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.
The complete proof is there : Cauchy?Schwarz inequality - Wikipedia, the free encyclopedia.
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,
$\displaystyle 0\le \left\| t \right\|^{2}\cdot \left\| v \right\|^{2}=\left\| u \right\|^{2}\cdot \left\| v \right\|^{2}-\left\langle u,v \right\rangle ^{2}\implies \left| \left\langle u,v \right\rangle \right|\le \left\| u \right\|\cdot \left\| v \right\|.\quad\blacksquare$