Here is an elementary proof of the Steinhaus transform (from which said metricity follows as a special case, as noted in **Suresh's** answer).

>**Lemma.** Let $p,q,r > 0$ such that $p \le q$. Then, $\frac{p}{q} \le \frac{p+r}{q+r}.$

>**Corollary.** Let $d(x,y)$ be a metric. Then, for arbitrary (but fixed) $a$, 
\begin{equation*}
 \delta(x,y) := \frac{2d(x,y)}{d(x,a)+d(y,a)+d(x,y)},
\end{equation*}
is a metric.

*Proof.* Only the triangle inequality for $\delta$ is nontrivial. Let $p=d(x,y)$, $q=d(x,y)+d(x,a)+d(y,a)$, and $r=d(x,z)+d(y,z)-d(x,y)$. Applying the lemma, we obtain
\begin{eqnarray*}
 \delta(x,y) &=& \frac{2d(x,y)}{d(x,a)+d(y,a)+d(x,y)} \le \frac{2d(x,z)+2d(y,z)}{d(x,a)+d(y,a)+d(x,z)+d(y,z)}\\
 &=& \frac{2d(x,z)}{d(x,a)+d(z,a)+d(x,z)+d(y,z)+d(y,a)-d(z,a)} + \frac{2d(y,z)}{d(y,a)+d(z,a)+d(y,z)+d(x,z)+d(x,a)-d(z,a)}\\
&\le& \delta(x,z)+\delta(y,z),
\end{eqnarray*}
where the last inequality again uses triangle inequality for $d$.