算术-几何平均值不等式

当 ${\displaystyle n=2}$  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${\displaystyle x_{1},\cdots ,x_{n},y_{1},\cdots ,y_{n}}$ ，

${\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{n}+y_{1}+\cdots y_{n}}{2n}}=\ {\frac {1}{2}}\left({\frac {x_{1}+\cdots +x_{n}}{n}}+{\frac {y_{1}+\cdots +y_{n}}{n}}\right)}$ 

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

${\displaystyle =\ {\sqrt[{2n}]{x_{1}\cdot x_{2}\cdots x_{n}y_{1}\cdot y_{2}\cdots y_{n}}}}$ 

假设${\displaystyle P_{n}}$ 成立，那么${\displaystyle P_{n-1}}$ 成立。证明：对于${\displaystyle n-1}$  个正实数${\displaystyle x_{1},\cdots ,x_{n-1}}$ ，设${\displaystyle \mathbf {A} _{n-1}={\frac {x_{1}+\cdots +x_{n-1}}{n-1}}}$ ，${\displaystyle \mathbf {G} _{n-1}={\sqrt[{n-1}]{x_{1}\cdot x_{2}\cdots x_{n-1}}}}$ ，那么由于${\displaystyle P_{n}}$ 成立， ${\displaystyle {\frac {x_{1}+\cdots +x_{n-1}+\mathbf {A} _{n-1}}{n}}\geq {\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n-1}\mathbf {A} _{n-1}}}}$ 。

但是 ${\displaystyle x_{1}+\cdots +x_{n-1}=(n-1)\mathbf {A} _{n-1}}$ ， ${\displaystyle x_{1}\cdot x_{2}\cdots x_{n-1}=\mathbf {G} _{n-1}^{n-1}}$ ，因此上式正好变成

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

也就是说${\displaystyle \mathbf {A} _{n-1}\geq \mathbf {G} _{n-1}}$ 

综上可以得到结论：对任意的自然数 ${\displaystyle n\geq 2}$ ，命题 ${\displaystyle P_{n}}$  都成立。这是因为由前两条可以得到：对任意的自然数 ${\displaystyle k}$ ，命题 ${\displaystyle P_{2^{k}}}$  都成立。因此对任意的 ${\displaystyle n\geq 2}$ ，可以先找 ${\displaystyle k}$  使得 ${\displaystyle 2^{k}\geq n}$ ，再结合第三条就可以得到命题 ${\displaystyle P_{n}}$  成立了。

由对称性不妨设 ${\displaystyle x_{n+1}}$  是 ${\displaystyle x_{1},x_{2}\cdots x_{n+1}}$  中最大的，由于 ${\displaystyle \mathbf {A} _{n+1}={\frac {n\mathbf {A} _{n}+x_{n+1}}{n+1}}}$  ，设 ${\displaystyle x_{n+1}=\mathbf {A} _{n}+b}$ ，则 ${\displaystyle b\geq 0}$ ，并且有 ${\displaystyle \mathbf {A} _{n+1}=\mathbf {A} _{n}+{\frac {b}{n+1}}}$ 。

根据二项式定理，

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

${\displaystyle =x_{1}x_{2}\cdots x_{n+1}=\mathbf {G} _{n+1}^{n+1}}$ 

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

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

所以 ${\displaystyle (n+1)\mathbf {A} _{n+1}=x_{1}+x_{2}\cdots +x_{n}+x_{n+1}\geq 2n\mathbf {G} _{n+1}-(n-1)\mathbf {G} _{n+1}=(n+1)\mathbf {G} _{n+1}}$ ，从而有${\displaystyle \mathbf {A} _{n+1}\geq \mathbf {G} _{n+1}}$ 。