# Cauchy-Schwartz Inequality

• Jun 28th 2009, 02:48 PM
Fel
Cauchy-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 PM
arbolis
• Jun 28th 2009, 03:37 PM
Fel
I've seen that already, but I couldn't understand it all that well.
• Jun 28th 2009, 04:33 PM
Bhajun
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 PM
Krizalid
If $v=0_v,$ then we have the equality, so there's nothing to prove there. Thus let's take care about the interesting case when it's $v\ne0_v.$

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

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

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