1. 设 a1,⋯,an 为正实数, 证明矩阵 \bex\sex1ai+ajn×n\eex 半正定.
证明: \beex \bea \sum_{i,j=1}^n \frac{1}{a_i+a_j}x_ix_j &=\sum_{i,j=1}^n x_ix_j\int_0^1 t^{a_i+a_j-1}\rd t\\ &=\int_0^1 \sum_{i=1}^n t^{a_i-\frac{1}{2}}x_i\cdot \sum_{j=1}^n t^{a_j-\frac{1}{2}}x_j\rd t\\ &=\int_0^1 y^2(t,x)\rd t\quad\sex{y(t,x)=\sum_{i=1}^n t^{a_i-\frac{1}{2}}x_i}\\ &\geq 0. \eea \eeex
