# Why is $\langle\alpha x+y, \alpha x+y\rangle = |\alpha|^2\|x\|^2 + (\alpha\langle x, y\rangle + \bar{\alpha}\langle y, x\rangle) + \|y\|^2.$

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