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: $$ g=dr^2+f^2(r, \theta;)d\theta^2, $$ such that the $\theta$ parameter runs from $0$ to $2\pi$. Assume that $f$ is a smooth function on $(0,+\infty)\times S^1$ such that $f(0, \theta)=0$.
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 $C=2\pi$. If $C<2\pi$, 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 $C>2\pi$, 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 $\gamma(t)$ such that $D_{\gamma'}\gamma'=0$.) One can show this directly for some simple cases, eg. a flat metric with a cone angle greater than $2\pi$.
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.