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I want to prove $|\langle x, y \rangle| \leq \|x\| \, \|y\|$ for all $x,y \in \mathbb{R}^n$ or $\mathbb{C}^n $. I know there exist tons of proofs for this inequality, but I want to prove 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 it's possible to rearrange this inequality such that $0 \leq at^2 + bt +c$.
  3. Now we can look at the discriminant $b^2 -4ac$ and formulate a new inequality such that the Cauchy-Schwarz inequality follows.
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Start by squaring your inequality in (1), giving $$ 0 \le \|tx+y\|^2 = \def\<#1>{\left<#1\right>}\<tx+y,tx+y> = t^2\|x\|^2 + 2t\<x,y> + \|y\|^2 $$ The discriminant is therefore given by $$ 4\<x,y>^2 - 4\|x\|^2\|y\|^2 $$ As the discriminant cannot be positive (note that a non-negative quadratic real polynomial has at most one root), we have $$ 4\<x,y>^2 - 4\|x\|^2\|y\|^2 \le 0 \iff \<x,y>^2 \le \|x\|^2\|y\|^2. $$ Taking square roots gives Cauchy-Schwarz.

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  • $\begingroup$ 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? $\endgroup$ – faoeoe Oct 16 '18 at 19:36
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Kuldeep Singh's Linear Algebra: Step by Step (2013) proves this at p 293.

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