權方和不等式

權方和不等式是一種分式不等式。

$a_{i},b_{i}>0$

${\begin{cases}\displaystyle \sum _{i=1}^{n}{\frac {a_{i}^{m+1}}{b_{i}^{m}}}\geq {\frac {(\displaystyle \sum _{i=1}^{n}a_{i})^{m+1}}{(\displaystyle \sum _{i=1}^{n}b_{i})^{m}}},m<-1,m>0\\\displaystyle \sum _{i=1}^{n}{\frac {a_{i}^{m+1}}{b_{i}^{m}}}\leq {\frac {(\displaystyle \sum _{i=1}^{n}a_{i})^{m+1}}{(\displaystyle \sum _{i=1}^{n}b_{i})^{m}}},-1<m<0\end{cases}}$

取等條件：${\frac {a_{1}}{b_{1}}}={\frac {a_{2}}{b_{2}}}=...={\frac {a_{n}}{b_{n}}}$

證明

利用伯努利不等式：

$s={\frac {1}{a_{1}+a_{2}+...+a_{n}}},t={\frac {1}{b_{1}+b_{2}+...+b_{n}}}$

m<-1 或 m>0時

$\sum _{i=1}^{n}{\frac {(sa_{i})^{m+1}}{(tb_{i})^{m}}}=\sum _{i=1}^{n}tb_{i}({\frac {sa_{i}}{tb_{i}}})^{m+1}\geq \sum _{i=1}^{n}tb_{i}[1+(m+1)({\frac {sa_{i}}{tb_{i}}}-1)]$

$=\sum _{i=1}^{n}tb_{i}[(m+1){\frac {sa_{i}}{tb_{i}}}-m]=\sum _{i=1}^{n}(m+1)sa_{i}-mtb_{i}=m+1-m=1$

$\sum _{i=1}^{n}{\frac {a_{i}^{m+1}}{b_{i}^{m}}}\geq {\frac {t^{m}}{s^{m+1}}}={\frac {(\displaystyle \sum _{i=1}^{n}a_{i})^{m+1}}{(\displaystyle \sum _{i=1}^{n}b_{i})^{m}}}$

-1<m<0時

$\sum _{i=1}^{n}{\frac {(sa_{i})^{m+1}}{(tb_{i})^{m}}}=\sum _{i=1}^{n}tb_{i}({\frac {sa_{i}}{tb_{i}}})^{m+1}\leq \sum _{i=1}^{n}tb_{i}[1+(m+1)({\frac {sa_{i}}{tb_{i}}}-1)]$

$=\sum _{i=1}^{n}tb_{i}[(m+1){\frac {sa_{i}}{tb_{i}}}-m]=\sum _{i=1}^{n}(m+1)sa_{i}-mtb_{i}=m+1-m=1$

$\sum _{i=1}^{n}{\frac {a_{i}^{m+1}}{b_{i}^{m}}}\leq {\frac {t^{m}}{s^{m+1}}}={\frac {(\displaystyle \sum _{i=1}^{n}a_{i})^{m+1}}{(\displaystyle \sum _{i=1}^{n}b_{i})^{m}}}$

參見

• 赫爾德不等式