权方和不等式此条目序言章节没有充分总结其内容要点。 (2016年10月23日)请考虑扩充序言，为条目所有重要方面提供易懂的概述。请在条目的讨论页讨论此问题。权方和不等式是一种分式不等式。 a i , b i > 0 {\displaystyle a_{i},b_{i}>0} { ∑ i = 1 n a i m + 1 b i m ≥ ( ∑ i = 1 n a i ) m + 1 ( ∑ i = 1 n b i ) m , m < − 1 , m > 0 ∑ i = 1 n a i m + 1 b i m ≤ ( ∑ i = 1 n a i ) m + 1 ( ∑ i = 1 n b i ) m , − 1 < m < 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： a 1 b 1 = a 2 b 2 = . . . = a n b n {\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] s = 1 a 1 + a 2 + . . . + a n , t = 1 b 1 + b 2 + . . . + b n {\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 ∑ i = 1 n ( s a i ) m + 1 ( t b i ) m = ∑ i = 1 n t b i ( s a i t b i ) m + 1 ≥ ∑ i = 1 n t b i [ 1 + ( m + 1 ) ( s a i t b i − 1 ) ] {\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)]} = ∑ i = 1 n t b i [ ( m + 1 ) s a i t b i − m ] = ∑ i = 1 n ( m + 1 ) s a i − m t b i = m + 1 − m = 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} ∑ i = 1 n a i m + 1 b i m ≥ t m s m + 1 = ( ∑ i = 1 n a i ) m + 1 ( ∑ i = 1 n b i ) m {\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 ∑ i = 1 n ( s a i ) m + 1 ( t b i ) m = ∑ i = 1 n t b i ( s a i t b i ) m + 1 ≤ ∑ i = 1 n t b i [ 1 + ( m + 1 ) ( s a i t b i − 1 ) ] {\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)]} = ∑ i = 1 n t b i [ ( m + 1 ) s a i t b i − m ] = ∑ i = 1 n ( m + 1 ) s a i − m t b i = m + 1 − m = 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} ∑ i = 1 n a i m + 1 b i m ≤ t m s m + 1 = ( ∑ i = 1 n a i ) m + 1 ( ∑ i = 1 n b i ) m {\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 ^ 1.0 1.1 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）.
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