Showing a function is a norm. I'm trying to prove that $\Vert v\Vert :=\langle v,v\rangle^{1/2} $ defines a norm, but I'm having trouble with the triangle inequality. 
$\Vert u+v\Vert=\langle u+v,u+v\rangle^{1/2}=(\langle u,u \rangle +2\langle u,v\rangle+\langle v,v \rangle)^{1/2}$
The latter expression is greater than or equal to $\Vert u \Vert + \Vert v \Vert$, but this is the opposite of what I am trying to prove. Where am I going wrong?

Try comparing their squares, $\|u+v\|^2$ versus $(\|u\|+\|v\|)^2$. 
