# Schwarz Inequality from baby Rudin

I am trying to construct a more intuitive proof of the Schwarz Inequality.

If $a_1, ...,a_n$ and $b_1,...,b_n$ are complex numbers then

$|\sum\limits_{j=1}^na_j\overline{b_j}|^2 \leq \sum\limits_{j=1}^n|a_j|^2+\sum\limits_{j=1}^n|b_j|^2$

Here is what I have so far:

Put $a_k = w_k + x_ki$ and $b_k = y_k + z_ki$. $|\sum\limits_{j=1}^n\left[\left(w_j+x_ji\right)\left(y_j-z_ji\right)\right]|^2\leq\sum\limits_{j=1}^n\left(w_j^2+x_j^2\right)+\sum\limits_{j=1}^n\left(y_j^2+z_j^2\right)$ $\bigg(\sum\left(w_j+y_j\right)\bigg)^2+\bigg(\sum\left(x_j-z_j\right)\bigg)^2\leq\sum\left(w_j^2+x_j^2\right)+\sum\left(y_j^2+z_j^2\right)$

Obviously this is now an inequality involving real numbers. I am looking for guidance on how to prove this last inequality.

• Some very intuitive proofs can be found here: math.stackexchange.com/questions/436559/… – littleO Apr 9 '17 at 14:39
• I need a proof for complex numbers specifically... not in a more generalized space. – Joe Apr 9 '17 at 14:44
• It may be a "more generalized space" but you can take a vector space over the field of complex numbers and it is the same thing that gets proved. Maybe what you seek is a more intuitive coordinate based proof. From my experience, there doesn't seem to really be any very intuitive versions of that sort. I personally hate the proof in Rudin's book, and for that matter all coordinate based proofs of the Schwarz Ineq. When I was going over that I tried to make sure I followed his reasoning and filled in the blanks, then left it at that. However, I don't know everything. – smokeypeat Jul 1 '17 at 4:48