柯西不等式做差证明

\[(\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^nb_j^2)≥(\sum_{i=1}^na_ib_i)^2 \]

\[\sum_{i=1}^n\sum_{j=1}^na_ib_j-\sum_{i=1}^n\sum_{j=1}^n(a_ib_ia_jb_j) \]

\[=\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^na_i^2b_j^2+a_j^2b_i^2-\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n2a_ib_ia_jb_j \]

\[\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n(a_ib_j-a_jb_i)^2>0 \]

柯西不等式证明