Let $a_1, a_2,....,a_n, b_1, b_2,...,b_n$, let $\frac{b_1}{a_1} = max \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ , $\frac{b_n}{a_n} = min \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ show that:
$$\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$$
Note that \begin{align*} &\ \sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2 - \left(\sum_{i=1}^na_ib_i \right)^2-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2\left(\sum_{i=1}^na_i^2 \right)^2\\ =&\ \sum_{i,j=1}^n\left[a_i^2b_j^2- a_ib_ia_jb_j-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\ =&\ \frac{1}{2}\sum_{i,j=1}^n\left[a_i^2b_j^2 + a_j^2b_i^2- 2a_ib_ia_jb_j-\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\ \le&\ \frac{1}{2}\sum_{i,j=1}^n\left[a_i^2b_j^2 + a_j^2b_i^2- 2a_ib_ia_jb_j-\left(\frac{b_i}{a_i}-\frac{b_j}{a_j}\right)^2 a_i^2a_j^2\right]\\ =&\ 0. \end{align*} The inequality follows immediately.