Quod Erat Demonstrandum

2012/03/30

WPS 不等式

Filed under: Pure Mathematics — johnmayhk @ 5:04 上午
Tags: ,

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}

由 Hölder’s inequality 出發,代

a_kb_k=x_kb_k^q=y_kk=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}

或曰

\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}}

此乃「權方和」不等式。(權:Weighted 方:Power)

權方和不等式也頗美麗,其效果好像把左邊本來難以拆開的東西,拆開了。相信它應用繁多,這裡只舉一例。

比如,設 a,b 為正數,證明

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

解:

\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

直接循上述方法,分別考慮(比方說) \mu=\frac{1}{3},\frac{1}{4},\frac{1}{5},\dots,我們可以輕易製造以下不等式:

\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

如果考慮 a,b,c,也不難得出

\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

之類,可作 pure mathematics 練習之用。但,還有幾天,特區便正式告別可愛的純數課程了(2 April 2012 考,默默祝福同學)。

延伸閱讀:

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

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