已知 $x_1,x_2,\cdots,x_n>0$，$n\in\mathbb N^{\ast}$，求证：\[\dfrac 1{x_1}+\dfrac 2{x_1+x_2}+\cdots+\dfrac n{x_1+x_2+\cdots+x_n}<2\left(\dfrac 1{x_1}+\dfrac1{x_2}+\cdots+\dfrac 1{x_n}\right).\]
解析 根据柯西不等式，有\[\dfrac {1^2}{x_1}+\dfrac {2^2}{x_2}+\cdots+\dfrac {n^2}{x_n}\geqslant \dfrac{(1+2+\cdots+n)^2}{x_1+x_2+\cdots+x_n}=\dfrac {n^2(n+1)^2}4\cdot \dfrac{1}{x_1+x_2+\cdots+x_n},\]于是\[\dfrac{n}{x_1+x_2+\cdots+x_n}\leqslant \dfrac{4}{n(n+1)^2}\cdot \left(\dfrac {1^2}{x_1}+\dfrac{2^2}{x_2}+\cdots+\dfrac{n^2}{x_n}\right), \]于是\[\begin{split} LHS&=\sum_{k=1}^n\dfrac{k}{x_1+x_2+\cdots+x_k}\\ &\leqslant \sum_{k=1}^n\dfrac{4}{k(k+1)^2}\left(\dfrac 1{x_1}+\dfrac{2^2}{x_2}+\cdots+\dfrac{k^2}{x_k}\right)\\ &=\sum_{k=1}^n\dfrac{4k^2}{x_k}\left(\sum_{i=k}^n\dfrac{1}{i(i+1)^2}\right)\\ &<\sum_{k=1}^n\dfrac{4k^2}{x_k}\sum_{i=k}^n\left(\dfrac{1}{2i^2}-\dfrac{1}{2(i+1)^2}\right)\\ &=\sum_{k=1}^n\dfrac{4k^2}{x_k}\cdot\left(\dfrac{1}{2k^2}-\dfrac{1}{2(n+1)^2}\right)\\ &<2\sum_{k=1}^n\dfrac{1}{x_k}\\ &=RHS,\end{split}\]因此原命题得证．