Prove Lagrange's Identity without induction Prove Lagrange's Identity without induction.
$$
\sum_{1\leq j <k\leq n}(a_jb_k-a_kb_j)^2=\left( \sum_{k=1}^na_k^2 \right)\left( \sum_{k=1}^n b_k^2 \right)-\left( \sum_{k=1}^na_kb_k \right)^2
$$
I tried expanding the left side but I could never get anywhere, I'm looking for some tips on how to get started on the right direction, not complete solutions.
Thanks!
 A: Hint:
You have to play with the indices. Develop the sum of squares, first removing the condition $j<k$ and replacing by $j\neq k$ (this will cancel out a factor $2$), then note the case $j=k$ brings in no supplementary term. When all conditions on $j$ and $k$ have  been removed, it is easy to factorise.
$$ \sum_{1\le j < k\le n}(a_jb_k- a_kb_j)^2 = \sum_{1\le j < k\le n}(a_jb_k)^2 -2\sum_{1\le j < k\le n} a_j a_k b_j b_k + \sum_{1\le j < k\le n}(a_kb_j)^2 $$
 Note that $\displaystyle\sum\limits_{1\le j < k\le n}(a_jb_k)^2 + \sum\limits_{1\le j < k\le n}(a_kb_j)^2 $ can be written as $\displaystyle\sum\limits_{1\le j\neq k\le n}(a_jb_k)^2 $. 
Also $\, \displaystyle 2\!\!\!\sum\limits_{1\le j < k\le n} a_j a_k b_jb_k =\!\!\!\sum\limits_{1\le j \neq k\le n} a_j a_k b_jb_k $ , so that our sum of squares is $\displaystyle\sum\limits_{1\le j\neq k\le n}(a_jb_k)^2 -\sum\limits_{1\le j \neq k\le n} a_j a_k b_jb_k $ 
Finally, note that in the last formula, for $ j = k$ the terms would cancel, since their contribution  would be equal to
$$\sum\limits_{1\le j\le n}(a_jb_j)^2 -\sum\limits_{1\le j \le n} a_j^2 b_j^2$$
Thus, we can incorporate the case $j=k$ to our original sum of squares, which will finally be equal to: 
 \begin{align*}\sum_{1\le j, k\le n}(a_jb_k)^2 -\sum_{1\le j ,k\le n} a_j b_j a_k b_k& =\sum_{1\le j\le n}a_j^2\sum_{1\le j\le n}b_j^2 -\Bigl(\sum_{1\le j \le n} a_j b_j\Bigr)\Bigl(\sum_{1\le j \le n} a_j b_j\Bigr)\\&=\sum_{1\le j\le n}a_j^2\sum_{1\le j\le n}b_j^2 -\Bigl(\sum_{1\le j \le n} a_j b_j\Bigr)^2
\end{align*} 
A: Both expressions are of the form $\sum_{\substack{i \leq j \\ k \leq \ell}} C_{i,j,k,\ell} a_i a_j b_k b_\ell$, and you can compute the coefficient of $a_i a_j b_k b_\ell$ on both sides, showing that it is equal in all cases.
