TITLE: Is the function $d(x,y) = \frac{\|x-y\|}{\|x\|\|y\|}$ a metric?
QUESTION [11 upvotes]: $d$ is defined for all $x,y \in \mathbb{R}^2 - \{0\}$.
It's clear that $d(x,y) = 0 \iff x=y$ and $d(x,y)=d(y,x)$
I am having issues with triangle inequality. I couldn't find a counterexample for which the triangle inequality doesn't hold. So I tried to prove it.
What I have so far is:
$$d(x,z) =  \frac{\|x-z\|}{\|x\|\|z\|} \leq  \frac{\|x-y\|}{\|x\|\|z\|}
+  \frac{\|y-z\|}{\|x\|\|z\|} $$
I'm stuck here.
I appreciate if you could give me some hints.
Thanks.

REPLY [4 votes]: $\newcommand{\Reals}{\mathbf{R}}$Identify the set of non-zero vectors in $\Reals^{2}$ with the set of non-zero complex numbers. The Euclidean norm corresponds with the complex modulus, so if $x$ and $y$ are non-zero, then
$$
\frac{\|x - y\|}{\|x\|\, \|y\|}
  = \left\|\frac{x - y}{xy}\right\|
  = \left\|\frac{1}{y} - \frac{1}{x}\right\|.
$$
That is, $d$ corresponds to the ordinary Euclidean distance after a bijection (the complex reciprocal map), and therefore satisfies the triangle inequality.