Problem:
let $A$ and $B$ be positive real numbers such that $$-A\le a_{i}\le A,-B\le b_{i}\le B~(i=1,2,\cdots,n)$$.
Show that $$n\sum_{i=1}^{n}a^2_{i}b^2_{i}-\left(\sum_{i=1}^{n}a_{i}b_{i}\right)^2\le 2B^2\left[n\sum_{i=1}^{n}a^2_{i}-\left(\sum_{i=1}^{n}a_{i}\right)^2\right]+2A^2\left[n\sum_{i=1}^{n}b^2_{i}-\left(\sum_{i=1}^{n}b_{i}\right)^2\right]$$
maybe it relate this well known polya-szego inequality:$$\left(\sum_{k=1}^{n}a^2_{k}\right)\left(\sum_{k=1}^{n}b^2_{k}\right)\le\dfrac{1}{4}\left(\sqrt{\dfrac{M_{1}M_{2}}{m_{1}m_{2}}}+\sqrt{\dfrac{m_{1}m_{2}}{M_{1}M_{2}}}\right)^2\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^2$$
We need to prove that $$\sum_{1\leq i<j\leq n}(a_ib_i-a_jb_j)^2\leq2A^2\sum_{1\leq i<j\leq n}(b_i-b_j)^2+2B^2\sum_{1\leq i<j\leq n}(a_i-a_j)^2,$$ for which it's enough to prove that $$(a_ib_i-a_jb_j)^2\leq2A^2(b_i-b_j)^2+2B^2(a_i-a_j)^2,$$ which follows from C-S: $$(a_ib_i-a_jb_j)^2=(a_ib_i-a_ib_j+a_ib_j-a_jb_j)^2=(a_i(b_i-b_j)+b_j(a_i-a_j))^2\leq$$ $$\leq(1^2+1^2)(a_i^2(b_i-b_j)^2+b_j^2(a_i-a_j)^2)\leq2A^2(b_i-b_j)^2+2B^2(a_i-a_j)^2.$$ Done!