A metric on $\mathbb{C^n}$ 
Possible Duplicate:
Show that $d$ is a metric on $\mathbb{C^n}$ 


On $\mathbb{C^n}$, define $||z||=(\sum_{j=1}^{n}|z_{j}|^{2})^{1/2}$ and for $z,w\in\mathbb{C^n}$ define $d(z,w)=||z-w||$. Show that $d$ is a metric on $\mathbb{C^n}$.

My attempt:
(1) (nonnegativity) It is clear that for any $z,w\in\mathbb{C^n}$, $d(z,w)=||z-w|| = (\sum_{j=1}^{n}|z_{j}-w_{j}|^{2})^{1/2}\geq 0$ since $|z_{j}-w_{j}|^{2}\geq0$. Also, $||z-w||=0$ iff $(\sum_{j=1}^{n}|z_{j}-w_{j}|^{2})^{1/2}= 0$ iff $z=w$.
(2) (symmetry) $d(z,w)=||z-w|| = (\sum_{j=1}^{n}|z_{j}-w_{j}|^{2})^{1/2}=(\sum_{j=1}^{n}|w_{j}-z_{j}|^{2})^{1/2}=d(w,z)$ by properties of modulus in $\mathbb{C^n}$. 
(3) (triangle inequality) $\forall w,z,v\in\mathbb{C^n}$, $$\begin{align*}d(z,w)&=||z-w||\\
&= \left(\sum_{j=1}^{n}|z_{j}-w_{j}|^{2}\right)^{1/2}\\
&=\left(\sum_{j=1}^{n}|z_{j}+v_{j}-v_{j}-w_{j}|^{2}\right)^{1/2}\\
&=\left(\sum_{j=1}^{n}|(z_{j}-v_{j})+(v_{j}-w_{j})|^{2}\right)^{1/2}\leq...
\end{align*}$$ 
I'm not sure how to split up the sum here. 

Part (3) follows from Cauchy-Schwarz inequality.
$\vert \langle z, w\rangle\vert^2\leq \langle z,z\rangle \langle w,w\rangle  $
