inequalities - Is the Jaccard distance a distance?

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 $d$.