How to prove that $\sum_{k \neq j} a_kb_ka_jb_j-\sum_{k \neq j} a_k^2b_j^2=2\sum_{1≤k

Question

$$\sum_{k \neq j} a_kb_ka_jb_j-\sum_{k \neq j} a_k^2b_j^2=2\sum_{1≤k<j≤n}a_kb_ka_jb_j-\sum_{1≤k<j≤n}(a_k^2b_j^2+a_j^2b_k^2)$$

How to get it?

Please help.Thank you.

Answer 1

Hint: For every family $(z_{k,j})$, $$ \sum_{k\ne j}z_{k,j}=\sum_{k\lt j}z_{k,j}+\sum_{j\lt k}z_{k,j}=\sum_{k\lt j}(z_{k,j}+z_{j,k}). $$ In particular, if $z_{k,j}=x_ky_j$, then $$ \sum_{k\ne j}x_ky_j=\sum_{k\lt j}(x_ky_j+x_jy_k), $$ while, if $z_{k,j}=x_kx_j$, then $z_{k,j}=z_{j,k}=x_kx_j$ hence $$ \sum_{k\ne j}x_kx_j=2\sum_{k\lt j}x_kx_j. $$