# 算术-几何平均值不等式

（重定向自算術幾何平均不等式

${\displaystyle \mathbf {A} _{n}\geq \mathbf {G} _{n}}$

## 例子

${\displaystyle n=4}$  的情况，设: ${\displaystyle x_{1}=3.5,\ x_{2}=6.2,\ x_{3}=8.4,\ x_{4}=5}$ ， 那么

${\displaystyle \mathbf {A} _{4}={\frac {3.5+6.2+8.4+5}{4}}=5.775,\ \mathbf {G} _{4}={\sqrt[{4}]{3.5\times 6.2\times 8.4\times 5}}\approx 5.4945}$

## 历史上的证明

### 柯西的证明

1821年，法国数学家柯西在他的著作《分析教程》中给出一个使用逆向归纳法的证明[1]

${\displaystyle n=2}$  时，${\displaystyle P_{2}}$ 显然成立。假设 ${\displaystyle P_{n}}$  成立，那么 ${\displaystyle P_{2n}}$  成立。证明：对于${\displaystyle 2n}$  个正实数${\displaystyle x_{1},\cdots ,x_{n},y_{1},\cdots ,y_{n}}$

${\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{n}+y_{1}+\cdots y_{n}}{2n}}=\ {\frac {1}{2}}\left({\frac {x_{1}+\cdots +x_{n}}{n}}+{\frac {y_{1}+\cdots +y_{n}}{n}}\right)}$
${\displaystyle \geq \ {\frac {1}{2}}\left({\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}+{\sqrt[{n}]{y_{1}\cdot y_{2}\cdots y_{n}}}\right)\geq \ {\sqrt {{\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}\cdot {\sqrt[{n}]{y_{1}\cdot y_{2}\cdots y_{n}}}}}}$
${\displaystyle =\ {\sqrt[{2n}]{x_{1}\cdot x_{2}\cdots x_{n}y_{1}\cdot y_{2}\cdots y_{n}}}}$

${\displaystyle \mathbf {A} _{n-1}^{n}\geq \mathbf {G} _{n-1}^{n-1}\mathbf {A} _{n-1}}$

### 归纳法的证明

${\displaystyle \mathbf {A} _{n+1}^{n+1}=(\mathbf {A} _{n}+{\frac {b}{n+1}})^{n+1}\geq \mathbf {A} _{n}^{n+1}+(n+1)\mathbf {A} _{n}^{n}{\frac {b}{n+1}}=\mathbf {A} _{n}^{n}(\mathbf {A} _{n}+b)=\mathbf {A} _{n}^{n}x_{n+1}\geq \mathbf {G} _{n}^{n}x_{n+1}}$
${\displaystyle =x_{1}x_{2}\cdots x_{n+1}=\mathbf {G} _{n+1}^{n+1}}$

${\displaystyle n}$  的情况下有不等式 ${\displaystyle \mathbf {A} _{n}\geq \mathbf {G} _{n}}$ ${\displaystyle x_{n+1}+(n-1)\mathbf {G} _{n+1}\geq n{\sqrt[{n}]{x_{n+1}\mathbf {G} _{n+1}^{n-1}}}}$  成立，于是：

${\displaystyle {\frac {x_{1}+x_{2}\cdots +x_{n}+x_{n+1}+(n-1)\mathbf {G} _{n+1}}{n}}\geq \mathbf {G} _{n}+{\sqrt[{n}]{x_{n+1}\mathbf {G} _{n+1}^{n-1}}}\geq 2{\sqrt[{2n}]{\mathbf {G} _{n}^{n}x_{n+1}\mathbf {G} _{n+1}^{n-1}}}=2\mathbf {G} _{n+1}}$

### 基于琴生不等式的证明

${\displaystyle \ln {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\frac {\ln {x_{1}}+\ln {x_{2}}+\cdots +\ln {x_{n}}}{n}}}$

### 基于排序不等式的证明

${\displaystyle b_{i}={\frac {a_{i}}{{\mathbf {G} }_{n}}}(i=1,2,3,...,n)}$ ，于是有 ${\displaystyle b_{1}b_{2}\cdots b_{n}=1}$ ，再作代换 ${\displaystyle b_{1}={\frac {c_{1}}{c_{2}}},b_{2}={\frac {c_{2}}{c_{3}}},\cdots ,b_{n}={\frac {c_{n}}{c_{1}}}}$ ，运用排序不等式得到：

${\displaystyle {\frac {c_{1}}{c_{2}}}+{\frac {c_{2}}{c_{3}}}+\cdots +{\frac {c_{n}}{c_{1}}}\geqslant {\frac {c_{1}}{c_{1}}}+{\frac {c_{2}}{c_{2}}}+...+{\frac {c_{n}}{c_{n}}}=n}$

## 推广

### 加权算术-几何平均不等式

${\displaystyle p_{1}x_{1}+p_{2}x_{2}\cdots +p_{n}x_{n}\geq x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}}$

### 矩阵形式

${\displaystyle {\begin{bmatrix}a_{11}&\cdots &a_{1k}\\\vdots &\ddots &\vdots \\a_{n1}&\cdots &a_{nk}\end{bmatrix}}}$

${\displaystyle A_{j}={\frac {1}{n}}\sum _{i=1}^{n}a_{ij}}$ ${\displaystyle G_{i}={\sqrt[{k}]{\prod _{j=1}^{k}a_{ij}}}}$ ，那么有：

${\displaystyle {\sqrt[{k}]{A_{1}A_{2}\cdots A_{k}}}\leqslant {\frac {G_{1}+G_{2}+\cdots +G_{n}}{n}}}$

### 极限形式

${\displaystyle \int _{0}^{1}f(x)dx\geq \exp(\int _{0}^{1}\ln f(x)dx)}$

### 算數-幾何-調和平均值不等式

${\displaystyle \mathbf {A} _{n}\geq \mathbf {G} _{n}\geq \mathbf {H} _{n}}$

${\displaystyle {\frac {{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+...+{\frac {1}{x_{n}}}}{n}}\geq {\sqrt[{n}]{{\frac {1}{x_{1}}}{\frac {1}{x_{2}}}\cdots {\frac {1}{x_{n}}}}}={\frac {1}{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}}$

${\displaystyle {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\geq {\frac {n}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+...+{\frac {1}{x_{n}}}}}}$

${\displaystyle \mathbf {G} _{n}\geq \mathbf {H} _{n}}$

${\displaystyle {\frac {1}{x_{1}}}={\frac {1}{x_{2}}}=\cdots ={\frac {1}{x_{n}}}}$

${\displaystyle x_{1}=x_{2}=\cdots =x_{n}}$

## 参考来源

1. ^ Augustin-Louis Cauchy, Cours d'analyse de l'École Royale Polytechnique, premier partie, Analyse algébrique,页面存档备份，存于互联网档案馆） Paris, 1821. p457.
2. ^ George Chrystal, Algebra:An Elementary Text-Book, Part II, Chapter XXIV.p46.
3. ^ P. H. Diananda , A Simple Proof of the Arithmetic Mean Geometric Mean Inequality ,The American Mathematical Monthly, Vol. 67, No. 10 (Dec., 1960), pp. 1007
• 匡继昌，《常用不等式》，山东科技出版社。
• 李胜宏，《平均不等式与柯西不等式》，华东师大出版社。
• 莫里斯·克莱因（Morris Kline），张理京 张锦炎 江泽涵 译，《古今数学思想》，上海科学技术出版社。
• 李兴怀，《学科奥林匹克丛书·高中数学》，广东教育出版社。