I have a proof that proves the Cauchy Schwarz inequality. I am at the stage where I am trying to show the inequality holds for when $\langle x, y\rangle \neq 0$ and $x$ and $y$ are linearly independent.
For $\alpha \in \mathbb{C}$, $\alpha x +y \neq 0$, then
$$0 < \|\alpha x+y\|^2 = \langle\alpha x+y, \alpha x+y\rangle = |\alpha|^2\|x\|^2 + (\alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle) + \|y\|^2.$$
How do we see that $\langle\alpha x+y, \alpha x+y\rangle = |\alpha|^2\|x\|^2 + (\alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle) + \|y\|^2$?
Recall that a complex inner product in complex linear in the first argument (i.e. $\langle kv, w\rangle = k\langle v, w\rangle$), complex antilinear in the second argument (i.e. $\langle v, kw\rangle = \bar{k}\langle v, w\rangle$), and $\langle v, v\rangle = \|v\|^2$. With all of this in mind, we have
\begin{align*} \langle\alpha x + y, \alpha x + y\rangle &= \langle\alpha x, \alpha x\rangle + \langle\alpha x, y\rangle + \langle y, \alpha x\rangle + \langle y, y\rangle\\ &= \alpha\bar{\alpha}\langle x, x\rangle + \alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle + \|y\|^2\\ &=|\alpha|^2\|x\|^2 + \alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle + \|y\|^2. \end{align*}