Composition of a function with a metric

In general, if $d$ is a metric, and $f$ is strictly increasing, $f(0)=0$, and concave on $[0,\infty)$, then $f\circ d$ is again an metric.

Here is a short proof for triangle inequality. Let $d_1=d(x,y),d_2=d(y,z),d_3=d(x,z)$, we have $d_1\leq d_2+d_3$. By increasing, we have $f(d_1)\leq f(d_2+d_3)$, and we want $f(d_2+f_3)\leq f(d_2)+f(d_3)$, which is equivalent to $\frac{f(d_2+d_3)-f(d_2)}{(d_2+d_3)-d_2}\leq \frac{f(d_3)-f(0)}{d_3-0}$. But this follows directly from concavity of $f$ since $0\leq d_2\leq d_2+d_3$.

Note, if $f$ is 2nd order differentiable, $f$ is concave iff $f''(x)\leq 0$ for all $x\in(0,\infty)$.

So by this criterion, a),c) will give you a metric.