Show that $\langle v,w\rangle ^2 =\langle v,v\rangle\langle w,w\rangle$ if and only if $v$ is parallel to $w$ Let $ \langle.,.\rangle$ be the usual scalar product, show that $\langle v,w\rangle ^2 =\langle v,v\rangle\langle w,w\rangle$ if and only if $v$ is parallel to $w$.
this was asked after proving that every scalar product satisfies the inequality, which i managed to do.
any help on this please]
just to clarify, I'm having troubles proving the forward part, so I have proved that if $v$ is parallel to $w$ then the equality holds, just can't do the other way.

This basically comes from the proof. You're expanding $\langle u-tw, u-tw\rangle$, and noticing that $0 \leq \langle u-tw, u-tw\rangle$, then picking a normalizing $t$ to give you the inequality. It's clear then that to get equality you need $u=tw$, no matter what $t$ you chose, i.e. linear dependence.
