权方和不等式

权方和不等式是一种分式不等式。

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

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

取等条件：${\displaystyle {\frac {a_{1}}{b_{1}}}={\frac {a_{2}}{b_{2}}}=...={\frac {a_{n}}{b_{n}}}}$

证明

利用伯努利不等式：[1]

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

m<-1 或 m>0时

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

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

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

-1<m<0时

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

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

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

参见

• 赫尔德不等式

参考资料

1. 杨海霞 李志. 权方和不等式的完善. 中学数学研究. 2010, (2) [2014-03-20]. （原始内容存档于2016-03-04）.