# 平均数不等式

${\displaystyle x_{1},x_{2},\ldots ,x_{n}\in \mathbb {R_{+}} \Rightarrow {\dfrac {n}{\displaystyle \sum _{i=1}^{n}{\dfrac {1}{x_{i}}}}}\leq {\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}\leq {\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}}{n}}\leq {\sqrt {\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}^{2}}{n}}}}$

${\displaystyle H_{n}\leq G_{n}\leq A_{n}\leq Q_{n}}$

${\displaystyle G_{n}={\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}={\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}$

${\displaystyle A_{n}={\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}}{n}}={\dfrac {x_{1}+x_{2}+\cdots +x_{n}}{n}}}$

${\displaystyle Q_{n}={\sqrt {\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}^{2}}{n}}}={\sqrt {\dfrac {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}{n}}}}$

## ${\displaystyle n=2}$时的情形

• 第一个不等号
 ${\displaystyle \left(x_{1}-x_{2}\right)^{2}=x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}\,\;}$ ${\displaystyle \geq 0\,\;}$ ${\displaystyle \left(x_{1}+x_{2}\right)^{2}=x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}\,\;}$ ${\displaystyle \geq 4x_{1}x_{2}\,\;}$ ${\displaystyle 1\,\;}$ ${\displaystyle \geq {\frac {4x_{1}x_{2}}{\left(x_{1}+x_{2}\right)^{2}}}\,\;}$ ${\displaystyle 1\,\;}$ ${\displaystyle \geq {\frac {2{\sqrt {x_{1}x_{2}}}}{x_{1}+x_{2}}}\,\;}$ ${\displaystyle {\sqrt {x_{1}x_{2}}}\,\;}$ ${\displaystyle \geq {\frac {2x_{1}x_{2}}{x_{1}+x_{2}}}={\frac {2}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}}}\,\;}$
• 第二个不等号
 ${\displaystyle \left(x_{1}-x_{2}\right)^{2}=x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}\,\;}$ ${\displaystyle \geq 0\,\;}$ ${\displaystyle \left(x_{1}+x_{2}\right)^{2}=x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}\,\;}$ ${\displaystyle \geq 4x_{1}x_{2}\,\;}$ ${\displaystyle \left({\frac {x_{1}+x_{2}}{2}}\right)^{2}\,\;}$ ${\displaystyle \geq x_{1}x_{2}\,\;}$ ${\displaystyle {\frac {x_{1}+x_{2}}{2}}\,\;}$ ${\displaystyle \geq {\sqrt {x_{1}x_{2}}}\,\;}$
• 第三个不等号
 ${\displaystyle ({\frac {x_{1}-x_{2}}{2}})^{2}={\frac {x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}}{4}}\,\;}$ ${\displaystyle \geq 0\,\;}$ ${\displaystyle {\frac {x_{1}^{2}+x_{2}^{2}}{2}}={\frac {x_{1}^{2}+x_{2}^{2}+x_{1}^{2}+x_{2}^{2}}{4}}\,\;}$ ${\displaystyle \geq {\frac {2x_{1}x_{2}+x_{1}^{2}+x_{2}^{2}}{4}}={\frac {(x_{1}+x_{2})^{2}}{4}}\,\;}$ ${\displaystyle {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}}{2}}}\,\;}$ ${\displaystyle \geq {\frac {x_{1}+x_{2}}{2}}\,\;}$

## 证明方法

${\displaystyle n=2}$时易证；

${\displaystyle S=a_{1}+a_{2}+\cdots +a_{k}}$${\displaystyle \left({\frac {a_{1}+a_{2}+\cdots +a_{k+1}}{k+1}}\right)^{k+1}=\left[{\frac {S}{k}}+{\frac {ka_{k+1}-S}{k\left(k+1\right)}}\right]^{k+1}}$，根据引理

${\displaystyle \left[{\frac {S}{k}}+{\frac {ka_{k+1}-S}{k\left(k+1\right)}}\right]^{k+1}\geq \left({\frac {S}{k}}\right)^{k+1}+(k+1)\left({\frac {S}{k}}\right)^{k}{\frac {ka_{k+1}-S}{k(k+1)}}=\left({\frac {S}{k}}\right)^{k}a_{k+1}\geq a_{1}a_{2}\cdots a_{k}a_{k+1}}$，当且仅当${\displaystyle ka_{k+1}-S=0}$${\displaystyle a_{1}=a_{2}=\cdots =a_{k}}$时，即${\displaystyle a_{1}=a_{2}=\cdots =a_{k}=a_{k+1}}$时取等号。

${\displaystyle n=1}$时命题为：若${\displaystyle a=1}$，则${\displaystyle a\geqslant 1}$，当且仅当${\displaystyle a=1}$时等号成立。命题显然成立。

${\displaystyle A_{n}\leq Q_{n}}$等价于${\displaystyle Q_{n}^{2}-A_{n}^{2}\geq 0}$。事实上，${\displaystyle Q_{n}^{2}-A_{n}^{2}}$等于${\displaystyle a_{1},a_{2},\ldots ,a_{n}}$方差，通过这个转化可以证出${\displaystyle Q_{n}^{2}-A_{n}^{2}\geq 0}$，证明如下。

${\displaystyle Q_{n}^{2}-A_{n}^{2}=Q_{n}^{2}-2A_{n}^{2}+A_{n}^{2}={\frac {a_{1}^{2}+a_{2}^{2}+\ldots +a_{n}^{2}}{n}}-{\frac {2a_{1}A_{n}+2a_{2}A_{n}+\ldots +2a_{n}A_{n}}{n}}+{\frac {nA_{n}^{2}}{n}}}$

${\displaystyle ={\frac {(a_{1}^{2}-2a_{1}A_{n}+A_{n}^{2})+(a_{2}^{2}-2a_{2}A_{n}+A_{n}^{2})+\ldots +(a_{n}^{2}-2a_{n}A_{n}+A_{n}^{2})}{n}}}$

${\displaystyle ={\frac {(a_{1}-A_{n})^{2}+(a_{2}-A_{n})^{2}+\ldots +(a_{n}-A_{n})^{2}}{n}}}$

${\displaystyle \geqslant 0}$

## 参见

1. ^ 普通高中课程标准实验教科书 数学 选修4-5 不等式选讲. 人民教育出版社. 2007: 52. ISBN 978-7-107-18675-2.