# Proof Cauchy-Schwarz inequality

I want to proof $$|\langle x, y \rangle| \leq \|x\| \, \|y\|$$ for all $$x,y \in \mathbb{R}^n$$ or $$\mathbb{C}^n$$. I know there exist a ton of proofs for this inequality, but it want to proof it through a specific schematic.

1. We know that the length of a Vector is $$\geq 0$$, hence for an arbitrary $$t \in \mathbb{R}$$ it follows that $$0 \leq \| tx + y \|$$.
2. I think its possible to rearrange this inequality such that $$0 \leq at^2 + bt +c$$.
3. Know we can look at the discriminant $$b^2 -4ac$$ formulate a new inequality such that the Cauchy-Schwarz inequality follows.
• – Clement C. Oct 16 '18 at 19:02

Start by squaring your inequality in (1), giving $$0 \le \|tx+y\|^2 = \def\<#1>{\left<#1\right>}\ = t^2\|x\|^2 + 2t\ + \|y\|^2$$ The discriminant is therefore given by $$4\^2 - 4\|x\|^2\|y\|^2$$ As the discriminant cannot be positive (note that a non-negative quadratic real polynomial has at most one root), we have $$4\^2 - 4\|x\|^2\|y\|^2 \le 0 \iff \^2 \le \|x\|^2\|y\|^2.$$ Taking square roots gives Cauchy-Schwarz.
• Thank you! I have two questions. 1) For $\mathbb{R}$ i understand the rearrangement, but for $\mathbb{C}$ this is only sesquilinear, isn't it? Why is it also for $\mathbb{C}$ correct ? 2) Why exactly ist the discriminant not positiv? – faoeoe Oct 16 '18 at 19:36