Denote by $k(I, r)$ the incircle of $\triangle ABC$. It is clear that the image of $d$ is the interval $\Delta = (0, r]$.

Varying $D$ and $E$ on $AI$, it follows that $f$ is a mid-point concave function on $\Delta$, i.e. $f(2x) + f(2y) \le f(x + y)$.

Note that the points on given distance (from the boundary) form a triangle homothetic to $\triangle ABC$ (with center of homothety $I$) and the mid-points of the segments with ends at these points run over the whole interior of this triangle. It follows that $f$ is an increasing function on $\Delta$.

Since $f$ is a mid-point concave and increasing function, then it is continuous. Indeed, if $f^{-}(x)$ and $f^{+}(x)$ are the left- and the right-hand limit of $f$ at $x$ ($f^{+}(r) := f(r)$), then $f^{-}(x) + f(x) \le 2f^{-}(x)$ and $f^{-}(x) + f^{+}(x) \le 2f(x)$ (why?) and hence $f^{-}(x) = f(x) = f^{+}(x)$. So $f$ is a convex and increasing function.

Then, since $d$ is a concave function, it is easy to see that $f \circ d$ is a concave function. This also follows from the fact that $f \circ d$ is a mid-point concave and continuous function (because of the continuity of $f$ and $d$).