Solution
= Solution
If $v=0$, equality is immediate. If $v\ne0$, the nonnegative squared <norm>
$$
0\leq\left\|u-\frac{\langle u,v\rangle}{\|v\|^2}v\right\|^2=\|u\|^2-\frac{\langle u,v\rangle^2}{\|v\|^2}
$$
proves the <Cauchy-Schwarz inequality>. Equality holds precisely when the displayed <vector> vanishes. Thus \b[equality is equivalent to <linear dependence> of $u,v$], including the cases with a <zero> <vector>.