On $\mathbb C^n$, define $||z||=(\sum_{j=1}^{n}|z_j|^2)^\frac{1}{2}$ and for $x,z\in\mathbb C^n$ define $d(z,w)=||z-w||.$ Prove that $d$ is a metric on $\mathbb C^n$.
My attempt:
I need to show that:
(1) $\forall x,z\in\mathbb C^n$, $d(x,z)\geq0$ and $d(x,z)=0$ iff $x=z$.
$d(x,z)=||x-z||=(\sum_{j=1}^{n}|x_j-z_j|^2)^\frac{1}{2}\geq0.$ For $x=z$, $(\sum_{j=1}^{n}|x_j-z_j|^2)^\frac{1}{2}=\sum_{j=1}^{n}0 = 0$.
(2) $\forall x,z\in\mathbb C^n$, $d(x,z)=d(z,x)$.
$d(x,z)=||x-z||=(\sum_{j=1}^{n}|x_j-z_j|^2)^\frac{1}{2} = (\sum_{j=1}^{n}|z_j-x_j|^2)^\frac{1}{2} = d(z,x)$ by properties of modulus in $\mathbb{C}$.
(3) $\forall x,y,z\in\mathbb C^n$, $d(x,z)\leq d(x,y)+d(y,z)$.
This is always the tough one. Do I need Cauchy-Schwarz or Hoelder's Inequality? Attempt: $d(x,z)=||x-y+y-z||=(\sum_{j=1}^{n}|x_j-y_j+y_j-z_j|^2)^\frac{1}{2} =(\sum_{j=1}^{n}|x_j-y_j|+|y_j-z_j|^2)^\frac{1}{2}$,...