Cauchy-Schwarz inequality proof help I can't figure out why this proof of the CS inequality starts out as shown below.  I understand everything afterwards.  If you expand it you get an expression that is the proof. Can someone enlighten me?
$$\sum_{i=1}^n \sum_{j=1}^n\left(a_ib_j - a_jb_i\right)^2$$
 A: This might be sort of intuitive: consider the projections $ p_{ij} \colon \mathbb R^n \to \mathbb R^2 $, $ (x_1, \dotsc, x_n) \mapsto (x_i, x_j) $. Your sum vanishes iff $ \forall i, j \colon p_{i,j}(a_1, \dotsc, a_n) \parallel p_{i,j}(b_1, \dotsc, b_n) $. That happens iff $ (a_1, \dotsc, a_n) \parallel (b_1, \dotsc, b_n) $, which is precisely when the equality in CS holds. The "less parallel" the vectors are, the further the inequality is from equality.
A: *

*Continuing Nasekatnaushi's interesting idea with my interpretation, "less parallel" means, for example, perpendicular. Now, since  $\langle b_i,b_j\rangle\cdot\langle b_j,-b_i\rangle =0$, it is natural to consider the procucts $\langle a_i,a_j\rangle\cdot\langle b_j,-b_i\rangle =a_ib_j-a_jb_i$ which are not in general zero and to consider the sum of their squares for the proof of CS inequality.

*In CS inequality, if we omit all terms not involving indicies $i$ and $j$, only $a_i^2b_j^2+a_j^2b_i^2\geq 2a_ia_jb_ib_j$ expression remains. This may give an idea for the start of the proof too.

