The aim of this question is to prove the cauchy schwarz inequality. Use the fact that $\|\mathbf{x}\|^2 = \mathbf{x} \cdot\mathbf{x} \geq 0$ to show the following:
a) $\|\|\mathbf{v}\|^2 \mathbf{u} - (\mathbf{u} \cdot \mathbf{v})\mathbf{v}\|^2 = \|\mathbf{v}\|^4 \|\mathbf{u}\|^2 - \|\mathbf{v}\|^2(\mathbf{u}\cdot\mathbf{v})^2$
I got this part. It was fairly straightforward.
b) Use part a) and the fact that $\|\mathbf{x}\|^2 = \mathbf{x} \cdot\mathbf{x} \geq 0$ to show that $|\mathbf{u}\cdot\mathbf{v}|\leq\|\mathbf{u}\|\|\mathbf{v}\|$
I'm stuck here. I know what the cauchy schwarz inequality is and I've seen various proofs of it which I get, but I'm not sure how we use part a) for this proof. Can anyone offer any insight?
Here it is a proof I like. Let $\langle\cdot,\cdot\rangle:V\times V\to\textbf{F}$ be an inner product where $V$ is a vector space. Let $u\in V$. If $u = 0$, then clearly $\langle u,v\rangle = 0 \leq \|u\|\|v\|$ for every $v\in V$. Suppose otherwise that $u\neq 0$. Then define $w\in V$ as follows: \begin{align*} w = \frac{\langle u,v\rangle}{\langle u,u\rangle}u \end{align*}
As it can be seen, $v = w + (v - w)$. Once $w\perp(v - w)$, one concludes that \begin{align*} \|v\|^{2} = \langle v,v\rangle = \|w\|^{2} + \|v - w\|^{2} \geq \|w\|^{2} = \frac{|\langle u,v\rangle|^{2}}{\|u\|^{2}} \Rightarrow |\langle u,v\rangle| \leq \|u\|\|v\| \end{align*}
Hopefully this helps!