来个n元的

创之谜

已知\(a_1\sim a_n\)为正实数.求证：\[\sum\limits_{k=1}^{n}\dfrac{k^2}{a_1+a_2+\cdots+a_k }<4\sum\limits_{k=1}^{n}\dfrac{k}{a_k }\]

kuing

由柯西有
\[\frac{(1+2+\cdots+k)^2}{a_1+a_2+\cdots+a_k}\leqslant\frac1{a_1}+\frac{2^2}{a_2}+\cdots+\frac{k^2}{a_k},\]
即
\[\frac{k^2}{a_1+a_2+\cdots+a_k}\leqslant\frac4{(k+1)^2}\left(\frac1{a_1}+\frac{2^2}{a_2}+\cdots+\frac{k^2}{a_k}\right),\]
求和得
\begin{align*}
\sum_{k=1}^n\frac{k^2}{a_1+a_2+\cdots+a_k}&\leqslant4\sum_{k=1}^n\frac1{(k+1)^2}\left(\frac1{a_1}+\frac{2^2}{a_2}+\cdots+\frac{k^2}{a_k}\right)\\
&=4\sum_{k=1}^n\sum_{i=k}^n\frac1{(i+1)^2}\cdot\frac{k^2}{a_k},
\end{align*}
因为
\[\sum_{i=k}^n\frac1{(i+1)^2}<\sum_{i=k}^n\frac1{i(i+1)}=\frac1k-\frac1{n+1}<\frac1k,\]
所以
\[\sum_{k=1}^n\frac{k^2}{a_1+a_2+\cdots+a_k}<4\sum_{k=1}^n\frac k{a_k}.\]