# Cauchy-Schwarz proof in my Cambridge notes

I have been given a couple of properties of inner product and a norm in these cambridge notes on metric and topological spaces, page 6.

And then lemma 3.5 (i) talks about cauchy-schwarz inequality. I have looked at their proof and they show that:

$<u,u>-\frac {<u,v>^2}{<v,v>} \geq 0$ and this should be equivalent to $||u||||v|| \geq |<u,v>|$. I fail to see it. Especially, given that they have not shown an explicit form of either the norm or an inner product, rather their properties.

• The statement of the lemma starts by defining $\lVert u\rVert \stackrel{\rm def}{=} \langle u,u\rangle$ (Lemma 3.5, p.6). How does that not suffice? – Clement C. Dec 29 '16 at 17:30
• $\langle u , u \rangle \langle v, v \rangle \geq \langle u , v \rangle ^2$ can"t you just take the square root of that? – Emil Dec 29 '16 at 17:33

By definition, $\|u\|^2 = \langle u,u \rangle$. Hence $$\langle u,u \rangle - \frac{\langle u,v \rangle^2}{\langle v,v \rangle} \geq 0 \implies \langle u,u \rangle \geq \frac{\langle u,v \rangle^2}{\langle v,v \rangle} \implies \langle u,u \rangle \langle v,v\rangle \geq \langle u,v\rangle^2 \implies \|u\|^2 \|v\|^2 \geq \langle u,v\rangle^2$$