Under what condition on $f$ is $f(d(x,y))$ a metric $f:\mathbb{R}\rightarrow \mathbb{R}$ and $(X,d)$ is a metric space. 
Under what condition on $f$ is $\rho:X\times X \rightarrow \mathbb{R} $ given by
$$\rho(x,y)=f(d(x,y))$$ also a metric?
My attempt (intuitively):
$f$ must be defined everywhere (well defined) i.e. it has to be continuous and differentiable. I am also thinking that for the triangle inequality to hold, $f$ must be monotonically increasing.
Does $f(x)=tan^{-1}(x)$ work?

Let's check the conditions that a metric must satisfy


*

*$d(x,y)\ge 0$

*$d(x,y) \iff x=y$

*$d(x,y)=d(y,x)$

*$d(x,y)+d(y,z)\ge d(x,z)$


Those are the conditions for a metric. So clearly we must have that $f(x)\ge 0$ when $x\ge 0$ to sate the first condition. For the second condition we must have that $f(0)=0$, but it is important also that that there is no non-zero $r$ such that $f(r)=0$ as otherwise we pick $x,y$ such that $d(x,y)=r$ and then the second condition is broken. The third is trivially true. For the last one we need to have 
$$f(d(x,z))=\rho(x,z)\le\rho(x,y)+\rho(y,z)=f(d(x,y))+f(d(y,z))$$
which means we must have the function being subadditive, that is $f(x+y)\leq f(x)+f(y).$
