# 平均数不等式

$H_n \le G_n \le A_n \le Q_n$

$G_n=\sqrt[n]{\prod_{i=1}^nx_i}=\sqrt[n]{x_1x_2\cdots x_n}$

$A_n = \dfrac{\displaystyle \sum_{i=1}^{n} x_i }{n} = \dfrac{x_1+x_2+\cdots+x_n}{n}$

$Q_n = \sqrt{\dfrac{\displaystyle \sum_{i=1}^{n} x_i^2}{n}} = \sqrt{\dfrac{x^2_1+x^2_2+\cdots+x^2_n}{n}}$

## $n = 2$ 时的情形

• 第一个不等号
 $\left( x_1 - x_2 \right) ^2 = x_1^2 - 2 x_1 x_2 + x_2^2\,\;$ $\ge 0\,\;$ $\left( x_1 + x_2 \right) ^2 = x_1^2 + 2 x_1 x_2 + x_2^2\,\;$ $\ge 4 x_1 x_2\,\;$ $1\,\;$ $\ge \frac{4 x_1 x_2}{\left( x_1 + x_2 \right) ^2}\,\;$ $1\,\;$ $\ge \frac{2 \sqrt{x_1 x_2}}{ x_1 + x_2 }\,\;$ $\sqrt{x_1 x_2}\,\;$ $\ge \frac{2 x_1 x_2}{ x_1 + x_2 } = \frac{2}{ \frac{1}{x_1} + \frac{1}{x_2} }\,\;$
• 第二个不等号
 $\left( x_1 - x_2 \right) ^2 = x_1^2 - 2 x_1 x_2 + x_2^2\,\;$ $\ge 0\,\;$ $\left( x_1 + x_2 \right) ^2 = x_1^2 + 2 x_1 x_2 + x_2^2\,\;$ $\ge 4 x_1 x_2\,\;$ $\left( \frac{x_1 + x_2}{2} \right)^2\,\;$ $\ge x_1 x_2\,\;$ $\frac{x_1 + x_2}{2}\,\;$ $\ge \sqrt{x_1 x_2}\,\;$
• 第三个不等号
 $(\frac{ x_1 - x_2 }{2}) ^2 = \frac{x_1^2 - 2 x_1 x_2 + x_2^2}{4}\,\;$ $\ge 0\,\;$ $\frac{x_1^2 + x_2^2}{2} = \frac{x_1^2 + x_2^2 + x_1^2 + x_2^2}{4}\,\;$ $\ge \frac{2 x_1 x_2 + x_1^2 + x_2^2}{4} = \frac{(x_1 + x_2)^2}{4}\,\;$ $\sqrt{\frac{x_1^2 + x_2^2}{2}}\,\;$ $\ge \frac{x_1 + x_2}{2}\,\;$

## 证明方法

（注：在此证明的，是对n维形式的均值不等式的证明方法。）

$S = a_1 + a_2 + \cdots + a_{k}$$\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

$\left(\frac{S}{k}+\frac{ka_{k+1}-S}{k\left(k+1\right)}\right)^{k+1} \ge \left ( {\frac{S}{k}} \right )^{k+1}+(k+1)\left ( {\frac{S}{k}} \right )^{k}\frac{ka_{k+1}-S}{k(k+1)} = \left(\frac{S}{k}\right)^{k}a_{k+1} \ge a_1a_2 \cdots a_ka_{k+1}$,当且仅当$ka_{k+1}-S=0$$a_1=a_2=\cdots=a_k$时，即$a_1=a_2=\cdots=a_k=a_{k+1}$时取等号。