# 切比雪夫總和不等式

${\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}}$${\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n}}$，則：

${\displaystyle n\sum _{k=1}^{n}a_{k}b_{k}\geq \left(\sum _{k=1}^{n}a_{k}\right)\left(\sum _{k=1}^{n}b_{k}\right)\geq n\sum _{k=1}^{n}a_{k}b_{n+1-k}}$

${\displaystyle {\frac {1}{n}}\sum _{k=1}^{n}a_{k}b_{k}\geq \left({\frac {1}{n}}\sum _{k=1}^{n}a_{k}\right)\left({\frac {1}{n}}\sum _{k=1}^{n}b_{k}\right)\geq {\frac {1}{n}}\sum _{k=1}^{n}a_{k}b_{n+1-k}}$

## 证明

${\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}}$ ${\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n}}$ ，由排序不等式可知，最大的和为顺序和：

${\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}}$

${\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}}$
${\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{2}+a_{2}b_{3}+\cdots +a_{n}b_{1}}$
${\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{3}+a_{2}b_{4}+\cdots +a_{n}b_{2}}$
${\displaystyle \vdots }$
${\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{n}+a_{2}b_{1}+\cdots +a_{n}b_{n-1}}$

${\displaystyle n(a_{1}b_{1}+\cdots +a_{n}b_{n})\geq (a_{1}+\cdots +a_{n})(b_{1}+\cdots +b_{n})}$

${\displaystyle {\frac {(a_{1}b_{1}+\cdots +a_{n}b_{n})}{n}}\geq {\frac {(a_{1}+\cdots +a_{n})}{n}}\cdot {\frac {(b_{1}+\cdots +b_{n})}{n}}}$

## 积分形式

${\displaystyle f}$ ${\displaystyle g}$ 区间${\displaystyle [0,1]}$ 上的可积的实函数，并且两者都是递增或两者都是递减的，则：

${\displaystyle \int fg\geq \int f\int g}$