均值不等式证明

　　设$x_{1},~x_{2},~\ldots,~x_{n}$为非负实数，其中有：

　　调和平均数 $$H_{n} = \frac{n}{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \cdots + \frac{1}{x_{n}}} = \frac{n}{\sum\limits_{i = 1}^{n}\frac{1}{x_{i}}}$$

　　几何平均数 $$G_{n} = \sqrt[n]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{n}} = \sqrt[n]{\prod\limits_{i = 1}^{n}x_{i}}$$

　　算数平均数 $$A_{n} = \frac{x_{1} + x_{2} + \cdots + x_{n}}{n} = \frac{\sum\limits_{i = 1}^{n}x_{i}}{n}$$

　　平方平均数 $$Q_{n} = \sqrt{\frac{x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}}{n}} = \sqrt{\frac{\sum\limits_{i = 1}^{n}x_{i}^{2}}{n}}$$

　　均值不等式：$H_{n} \leq G_{n} \leq A_{n} \leq Q_{n}$，即 $$\frac{n}{\sum\limits_{i = 1}^{n}\frac{1}{x_{i}}} \leq \sqrt[n]{\prod\limits_{i = 1}^{n}x_{i}} \leq \frac{\sum\limits_{i = 1}^{n}x_{i}}{n} \leq \sqrt{\frac{\sum\limits_{i = 1}^{n}x_{i}^{2}}{n}}$$ 当$x_{1} = x_{2} = \cdots = ~x_{n}$时，取等号。

 

证明

　　先用归纳法证明$G_{n} \leq A_{n}$。

　　当$n = 2$时，由不等式 $a + b \geq 2 \cdot \sqrt{ab}$ 得到 $\frac{a + b}{2} \geq \sqrt{ab}$，成立。

　　假设当$n = k$时成立，当$n = k + 1$时，有 $$\begin{align*} A_{k + 1} &= \frac{\left( {k + 1} \right) \cdot A_{k + 1} + \left( {k - 1} \right) \cdot A_{k + 1}}{2k} \\ &= \frac{x_{1} + x_{2} + \cdots + x_{k} + x_{k + 1} + A_{k + 1} + A_{k + 1} + \cdots + A_{k + 1}}{2k} \\ &= \frac{\left( {x_{1} + x_{2} + \cdots + x_{k}} \right) + \left( {x_{k + 1} + A_{k + 1} + \cdots + A_{k + 1}} \right)}{2k} \\ &\geq \frac{1}{2k}\left( {k \cdot \sqrt[k]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k}} + k \cdot \sqrt[k]{x_{k + 1} \cdot A_{k + 1}^{k - 1}}} \right) \\ &= \frac{1}{2}\left( {\sqrt[k]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k}} + \sqrt[k]{x_{k + 1} \cdot A_{k + 1}^{k - 1}}} \right) \end{align*}$$ 由不等式$a + b \geq 2 \cdot \sqrt{ab}$得 $$\begin{align*} A_{k + 1} &\geq \sqrt[{2k}]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k} \cdot x_{k + 1} \cdot A_{k + 1}^{k - 1}} \\ A_{k + 1}^{2k} &\geq x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k} \cdot x_{k + 1} \cdot A_{k + 1}^{k - 1} \\ A_{k + 1}^{k + 1} &\geq x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k} \cdot x_{k + 1} \\ A_{k + 1} &\geq \sqrt[{k + 1}]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k} \cdot x_{k + 1}} \end{align*}$$ 即 $$\frac{x_{1} + x_{2} + \cdots + x_{k + 1}}{k + 1} \geq \sqrt[{k + 1}]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{k + 1}}$$ 得证。

　　下面证明$H_{n} \leq G_{n}$。

　　很简单，只需要将不等式$G_{n} \leq A_{n}$中的$x_{i}$都替换成$\frac{1}{x_{i}}$，得到 $$\frac{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \cdots + \frac{1}{x_{n}}}{n} \geq \sqrt[n]{\frac{1}{x_{1}} \cdot \frac{1}{x_{2}} \cdot \cdots \cdot \frac{1}{x_{n}}}$$ 即 $$\frac{n}{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \cdots + \frac{1}{x_{n}}} \leq \sqrt[n]{x_{1} \cdot x_{2} \cdot \cdots \cdot x_{n}}$$ 得证。

