# 浅谈Cauchy不等式

## 形式

$\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2 \geq \sum_{i=1}^{n}a_i^{2}b_i^2$

$iff:b_i=0 || \exists k \in \mathbb {R},a_i=k \cdot b_i(i \in \mathbb{N^+})$

## 证明

#### 法一：参数配方

##### 证明：

$f(t)=\sum_{i=1}^{n}b_i^2\cdot t^2-2\sum_{i=1}^{n}a_ib_it+\sum_{i=1}^{n}a_i^2$

$f(t)=\sum_{i=1}^{n}b_i^2\cdot t^2-2\sum_{i=1}^{n}a_ib_it+\sum_{i=1}^{n}a_i^2$

$=\sum_{i=1}^{n}(b_i^2t^2-2a_ib_it+a_i^2)$

$=\sum_{i=1}^{n}(b_i^2t^2+a_i^2-2a_ib_it)$

$=\sum_{i=1}^{n}(b_it-a_i)^2$

$f(t) \geq 0$

$\Delta t=b^2-4ac$

$=4\sum_{i=1}^{n}a_i^2b_i^2-4\times \sum_{i=1}^{n}b_i^2 \times \sum_{i=1}^{n}a_i^2 \leq 0$

$4\sum_{i=1}^{n}a_i^2b_i^2 \leq 4\times \sum_{i=1}^{n}b_i^2 \times \sum_{i=1}^{n}a_i^2$

$\sum_{i=1}^{n}a_i^2 \times \sum_{i=1}^{n}b_i^2 \geq \sum_{i=1}^{n}a_i^2b_i^2$

$f(t)=\sum_{i=1}^{n}(b_it-a_i)^2$

$f(t)=0$，即

$a_i=b_it$

$f(t)_{min}=0​$

$\Delta t \leq 0$

$\exists k \in \mathbb {R},a_i=k \cdot b_i(i \in \mathbb{N^+}）$

#### 法二：均值不等式证明

$ab \leq \frac{a^2+b^2}{2}$

$a,b \in \mathbb {R^+}$

$iff:a=b$

##### 证明：

$\sum_{i=1}^{n}a_i^2\sum_{i=1}^{n}b_i^2 \geq \sum_{i=1}^{n}a_i^{2}b_i^2$

$\sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2} \geq |\sum_{i=1}^{n}a_ib_i|$

$|\sum_{i=1}^{n}a_ib_i| \leq \sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2}$

$\frac{|\sum_{i=1}^{n}a_ib_i|}{\sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2}}\leq 1$

$|a_1+a_2+a_3+\cdots+a_n| \leq |a_1|+|a_2|+|a_3|+ \cdots + |a_n|$

$|\sum_{i=1}^{n}a_ib_i| \leq \sum_{i=1}^{n}|a_ib_i|$

$\frac{|\sum_{i=1}^{n}a_ib_i|}{\sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2}} \leq \frac{\sum_{i=1}^{n}|a_ib_i|}{\sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2}}$

$\frac{\sum_{i=1}^{n}|a_ib_i|}{\sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2}}$

$=\sum_{i=1}^{n}\frac{|a_i|}{\sqrt{\sum_{i=1}^{n}a_i^2}}\cdot \frac{|b_i|}{\sqrt{\sum_{i=1}^{n}b_i^2}}$

$ab \leq \frac{a^2+b^2}{2}$

$\sum_{i=1}^{n}\frac{|a_i|}{\sqrt{\sum_{i=1}^{n}a_i^2}}\cdot \frac{|b_i|}{\sqrt{\sum_{i=1}^{n}b_i^2}}$

$\leq \frac{1}{2}\cdot \sum_{i=1}^{n}(\frac{a_i^2}{\sum_{i=1}^{n}a_i^2}+ \frac{b_i^2}{\sum_{i=1}^{n}b_i^2})$

$\leq \frac{1}{2}\cdot (\frac{\sum_{i=1}^{n}a_i^2}{\sum_{i=1}^{n}a_i^2}+ \frac{\sum_{i=1}^{n}b_i^2}{\sum_{i=1}^{n}b_i^2})$

$\leq \frac{1}{2} \times 2 = 1$

$\frac{|\sum_{i=1}^{n}a_ib_i|}{\sqrt {\sum_{i=1}^{n}a_i^2 \sum_{i=1}^{n}b_i^2}}\leq 1$

#### 法三：n维向量证法

$|\vec a \cdot \vec b| = |\vec a|\cdot |\vec b| \cdot cos \theta$

$|\vec a \cdot \vec b| \leq |\vec a|\cdot |\vec b|$

$|\vec a \cdot \vec b|^2 \leq |\vec a|^2\cdot |\vec b|^2$

$\vec a,\vec b$$n$维向量时，用坐标的形式展开即可证明。

$\vec a=k\vec b$，即$a$$b$共线时，等号成立。

## 申明与感谢

• 内容采用“知识共享署名-非商业性使用-相同方式共享 4.0 国际许可协议”进行许可。请您在转载时注明来源及链接。
• 感谢@thorn的审稿。
posted @ 2020-01-17 00:31  BeyondLimits  阅读(541)  评论(2编辑  收藏  举报