I was told that the following fact is true.
Let $X$ be a finite dimensional Alexandrov space with non-negative curvature.
Then the function $$x\mapsto dist(x, \partial X)$$ is concave (namely its resriction to any shortest path is concave), where $\partial X$ is the boundary of $X$. I would be happy to have a reference to a proof or a proof.
Remark. In the special case when $X$ is a smooth Riemannian manifold with locally geodesically convex boundary (hence an Alexandrov space) this is true: a reference was given in a comment to this post Concavity of distance to the boundary of Riemannian manifold . However the argument given in the reference is based on the use of the Rauch comparison theorem (and hence Jacobi fields) and does not seem to immediately generalize to Alexandrov spaces.
