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?

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Does Cauchy--Schwarz ring any bells? – Branimir Ćaćić Mar 26 '13 at 7:31
You claim that $\sqrt{u^2+2u\cdot v+v^2}\ge \sqrt{u^2}+\sqrt{v^2}$. I dispute. – Brady Trainor Mar 26 '13 at 7:32
Try $u=1$, $v=1$. – Brady Trainor Mar 26 '13 at 7:34
Remember, powers distribute across multiplication, multiplication distributes across addition, but powers (and roots) do not distribute across addition. – Brady Trainor Mar 26 '13 at 7:36
Oops, should have stuck with my first counter example, $u=(1,0)$, $v=(0,1)$, but I would guess a counterexample with scalars can be constructed as well. – Brady Trainor Mar 26 '13 at 7:51

Try comparing their squares, $\|u+v\|^2$ versus $(\|u\|+\|v\|)^2$.
I just conclude $$\Vert u+v\Vert^2=\langle u+v,u+v\rangle=(\langle u,u \rangle +2\langle u,v\rangle+\langle v,v \rangle) = \Vert u\Vert^2 + 2\langle u,v\rangle + \Vert v\Vert^2$$ Then we should understand that $\langle u,v\rangle\leqslant \Vert u\Vert\cdot\Vert v\Vert$. And we have $$\Vert u+v\Vert^2 = \Vert u\Vert^2 + 2\langle u,v\rangle + \Vert v\Vert^2\leqslant \Vert u\Vert^2 + 2\Vert u\Vert\cdot\Vert v\Vert + \Vert v\Vert^2=(\Vert u\Vert+ \Vert v\Vert)^2.$$ So $$\Vert u+v\Vert\leqslant \Vert u\Vert+ \Vert v\Vert.$$