In Rudin's Real and Complex Analysis the Schwarz inequality: $|(x,y)| \le ||x|| \ ||y||$ is proven in the following manner:
Put $A = ||x||^2, B=|(x,y)|$ and $C = ||y||^2$. There is a complex number $\alpha$ such that $| \alpha| = 1 $ and $\alpha (y,x) = B$. For any real r, we then have:
$$(x-r\alpha y,x - r\alpha y) = (x,x) - r\alpha(y,x) - r\overline{\alpha}(x,y) + r^2(y,y).$$
The expression on the left is real and not negative. Hence:
$$A-2Br +Cr^2 \ge 0$$ for every real r. If $C = 0$, we must have $B = 0$. If $C >0$, take r = B/C and obtain $ B^2 \le AC $.
Could someone explain to me why there is no loss of generality when we take the complex number $\alpha$ such that $| \alpha| = 1 $ and $\alpha (y,x) = B$ ?
Did we not construct the inequality in order to get the result?
You are troubled that they implicitly assumed the existence of an $\alpha$ such that $\alpha\cdot (y,x)=B$; such an $\alpha$ always exists provided that $(y,x)\neq 0$. Simply let $\alpha = \frac{B}{(y,x)}$. Then $$ |\alpha|=\frac{|B|}{|(y,x)|}=\frac{|(x,y)|}{|(y,x)|}=1 $$ So, the only loss of generality that occurs is that the above proof doesn't quite cover the case where $(y,x)=0$. However, the Schwarz inequality still holds in this case, and the proof is much easier.