Prove the Cauchy-Schwartz inequality |<u, v>| <= |u| |v|. (<= means less than or equal to)

I attempted it a couple times, but didn't get anywhere. Help would be great, thanks.

Printable View

- Jun 28th 2009, 02:48 PMFelCauchy-Schwartz Inequality
Prove the Cauchy-Schwartz inequality |<u, v>| <= |u| |v|. (<= means less than or equal to)

I attempted it a couple times, but didn't get anywhere. Help would be great, thanks. - Jun 28th 2009, 03:30 PMarbolis
The complete proof is there : Cauchy?Schwarz inequality - Wikipedia, the free encyclopedia.

- Jun 28th 2009, 03:37 PMFel
I've seen that already, but I couldn't understand it all that well.

- Jun 28th 2009, 04:33 PMBhajun
Please post an attempt at a solution or clarify what you don't understand in the proof provided (preferrably the former).

- Jun 28th 2009, 06:05 PMKrizalid
If $\displaystyle v=0_v,$ then we have the equality, so there's nothing to prove there. Thus let's take care about the interesting case when it's $\displaystyle v\ne0_v.$

Under this assumption, put $\displaystyle t=u-\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot v$ and then

$\displaystyle \begin{aligned}

\left\langle t,v \right\rangle &=\left\langle u-\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot v,v \right\rangle \\

& =\left\langle u,v \right\rangle -\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot \left\langle v,v \right\rangle \\

& =0.

\end{aligned}$

From here we have $\displaystyle 0\le\|t\|^2=\left\langle t,t \right\rangle =\left\| u \right\|^{2}-\frac{\left\langle u,v \right\rangle ^{2}}{\left\| v \right\|^{2}}.$ Hence,

$\displaystyle 0\le \left\| t \right\|^{2}\cdot \left\| v \right\|^{2}=\left\| u \right\|^{2}\cdot \left\| v \right\|^{2}-\left\langle u,v \right\rangle ^{2}\implies \left| \left\langle u,v \right\rangle \right|\le \left\| u \right\|\cdot \left\| v \right\|.\quad\blacksquare$