　　最后证明$A_{n} \leq Q_{n}$。

　　先引入柯西不等式 $${\sum\limits_{i = 1}^{n}a_{i}^{2}} \cdot {\sum\limits_{i = 1}^{n}b_{i}^{2}} \geq \left( {\sum\limits_{i = 1}^{n}{a_{i} \cdot b_{i}}} \right)^{2}$$ 当$\frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \cdots = \frac{a_{n}}{b_{n}}$或$a_{1} = a_{2} = \cdots = a_{n} = 0$或$b_{1} = b_{2} = \cdots = b_{n} = 0$时，取等号。

　　归纳法证明柯西不等式。

　　当$n = 2$时， $$\begin{align*} \left( {a_{1}^{2} + a_{2}^{2}} \right) \cdot \left( {b_{1}^{2} + b_{2}^{2}} \right) &= a_{1}^{2} \cdot b_{1}^{2} + a_{1}^{2} \cdot b_{2}^{2} + a_{2}^{2} \cdot b_{1}^{2} + a_{2}^{2} \cdot b_{2}^{2} \\ &= a_{1}^{2} \cdot b_{1}^{2} + a_{2}^{2} \cdot b_{1}^{2} + 2 \cdot a_{1}a_{2}b_{1}b_{2} + a_{1}^{2} \cdot b_{2}^{2} + a_{2}^{2} \cdot b_{2}^{2} - 2 \cdot a_{1}a_{2}b_{1}b_{2} \\ &= \left( {a_{1} \cdot b_{1} + a_{2} \cdot b_{2}} \right)^{2} + \left( {a_{1} \cdot b_{2} - a_{2} \cdot b_{1}} \right)^{2} \\ &\geq \left( {a_{1} \cdot b_{1} + a_{2} \cdot b_{2}} \right)^{2} \end{align*}$$ 成立。

　　假设当$n = k$时成立，当$n = k + 1$时，有 $$\begin{align*} {\sum\limits_{i = 1}^{k + 1}a_{i}^{2}} \cdot {\sum\limits_{i = 1}^{k + 1}b_{i}^{2}} &= \left( {\left( \sqrt{\sum\limits_{i = 1}^{k}a_{i}^{2}} \right)^{2} + a_{k + 1}^{2}} \right) \cdot \left( {\left( \sqrt{\sum\limits_{i = 1}^{k}b_{i}^{2}} \right)^{2} + b_{k + 1}^{2}} \right) \\ &\geq \left( {\sqrt{{\sum\limits_{i = 1}^{k}a_{i}^{2}} \cdot {\sum\limits_{i = 1}^{k}b_{i}^{2}}} + a_{k + 1} \cdot b_{k + 1}} \right)^{2} \\ &\geq {\sum\limits_{i = 1}^{k}{a_{i} \cdot b_{i}}} + a_{k + 1} \cdot b_{k + 1} \\ &= {\sum\limits_{i = 1}^{k + 1}{a_{i} \cdot b_{i}}} \end{align*}$$ 得证。

　　由柯西不等式 $$\begin{align*} \left( {\frac{1}{n} \cdot {\sum\limits_{i = 1}^{n}x_{i}}} \right)^{2} &= \frac{1}{n^{2}}\left( {x_{1} \cdot 1 + x_{2} \cdot 1 + \cdots + x_{n} \cdot 1} \right)^{2} \\ & \leq \frac{1}{n^{2}} \cdot \left( {x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}} \right) \cdot \left( {1^{2} \cdot 1^{2} \cdot \cdots \cdot 1^{2}} \right) \\ &= \frac{x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}}{n} \end{align*}$$

即 $$\frac{1}{n} \cdot {\sum\limits_{i = 1}^{n}x_{i}} = \frac{x_{1} + x_{2} + \cdots + x_{n}}{n} \leq \sqrt{\frac{x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2}}{n}}$$ 得证。

 

参考资料

　　【不等式】均值不等式及其应用：https://zhuanlan.zhihu.com/p/33706065

　　柯西不等式的几种证明方法：https://zhuanlan.zhihu.com/p/397034475

　　均值不等式：https://baike.baidu.com/item/%E5%9D%87%E5%80%BC%E4%B8%8D%E7%AD%89%E5%BC%8F/8046796