# Quod Erat Demonstrandum

## 2012/03/30

### WPS 不等式

Filed under: Pure Mathematics — johnmayhk @ 5:04 上午
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$a_1,a_2,\dots ,a_n,b_1,b_2, \dots ,b_n,p,q$ 為正數，其中 $\frac{1}{p}+\frac{1}{q}=1$。我們有 Hölder’s inequality

$\displaystyle \sum_{k=1}^na_kb_k \le (\displaystyle \sum_{k=1}^na_k^p)^{1/p}(\displaystyle \sum_{k=1}^nb_k^q)^{1/q}$

$a_kb_k=x_k$$b_k^q=y_k$$k=1,2,\dots ,n$

$\displaystyle \sum_{k=1}^nx_k \le (\displaystyle \sum_{k=1}^n\frac{x_k^p}{y_k^{p/q}})^{1/p}(\displaystyle \sum_{k=1}^ny_k)^{1/q}$

$\displaystyle (\sum_{k=1}^nx_k)^p \le (\displaystyle \sum_{k=1}^n\frac{x_k^p}{y_k^{p/q}})(\displaystyle \sum_{k=1}^ny_k)^{p/q}$

$\frac{p}{q}=\mu$$\mu \ge 0$

$\frac{1}{p}+\frac{1}{q}=1$，故 $p=\mu+1$，於是上式化為

$\displaystyle (\sum_{k=1}^nx_k)^{\mu+1} \le (\displaystyle \sum_{k=1}^n\frac{x_k^{\mu+1}}{y_k^{\mu}})(\displaystyle \sum_{k=1}^ny_k)^{\mu}$

$\displaystyle \sum_{k=1}^n\frac{x_k^{\mu+1}}{y_k^{\mu}} \ge \frac{\displaystyle (\sum_{k=1}^nx_k)^{\mu+1}}{(\displaystyle \sum_{k=1}^ny_k)^{\mu}}$

$\frac{a}{\sqrt{a^2+3b^2}}+\frac{b}{\sqrt{b^2+3a^2}}\ge 1$

$\frac{a}{\sqrt{a^2+3b^2}}+\frac{b}{\sqrt{b^2+3a^2}}$

$=\frac{a^{3/2}}{\sqrt{a^3+3ab^2}}+\frac{b^{3/2}}{\sqrt{b^3+3a^2b}}$

$\ge \frac{(a+b)^{3/2}}{\sqrt{a^3+3ab^2+3a^2b+b^3}}$（由權方和不等式）

$=1$

$\frac{a}{\sqrt[3]{a^3+3ab^2+4b^3}}+\frac{b}{\sqrt[3]{b^3+3ba^2+4a^3}}\ge 1$

$\frac{a}{\sqrt[4]{a^4+10ab^3+5b^4}}+\frac{b}{\sqrt[4]{b^4+10ba^3+5a^4}}\ge 1$

$\frac{a}{\sqrt[5]{a^5+15ab^4+10a^2b^3+6b^5}}+\frac{b}{\sqrt[5]{b^5+15ba^4+10b^2a^3+6a^5}}\ge 1$

$\dots$

$\frac{a}{\sqrt{a^2+3b^2+3c^2+2bc}}+\frac{b}{\sqrt{b^2+3a^2+3c^2+2ac}}+\frac{c}{\sqrt{c^2+3a^2+3b^2+2ab}}\ge 1$

http://w3.math.sinica.edu.tw/math_media/d331/33108.pdf