Talk:Titu's lemma

Sedrakyan's inequality in China

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

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

I heard 權方和不等式 from Chinese forum and I created that page in 2014. I don't know what it calls in English. I wonder that they have different routes to the same goal. --Tttfffkkk (talk) 11:39, 9 April 2021 (UTC)