# 切比雪夫总和不等式

## 维基百科，自由的百科全书

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

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

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

## 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：

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