Talk:Cauchy–Schwarz inequality

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A proof was requested by an anon

A proof appears in the article inner product space. A moment ago, I added that same proof to this page. Michael Hardy 22:34, 11 Nov 2004 (UTC)

An alternate proof, which i learnt is as follows: cos x = u . v / ||u|| ||v|| as cos x is between -1 and 1, the absolute value of the denomenator must be larger or equal to the numerator, hence u . v <= ||u|| ||v|| TheDarkLeaf 17:30, 19 June 2005 (AEST)

But first you need to establish that the cosine does play that role. You can give an easy intuitive geometric argument, but whether it works in, e.g., infinite-dimensional spaces may be dubious. Michael Hardy 23:24, 19 Jun 2005 (UTC)

While it is likely that the only people who might be interested in this page would already be familiar with the various mathmatical symbols used, it wouldn't hurt to add a little to the end or somehow link them to their explaination page (dunno what they are or how to do that). -FjordPrefect

There are conspicuous links to inner product space and related articles; that's where those explanations should be sought. Michael Hardy 23:58, 12 August 2005 (UTC)

How can you justify allowing lambda to be <x,y>*<y,y>^-1? I just do not see how this is a general proof. —Preceding unsigned comment added by 216.106.49.131 (talk) 19:00, 23 November 2007 (UTC)

The expression has been shown to be valid for all complex λ, and y is assumed non-zero, so obviously we can take . --Zundark (talk) 19:24, 23 November 2007 (UTC)

[edit] TOD: Specific Cases

As promised in the first sentence we need to add specific information on how Cauchy-Schwartz applies to:

• Infinite series
• Integration
• Variances / covariances. Perhaps a link to the Cramér–Rao bound whose proof relies on Cauchy-Schwartz.

Eug (talk) 03:41, 15 April 2008 (UTC)