Confusing Cauchy-Schwarz Inequality Proof Teacher proved it like this:

Very elegant (way simpler than most of the ones I find online), but I'm still not convinced -- particularly the last two steps or so, where the absolute value on the left-hand side seems to disappear. 
Any thoughts? 
 A: This proof assumes that $\vec a.\vec b$ can be written as $\left\lVert\vec a\right\rVert.\left\lVert\vec b\right\rVert.\cos\theta$ for some number $\theta$. This is the same thing as asserting that $\left\lvert\vec a.\vec b\right\rvert\leqslant\left\lVert\vec a\right\rVert.\left\lVert\vec b\right\rVert$, and this is precisely the Cauchy-Schwarz inequality, which is what you want to prove. There is therefore a circular reasoning here.
A: given that $a\cdot b= \|a\|\cdot \|b\|\cos(\theta)$ we have that
$$|a\cdot b|= \|a\|\cdot \|b\||\cos(\theta)|$$ (since the length of a vector is always non-negative).
Noting that $|\cos(\theta)|\leq 1$ we can have that:


*

*$|\cos(\theta)|=1$ which implies that $|a\cdot b|\leq\|a\|\cdot \|b\|$ is satisfied with equality.

*$|\cos(\theta)|<1$ which implies that $|a\cdot b|<\|a\|\cdot \|b\|$
The latter proves the statement.
$$|a\cdot b| \leq \|a\|\cdot \|b\|$$

EDIT: As pointed out by @user in the comments down below, there's no sense in moving the $|\cos(\theta)|$ to the denominator as your teacher suggested, cause nothing prevents you to have $\theta=\frac{\pi}2$ implying that $\cos(\theta)=0$. It's better to keep it on the other side and deriving the conclusion from there.
Disclaimer I think you should,nevertheless, consider what @jose-carlos-santos and @giuseppe-negro are pointing out, that actually this is a circular reasoning, and hence not a valid proof.
