Talk:Cauchy-Bunyakovsky-Schwarz Inequality

this inequalite is the same that this Cauchy's Inequality ?? because i learn that Cauchy's Inequality is the same that Cauchy–Schwarz and Cauchy–Schwarz-(other name) -- Gamma 15:20, 1 January 2009 (UTC)

Yes it is but I don't understand the notation on this page. I wrote a new page using conventional notation and concepts. --prime mover (talk) 16:02, 1 January 2009 (UTC)

I'm not sure about the original exposition of this. I believe there's a misunderstanding as to the meaning of $\left\|{x}\right\|$ because the usual statement of it is $\left|{\left \langle {x, y} \right \rangle}\right| \le \left\|{x}\right\| \times \left\|{y}\right\|$.

More explanation is needed as to the precise meaning of all objects involved, because at the moment it's not at all solid. --prime mover (talk) 19:46, 31 March 2009 (UTC)

This proof is better. Wok 03:24, 20 February 2011 (CST)

Vectors
Under which version of the C-S inequality does this form fall under? Is it implied in one of the guys up, or does it need to be added?

for $\mathbf{x}$, $\mathbf{y} \in \R^n$.

--GFauxPas 21:08, 26 February 2012 (EST)


 * The dot product is an inner product on $\R^n$; hence you can consider it under the (semi-)inner product space version. --Lord_Farin 02:07, 27 February 2012 (EST)
 * Alternatively, you could take Cauchy's inequality with $1\le i \le n$. --Lord_Farin 02:09, 27 February 2012 (EST)

POTW Candidate
No, not a multiple poof with lots of parts, that's a nightmare to maintain. --prime mover (talk) 06:14, 19 November 2014 (UTC)