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.
 A: 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.
$$\arctan(u+v)=\int_0^{u+v}{dt\over 1+t^2}=\int_0^u{dt\over 1+t^2}+\int_0^v{dt'\over1+(u+t')^2}\leq\arctan u+\arctan v\ .$$
A: Completely orthogonal idea. It is trivial to show that $d_1(x,y)=|x-y|$ is a metric. Function $f(x)=\arctan(x)$ is strictly increasing, continuous (actually real-analytic) bounded function positive for $x\geq 0$. Try to prove that $d(x,y)=f(d_1(x,y))$ must be a metric for any such function $f$. Edit: Please see the comment bellow. The conditions I listed originally are not sufficient. 
A: Suppose that $\alpha, \beta \ge 0$.
For $\alpha\beta < 1$, we have
$$\arctan\alpha + \arctan\beta = \arctan\left(\tan\left(\arctan\alpha + \arctan\beta\right)\right)$$
$$=\arctan \frac{\alpha+\beta}{1-\alpha\beta} \ge \arctan (\alpha + \beta)$$
by the angle sum formula for tangent.  On the other hand, for $\alpha\beta \ge 1$ with (say) $\alpha > 1$, we find that $\beta \ge \frac{1}{\alpha}$, so
$$\arctan \alpha + \arctan\beta \ge \arctan \alpha + \arctan \frac{1}{\alpha} \ge \arctan\alpha + \frac{\pi}{2} - \arctan\alpha$$ $$\ge \frac{\pi}{2} > \arctan(\alpha+\beta)$$
since $\arctan x < \frac{\pi}{2}$ for all $x$.
Substituting $\alpha = |x-z|$, $\beta=|z-y|$, we find
$$\arctan|x-z| + \arctan|z-y| \ge \arctan |x-y|$$
A: first let $= \tan t_1 = m_1 \ge 0, \tan t_2 = m_2 \ge 0, 0 \le t_1, t_2 < \pi/2$
then $$\tan(t_1+t_2) = \dfrac{\tan t_1 + \tan t_2}{1 - \tan t_1 \tan t_2}
=  \dfrac{m_1 + m_2}{1 - m_1 m_2} \ge m_1 + m_2 $$ therefore 
$$\tan^{-1}(m_1 + m_2) \le t_1 + t_2 = \tan^{-1}m_1 + \tan^{-1} m_2$$ now substituting $m_1 = |x-y|, m_2 = |y-z|$ and using $\tan^{-1}$ is an increasing function with the triangle inequality $|x-z| \le |x-y| + |y-z|$
$$ \tan^{-1} |x-z| \le \tan^{-1}(|x-y| + |y-z|) \le \tan^{-1}|x-y|  + \tan^{-1} |y-z|$$ and that proves the triangle inequality.
