# 切比雪夫總和不等式

$a_1 \geq a_2 \ge \cdots \geq a_n$

$b_1 \geq b_2 \ge \cdots \geq b_n$

$n \sum_{k=1}^n a_kb_k \ge \left(\sum_{k=1}^n a_k\right)\left(\sum_{k=1}^n b_k\right) \ge n \sum_{k=1}^n a_kb_{n+1-k}$

$\frac1n \sum_{k=1}^n a_kb_k \ge \left(\frac1n \sum_{k=1}^n a_k\right)\left(\frac1n \sum_{k=1}^n b_k\right) \ge \frac1n \sum_{k=1}^n a_kb_{n+1-k}$

## 证明

$a_1 \geq a_2 \geq \cdots \geq a_n \,$

$b_1 \geq b_2 \geq \cdots \geq b_n. \,$

$a_1 b_1 + \cdots + a_n b_n \,$

$a_1 b_1 + \cdots + a_n b_n = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \,$
$a_1 b_1 + \cdots + a_n b_n \geq a_1 b_2 + a_2 b_3 + \cdots + a_n b_1 \,$
$a_1 b_1 + \cdots + a_n b_n \geq a_1 b_3 + a_2 b_4 + \cdots + a_n b_2 \,$
$\vdots \,$
$a_1 b_1 + \cdots + a_n b_n \geq a_1 b_n + a_2 b_1 + \cdots + a_n b_{n-1} \,$

$n (a_1 b_1 + \cdots + a_n b_n) \geq (a_1 + \cdots + a_n) (b_1 + \cdots + b_n);$

$\frac {(a_1 b_1 + \cdots + a_n b_n)} {n} \geq \frac {(a_1 + \cdots + a_n)}{n} \cdot \frac {(b_1 + \cdots + b_n)}{n}.$

## 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

$\int fg \geq \int f \int g.\,$

## 外部鏈結

Mathworld: Chebyshev Sum Inequality