设$a_1,a_2,\cdots,a_n\in\mathbb{R}$.证明: $\sum_{i,j=1}^n\frac{a_ia_j}{i+j}\geq 0$.
注意到
\begin{align*}
\sum_{i,j=1}^n\frac{a_ia_j}{i+j}
&=\sum_{i,j=1}^na_ia_j\int_0^1x^{i+j-1}\,\mathrm{d}x\\
&=\int_0^1\frac{1}{x}\sum_{i,j=1}^n\left(a_ix^i\right)\left(a_jx^j\right)\,\mathrm{d}x
=\int_0^1\frac{1}{x}\left(\sum_{i=1}^na_ix^i\right)^2\geq 0.
\end{align*}
原文地址:https://www.cnblogs.com/Eufisky/p/12101366.html
时间: 2024-10-22 15:28:09