# 切比雪夫總和不等式

${\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}}$

${\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n}}$

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

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

## 证明

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

${\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n}.\,}$

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

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

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

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

## 积分形式

fg 是区间 [0,1]上的可积的实函数，并且两者都是递增（或递减）的，则有

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

## 外部連結

Mathworld: Chebyshev Sum Inequality