平均数不等式

• 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

${\displaystyle ({\frac {x_{1}-x_{2}}{2}})^{2}={\frac {x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2}}{4}}\,\;}$	${\displaystyle \geq 0\,\;}$
${\displaystyle {\frac {x_{1}^{2}+x_{2}^{2}}{2}}={\frac {x_{1}^{2}+x_{2}^{2}+x_{1}^{2}+x_{2}^{2}}{4}}\,\;}$	${\displaystyle \geq {\frac {2x_{1}x_{2}+x_{1}^{2}+x_{2}^{2}}{4}}={\frac {(x_{1}+x_{2})^{2}}{4}}\,\;}$
${\displaystyle {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}}{2}}}\,\;}$	${\displaystyle \geq {\frac {x_{1}+x_{2}}{2}}\,\;}$

关于均值不等式的证明方法有很多，数学归纳法（第一数学归纳法或反向归纳法）、拉格朗日乘数法、琴生不等式法、排序不等式法、柯西不等式法等等，都可以证明均值不等式，在这里简要介绍数学归纳法证明n维形式的均值不等式的方法：

用数学归纳法证明，需要一个辅助结论。

引理：设${\displaystyle A\geq 0}$，${\displaystyle B\geq 0}$，则${\displaystyle \left(A+B\right)^{n}\geq A^{n}+nA^{n-1}B}$，且仅当${\displaystyle B=0}$时取等号。

引理的正确性较明显，条件${\displaystyle A\geq 0}$，${\displaystyle B\geq 0}$可以弱化为${\displaystyle A\geq 0}$，${\displaystyle A+B\geq 0}$，可以用数学归纳法证明。

原题等价于：${\displaystyle \left({\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}\right)^{n}\geq a_{1}a_{2}\cdots a_{n}}$，当且仅当${\displaystyle a_{1}=a_{2}=\cdots =a_{n}}$时取等号。

当${\displaystyle n=2}$时易证；

假设当${\displaystyle n=k}$时命题成立，即${\displaystyle \left({\frac {a_{1}+a_{2}+\cdots +a_{k}}{k}}\right)^{k}\geq a_{1}a_{2}\cdots a_{k}}$，当且仅当${\displaystyle a_{1}=a_{2}=\cdots =a_{k}}$时取等号。

那么当${\displaystyle n=k+1}$时，不妨设${\displaystyle a_{k+1}}$是${\displaystyle a_{1}}$、${\displaystyle a_{2}\cdots a_{k+1}}$中最大者，则${\displaystyle ka_{k+1}\geq a_{1}+a_{2}+\cdots +a_{k}}$

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

此外，人教版高中数学教科书《选修4-5 不等式选讲》也介绍了一个运用数学归纳法的证明方法[1]。

先运用数学归纳法证明一个引理：若${\displaystyle n}$（${\displaystyle n}$是正整数）个正数${\displaystyle a_{1},a_{2},...,a_{n}}$的乘积${\displaystyle a_{1}a_{2}...a_{n}=1}$，则它们的和${\displaystyle a_{1}+a_{2}+...+a_{n}\geqslant n}$，当且仅当${\displaystyle a_{1}=a_{2}=...=a_{n}=1}$时等号成立。

此引理证明如下：

当${\displaystyle n=1}$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。

假设当${\displaystyle n=k}$时命题成立，则现在证明当${\displaystyle n=k+1}$时命题也成立。

若这${\displaystyle k+1}$个数全部是1，即${\displaystyle a_{1}=a_{2}=...=a_{k+1}=1}$，则命题显然成立。

若这${\displaystyle k+1}$个数不全是1，则易证明必存在${\displaystyle i\neq j}$使${\displaystyle a_{i}>1,a_{j}<1}$。不妨设${\displaystyle a_{1}>1,a_{2}<1}$。由归纳假设，因为${\displaystyle (a_{1}a_{2})a_{3}...a_{k+1}=1}$，所以${\displaystyle a_{1}a_{2}+a_{3}+...+a_{k+1}\geqslant k}$，记此式为①式。由${\displaystyle a_{1}>1,a_{2}<1}$，知${\displaystyle a_{1}-1>0,1-a_{2}>0}$，则${\displaystyle (a_{1}-1)(1-a_{2})=a_{1}+a_{2}-a_{1}a_{2}-1>0}$，整理得${\displaystyle a_{1}+a_{2}>1+a_{1}a_{2}}$，记此式为②式。①+②得${\displaystyle a_{1}+a_{2}+a_{1}a_{2}+a_{3}+...+a_{k+1}>k+1+a_{1}a_{2}}$，整理得${\displaystyle a_{1}+a_{2}+...+a_{k+1}>k+1}$（此时等号不成立），命题成立。

综上，由数学归纳法，引理成立。

现在为了证明平均值不等式，考虑${\displaystyle n}$个正数${\displaystyle {\frac {a_{1}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}},{\frac {a_{2}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}},...,{\frac {a_{n}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}}$，它们的积为1，由引理，它们的和${\displaystyle {\frac {a_{1}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}+{\frac {a_{2}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}+...+{\frac {a_{n}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}={\frac {a_{1}+a_{2}+...+a_{n}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}\geqslant n}$，当且仅当${\displaystyle {\frac {a_{1}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}={\frac {a_{2}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}=...={\frac {a_{n}}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}}$即${\displaystyle a_{1}=a_{2}=\cdots =a_{n}}$时等号成立。

整理即得：${\displaystyle {\frac {a_{1}+a_{2}+...+a_{n}}{n}}\geqslant {\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}$，当且仅当${\displaystyle a_{1}=a_{2}=\cdots =a_{n}}$时等号成立。于是${\displaystyle G_{n}\leq A_{n}}$得证。

利用${\displaystyle G_{n}\leq A_{n}}$，易证${\displaystyle H_{n}\leq G_{n}}$。考虑${\displaystyle n}$个正数${\displaystyle {\frac {1}{a_{1}}},{\frac {1}{a_{2}}},...,{\frac {1}{a_{n}}}}$，有${\displaystyle {\frac {{\frac {1}{a_{1}}}+{\frac {1}{a_{2}}}+...+{\frac {1}{a_{n}}}}{n}}\geqslant {\sqrt[{n}]{{\frac {1}{a_{1}}}\cdot {\frac {1}{a_{2}}}\cdot ...\cdot {\frac {1}{a_{n}}}}}={\frac {1}{\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}}$，当且仅当${\displaystyle {\frac {1}{a_{1}}}={\frac {1}{a_{2}}}=...={\frac {1}{a_{n}}}}$即${\displaystyle a_{1}=a_{2}=\cdots =a_{n}}$时等号成立。两边取倒数整理得${\displaystyle {\frac {n}{{\frac {1}{a_{1}}}+{\frac {1}{a_{2}}}+...+{\frac {1}{a_{n}}}}}\leqslant {\sqrt[{n}]{a_{1}a_{2}...a_{n}}}}$，当且仅当${\displaystyle a_{1}=a_{2}=\cdots =a_{n}}$时等号成立，即${\displaystyle H_{n}\leq G_{n}}$。

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

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

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

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

${\displaystyle \geqslant 0}$，

当且仅当${\displaystyle a_{1}=a_{2}=\cdots =a_{n}=A_{n}}$时等号成立。

利用琴生不等式法也可以很简单地证明均值不等式，同时还有柯西归纳法等方法。

• 算术-几何平均值不等式
• 幂平均不等式

1. . 人民教育出版社. 2007: 52. ISBN 978-7-107-18675-2.