How to prove triangle inequality for $\|u\| = \langle{u,u}\rangle ^{0.5}$

Question

I want to prove that $\|u\| = \langle{u,u}\rangle ^{0.5}$ satisfies the 4 conditions for being a norm. I've already proved the first 3 conditions but I'm stuck on the triangle inequality i.e. $$\|x + y\| \leq \|x\| + \|y\|$$
Any ideas would be appreciated.

Answer 1

Hint: $ \|u+v\|^2=\|u\|^2+\langle u,v\rangle+\langle v,u\rangle+\|v\|^2.$

Edit: We have: \begin{align} \|u+v\|^2&=\|u\|^2+\langle u,v\rangle+\langle v,u\rangle+\|v\|^2 \\ &= \|u\|^2+\langle u,v\rangle+\overline{\langle u,v\rangle}+\|v\|^2 \\ &=\|u\|^2+2\Re\langle u,v\rangle+\|v\|^2 \\ &\le \|u\|^2+2\left|\langle u,v\rangle\right|+\|v\|^2 \\ \tag1 &\le \|u\|^2+2\|u\|\|v\|+\|v\|^2 \\ &= (\|u\|+\|v\|)^2, \end{align}

where we have used the Cauchy-Schwarz inequality in $(1)$. Taking square roots, since both sides are positive, we yield the desired result: $$\|u+v\|\le\|u\|+\|v\|.$$