# 初等代數/不等式

## 一些初等的不等式

### 均值不等式

#### 證明

1. 先证明 n=2 时，即 ${\displaystyle {\frac {a_{1}+a_{2}}{2}}\geq {\sqrt {a_{1}a_{2}}}}$

${\displaystyle {\frac {(a+b)}{2}}-(ab)^{0.5}={\Bigg [}{\frac {a+b-2(ab)^{0.5}}{2}}{\Bigg ]}={\Bigg [}{\frac {(a^{0.5}-b^{0.5})^{2}}{2}}{\Bigg ]}\geq 0}$

${\displaystyle \therefore {\frac {a_{1}+a_{2}}{2}}\geq {\sqrt {a_{1}a_{2}}}}$

2. 由此可推得当 n=2^k (k为自然数)时成立。即 n=2,4,8,16,32…… 时成立。（对对比较后再逐对比较，容易证明）

3. 当 n 为任意自然数的证明比较复杂巧妙。是由 2. 推得：

${\displaystyle n+p=2^{k}}$${\displaystyle A={\frac {a_{1}+a_{2}+a_{3}\cdots +a_{n}}{n}}}$

${\displaystyle {\frac {a_{1}+a_{2}+\cdots +a_{n}+\overbrace {A+A+A\cdots } ^{p}}{2^{k}}}\geqslant {\sqrt[{2^{k}}]{a_{1}a_{2}\cdots a_{n}\cdot AAA\cdots }}}$

${\displaystyle {\frac {a_{1}+a_{2}+\cdots +a_{n}+p\cdot A}{2^{k}}}\geqslant {\sqrt[{2^{k}}]{a_{1}a_{2}\cdots a_{n}\cdot A^{p}}}}$

${\displaystyle \left({\frac {nA+pA}{n+p}}\right)^{n+p}\geqslant a_{1}a_{2}\cdots a_{n}\cdot A^{p}}$

${\displaystyle A^{n+p}\geqslant a_{1}a_{2}\cdots a_{n}\cdot A^{p}}$

${\displaystyle A^{n}\cdot A^{p}\geqslant a_{1}a_{2}\cdots a_{n}\cdot A^{p}}$

${\displaystyle A\geqslant {\sqrt[{n}]{a_{1}a_{2}\cdots a_{n}}}}$

${\displaystyle {\frac {a_{1}+a_{2}+a_{3}+\cdots +a_{n}}{n}}\geqslant {\sqrt[{n}]{a_{1}a_{2}\cdots a_{n}}}}$

${\displaystyle \therefore }$當n為任意自然數時，該命題均成立得證