Is a $0$-hyperbolic group free? In his article, Abderezak Ould Houcine asks the following question: If $G$ is a hyperbolic group, let $\delta_0(G)$ denote the infinimum of $\delta$ for which $G$ is $\delta$-hyperbolic. When $\delta_0(G)=0$?
Of course, free groups are such groups. But are they other ones? 
If $\Gamma$ is a (Cayley) graph and $C$ a cycle of minimum length $\ell$, then taking three points $x,y,z \in C$ satisfying $d(x,y)=d(y,z)=d(z,x)= \ell/3$, the triangle induced by $C$ is geodesic and it is clearly not $0$-thin. So it seems that the only $0$-hyperbolic graphs are trees, and consequently, the only $0$-hyperbolic groups would be free.
Am I missing something? Do you have an example of non-free $0$-hyperbolic group?

If a graph is not a tree, then it contains a shortest contactible loop, which is an isometric image of a circle. A circle of length $L$ has $\delta\ge L/4$ (using the $4$-point form of hyperbolicity, with $x,z,y,v$ placed at equal distances around the circle). So yes, a $0$-hyperbolic graph is a tree. And if it's the Cayley graph of a group, the group is free.
It's possible that the remark was really intended for the metric $\ell_S$ considered in the paper later; this metric is defined only on the set of vertices, which  seems to allow non-free groups to have $0$-hyperbolicity in this metric.
