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In Luenberger book Cauchy-Schwarz Inequality is defined like this: For all $x,y$ in an inner product space $|(x|y)| \le \|x\|\|y\|$. Equality holds if and only if $x = \lambda y$ or $y = \theta$.

Proof starts for all scalars $\lambda$,

$ 0 \le (x-\lambda y | x-\lambda y) = (x|x) - \lambda(y|x) - \bar{\lambda}(x|y) + |\lambda|^2 (y|y) $

I understand this expansion. But then, it will select a particular $\lambda = (x|y)/(y|y)$, and obtains

$ 0 \le (x|x) - \frac{|(x|y)|^2}{(y|y)} $

I dont understand how he chose that particular $\lambda$. I guess I understand why, he chose it to get rid of it in the main equation, but is it okay to chose any $\lambda$ that will clean up the equation like this?

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    In my opinion, anybody interested in Cauchy-Schwarz inequality in general, and the question asked here in particular, should read and ponder [this page](http://www.dpmms.cam.ac.uk/~wtg10/csineq.html) by Tim Gowers on the subject. Those interested in going further might want to have a look (and probably, much more than a look) at the gem of a book titled [The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities](http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_index.html) by J. Michael Steele.2012-02-12

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I dont understand how he chose that particular $\lambda$. I guess I understand why, he chose it to get rid of it in the main equation, but is it okay to chose any $\lambda$ that will clean up the equation like this?

Yes, the inequality that you display holds for all $\lambda$ and so "it is okay to choose any $\lambda$" that you like. Luenberger's choice (it might well be the one used originally by Cauchy and/or Schwarz) "cleans up the equation" as you note, and provides motivation for its use. But if you have another value for $\lambda$ in mind that allows you to reach the conclusion $0 \le (x|x) - \frac{|(x|y)|^2}{(y|y)},$ (which is just a re-arrangement of the Cauchy-Schwarz Inequality), by all means, go for it.

See Appendix B of this Lecture Note for a more prolix proof of the Cauchy-Schwarz Inequality than the one in the Luenberger book.

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As said before, we always have $0\leq\|x-\lambda y\|^2=(x-\lambda y|x-\lambda y)$ Of course the choice $\lambda=\frac{(x|y)}{\|y\|^2}$ is neither arbitrary or accidental luck. Geometrically the specific $\lambda$ chosen is the one that gives the orthogonal projection of $x$ along the vector $y$.

The product $(x|y)$ removes the part of $x$ that is orthogonal to $y$, and dividing by $\|y\|^2$ normalizes $y$.

Consider the picture below where

  • the black arrow is to picture $x$

  • the blue arrow is to picture $y$

  • the red arrow is to picture $\lambda y$

  • the green arrow is to picture $x-\lambda y$

OrthogonalProjOfXontoY

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    @JonasMeyer Yes, of course. Thanks Jonas :)2012-02-12
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You have a wrong idea. Actually, you don't need to select $\lambda$. For every $\lambda$,

$0 \le (x-\lambda y | x-\lambda y)$, that means

$0 \le (x|x)-\lambda(y|x)-\bar{\lambda}(x|y) + |\lambda|^2 (y|y)$,Consider it a function with the independent variable $\lambda$. So if $\lambda$ is R

$\Delta= 4(x|y)^2-4(x|x)(y|y) \le 0$. If $\lambda is C$, you need to use other functions.