平均數不等式

平均數不等式者，又云平均值不等式、均值不等式，乃數學之朋不等式，亦基本不等式之推廣也。曰：

$x_{1},x_{2},\ldots ,x_{n}\in \mathbb {R_{+}} \Rightarrow {\dfrac {n}{\displaystyle \sum _{i=1}^{n}{\dfrac {1}{x_{i}}}}}\leq {\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}\leq {\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}}{n}}\leq {\sqrt {\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}^{2}}{n}}}$

其乃$H_{n}\leq G_{n}\leq A_{n}\leq Q_{n}$。

中： $H_{n}={\dfrac {n}{\displaystyle \sum _{i=1}^{n}{\dfrac {1}{x_{i}}}}}={\dfrac {n}{{\dfrac {1}{x_{1}}}+{\dfrac {1}{x_{2}}}+\cdots +{\dfrac {1}{x_{n}}}}}$

$G_{n}={\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}={\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}$

$A_{n}={\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}}{n}}={\dfrac {x_{1}+x_{2}+\cdots +x_{n}}{n}}$

$Q_{n}={\sqrt {\dfrac {\displaystyle \sum _{i=1}^{n}x_{i}^{2}}{n}}}={\sqrt {\dfrac {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}{n}}}$

當且僅為$x_{1}=x_{2}=\cdots =x_{n}$，等號成立。

即對此正數：調和平均數$\leq$幾何平均數$\leq$算術平均數$\leq$平方平均數。簡記云：“調幾算方”。

時$n=2$

一不等云：

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

二不等云：

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

三不等云：

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

證明方法

平均數不等式之證法多端，如歸納法、拉格朗日乘子法、琴生（Jensen）不等式法、排序不等式法、柯西－施瓦茨不等式法等。茲取琴生不等式為例，統證四均之序。設 $a_{1},\ldots ,a_{n}>0$.

取 $f_{1}(x)=\ln x$，於正實數域凹且遞增，有

$\ln {G_{n}}=\ln({\sqrt[{n}]{a_{1}\cdots a_{n}}})={\frac {\ln {a_{1}}+\cdots +\ln {a_{n}}}{n}}\leq \ln {\left({\frac {a_{1}+\cdots +a_{n}}{n}}\right)}=\ln {A_{n}}$

故 $G_{n}\leq A_{n}$.

取 $f_{2}(x)={\frac {1}{x}}$，於正實數域凸且遞減，得

${\frac {1}{G_{n}}}={\sqrt[{n}]{{\frac {1}{a_{1}}}\cdot {\frac {1}{a_{2}}}\cdot \cdots \cdot {\frac {1}{a_{n}}}}}\leq {\frac {{\frac {1}{a_{1}}}+\cdots +{\frac {1}{a_{n}}}}{n}}={\frac {1}{H_{n}}}$

故 $H_{n}\leq G_{n}$.

又取 $f_{3}(x)=x^{2}$，於正實數域凸且遞增，有

$A_{n}^{2}=({\frac {a_{1}+\cdots +a_{n}}{n}})^{2}\leq {\frac {a_{1}^{2}+\cdots +a_{n}^{2}}{n}}=Q_{n}^{2}$

故 $A_{n}\leq Q_{n}$.

綜上之所述，$H_{n}\leq G_{n}\leq A_{n}\leq Q_{n}$，等號惟於 $a_{1}=a_{2}=\cdots =a_{n}$ 時成立