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Let $V$ be a vector space over either $\mathbb{R}$ or $\mathbb{C}$ Let $\langle.,.\rangle$ be an inner product on $V$.

Prove the Cauchy Schwarz Inequality

$$\forall x,y\in V, |\langle x,y\rangle|^2 \leq |\langle x,x\rangle||\langle y,y\rangle|$$

all the proofs I can come up with right now involves some sort of norm on $V$. I don't think I can just assume that $V$ has a norm.

So are there ways to prove it using nothing but the definition of inner product?

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    $\begingroup$ The assumption in the CS-inequality is that the norm is the one induced by the IP. I.e. $\|x\| = \sqrt{\langle x,x\rangle}$. $\endgroup$ – user137731 Oct 1 '16 at 23:15
  • $\begingroup$ There are many many MSE post on this. $\endgroup$ – Jacky Chong Oct 1 '16 at 23:16
  • $\begingroup$ @Bye_World, I know, but the thing is I am supposed to use cauchy schwarz to prove that an inner product induces a norm, so I cannot use that norm to prove cauchy schwarz $\endgroup$ – Phantom Oct 1 '16 at 23:17
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Let $\;c\in\Bbb R\;$ , so

$$0\le\langle x+cy,\,x+cy\rangle=\langle x,x\rangle+2c\,\text{Re}\,\langle x,y\rangle+c^2\langle y,y\rangle\le\langle x,x\rangle+2c\,|\langle x,y\rangle|+c^2\langle y,y\rangle$$

The above is a non-negative real quadratic in the real variable $\;c\;$ , so its discriminant must be non-positive:

$$\Delta=4|\langle x,y\rangle|^2-4\langle x,x\rangle\langle y,y\rangle\le0\implies|\langle x,y\rangle|^2\le\langle x,x\rangle\langle y,y\rangle$$

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Let $\alpha=\frac{\langle x,y\rangle}{|\langle x,y\rangle|}$, then $$ \begin{align} 0 &\le\left\langle\frac{x}{\|x\|}-\alpha\frac{y}{\|y\|},\frac{x}{\|x\|}-\alpha\frac{y}{\|y\|}\right\rangle\\[9pt] &=\frac{\langle x,x\rangle}{\|x\|^2}+\frac{\langle y,y\rangle}{\|y\|^2}-\frac{2\mathrm{Re}\left(\overline\alpha\langle x,y\rangle\right)}{\|x\|\,\|y\|}\\ &=2-2\mathrm{Re}\left(\frac{\overline{\langle x,y\rangle}}{|\langle x,y\rangle|}\frac{\langle x,y\rangle}{\|x\|\,\|y\|}\right)\\ &=2-2\frac{|\langle x,y\rangle|}{\|x\|\,\|y\|}\\ \end{align} $$ Therefore, $$ |\langle x,y\rangle|\le\|x\|\,\|y\| $$ where $\|x\|^2=\langle x,x\rangle$.

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Our papers [Novi Sad J. Math. 47(2017), 177--188] and [Rostock. Math. Colloq. 71\,(2016), 28--40 offer some different proofs.

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  • $\begingroup$ I'd consider this as a "link-only" answer... please post the argument's details below the citations. $\endgroup$ – Parcly Taxel Dec 12 '17 at 10:36

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