This is the simplest case of a question that's been bugging me for a while: say we have a Riemannian metric in polar coordinates on a (2-d) surface:
g=dr^{2}+f^{2}(r, θ)dθ^{2}, such that the θ parameter runs from 0 to 2π. Assume that f is a smooth function on (0,∞)X S^{1} such that f(0, θ)=0.

Define the cone angle at the pole to be $ C=\lim_{r\rightarrow 0^{+}} \frac{L(\partial B(r))}{r} $, where B(r) is the geodesic disc of radius r centered at the origin. Then it's fairly easy to see(by switching into Cartesian coordinates) that a necessary condition for the metric to be smooth is that C=2π. If C<2π, there is a cone point at the origin. One can write out a cone metric, and show that the triangle inequality holds, so there is a singular metric, but which still induces a metric space structure.

Now, if C>2π, it seems pretty clear that we'll end up with a space which violates the triangle inequality; it will be shorter to take a broken segment through the origin than to follow the shortest geodesic(in the sense of a curve γ(t) such that D_{γ'}γ'=0.) One can show this directly for some simple cases, eg a flat metric with a cone angle greater than 2π.

But there must be an elementary proof of the general case! I can't seem to find one though, and I spent the afternoon playing around with the Topogonov and Rauch comparison estimates to no avail. The basic problem I'm having is that the cone angle condition is essentially a condition on metric balls, but we expect a violation of the triangle inequality, which is a condition on distances.

This is not really related to anything I'm working on, but it's driving me crazy, so I'd appreciate any insight.