Prove the following identity ${(\sum_1^na_jb_j)}^2 = {(\sum_1^na^2_j)}{(\sum_1^nb^2_j)}^2-\sum_1^n\sum_1^n{(a_kb_j-a_jb_k)}^2$ 
Prove the following identity
  $$\left(\sum_{j=1}^n a_j b_j\right)^2 = \left(\sum_{j=1}^n a^2_j\right)\left(\sum_{j=1}^nb^2_j\right) - \sum_{k=1}^{n-1}\sum_{j=k+1}^{n}(a_{k}b_j-a_jb_{k})^2$$

How should I approach and prove this identity? My thought process currently is to expand the LHS or make the RHS equal back to the LHS, however, my algebra is a little rusty in regards to summation. It would be much appreciated if you guys can give me helpful hints and nudge me in the correct path. And do explain your steps so that I may learn the underwhelming process in proving such identities.  
 A: \begin{align}
&\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)-\left(\sum_{i=1}^{n}a_ib_i\right)^2\\
=&\sum_{i=1}^{n}\sum_{j=1}^{n}a_i^2b_j^2-\sum_{i=1}^{n}\sum_{j=1}^{n}a_ib_ia_jb_j\\
=&\frac12\sum_{i=1}^{n}\sum_{j=1}^{n}\left(a_i^2b_j^2+a_j^2b_i^2-2a_ib_ia_jb_j\right)\\
=&\frac12\sum_{i=1}^{n}\sum_{j=1}^{n}(a_ib_j-a_jb_i)^2\\
=&\sum_{1\leq i<j\leq n}(a_{i}b_j-a_jb_{i})^2.
\end{align}
A: It's much easier to start with
$$ \sum_{k,j} (a_k b_j - a_j b_k)^2, $$
expand the brackets, and then split the $$ a_k^2 b_j^2 $$ terms from the $a_jb_j a_k b_k$ ones. And use that
$$ \sum_{i,j} f(i) g(j) = \left( \sum_i f(i) \right) \left( \sum_j g(j) \right). $$
A: It is the usual generalization of Lagrange's four squares identity that can be used to prove Cauchy-Schwarz' inequality. In the LHS we have terms of the type $a_j^2 b_j^2$ or $a_j b_j a_i b_i$ only, we just need to check that they appear in the RHS with the same multiplicity. 
The identity also follows from:
$$ |v\times w|^2 + |v\cdot w|^2 = |v|^2 |w|^2(\sin^2\theta+\cos^2\theta) = |v|^2 |w|^2 $$
with $v=(a_1,\ldots,a_n)$ and $w=(b_1,\ldots,b_n)$.
