Cauchy-Schwartz Inequality

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

$\begin{aligned}<br /> \left\langle t,v \right\rangle &=\left\langle u-\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot v,v \right\rangle \\ <br /> & =\left\langle u,v \right\rangle -\frac{\left\langle u,v \right\rangle }{\left\| v \right\|^{2}}\cdot \left\langle v,v \right\rangle \\ <br /> & =0. <br /> \end{aligned}$

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,

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