Prove $d(x,y)=\arctan|x-y|$ is a metric I have to show that $d$ is a metric on the real numbers, and the first three axioms are straight forward, the triangle inequality poses a problem.  I know we need to get 
$$
\begin{align*}
d(x,y)
&= \arctan|x-y| \\
&\le \arctan|x-z|+\arctan|z-y| \\
&= d(x,z)+d(z,y),
\end{align*}
$$
so what I've tried is 
$$\arctan|x-y| = \arctan|x-z+z-y| \le \arctan(|x-z|+|z-y|),$$
but I'm not even sure if this accomplishes anything because I don't know how to split it up.

Lemma. When $d$ is a metric on a set $X$ and $f:\>{\mathbb R}_{\geq0}\to{\mathbb R}_{\geq0}$ is a monotonically increasing function with $f(0)=0$ and $f(t)>0$ for $t>0$ which is sublinear, i.e.,
$$f(u+v)\leq f(u)+f(v)\qquad\forall u, \>v\geq0\ ,$$
then $d':=f\circ d$ is again a metric on $X$.
Proof. Only the triangle inequality needs verification:
$$d'(x,z)=f\bigl(d(x,z)\bigr)\leq f\bigl(d(x,y)+d(y,z)\bigr)\leq f\bigl(d(x,y)\bigr)+f\bigl(d(y,z)\bigr)=d'(x,y)+d'(y,z)\ .$$
Therefore it remains to prove that $t\mapsto \arctan t$ is sublinear for $t\geq0$. This can be seen as follows.
$$\arctan(u+v)=\int_0^{u+v}{dt\over 1+t^2}=\int_0^u{dt\over 1+t^2}+\int_0^v{dt'\over1+(u+t')^2}\leq\arctan u+\arctan v\ .$$
