\leq a_{2}\leq \cdots \leq a_{n}\quad }$ and ${\displaystyle \quad b_{1}\geq b_{2}\geq \cdots \geq b_{n},}$

then

${\displaystyle {1 \over n}\sum _{k=1}^{n}a_{k}b_{k}\leq \left({1 \over n}\sum _{k=1}^{n}a_{k}\right)\!\!\left({1 \over n}\sum _{k=1}^{n}b_{k}\right)\!.}$