I am trying to solve the following problem in Kolmogorov's analysis textbook.
Verify that $$ (\sum\limits_{k=1}^n a_k b_k)^2 = \sum\limits_{k=1}^n a_k^2 \sum\limits_{k=1}^n b_k^2 - \frac{1}{2} \sum\limits_{i=1}^n \sum\limits_{j=1}^n (a_i b_j - b_i a_j)^2. $$ Deduce the Cauchy-Schwarz inequality, $$ (\sum\limits_{k=1}^n a_k b_k)^2 \leq \sum\limits_{k=1}^n a_k^2 \sum\limits_{k=1}^n b_k^2. $$ from this identity.
The initial inequality is giving me some trouble, as there does not appear to be an algebraic trick. I have tried expanding out the summand on the right and breaking apart the sum. Since the left-hand is a summation only over $k$, it must be the case that the $i$ and $j$ sums vanish somehow. There doesn't seem to be a good way to factor the right-hand side in order to make this true.
Any help on this would be greatly appreciated.