All Gromov-hyperbolic spaces are CAT(0)

Question

I want to show that every $\delta-$hyperbolic space is CAT(0), by the definition I'm using $X$ is $\delta-$hyperbolic if the following inequality is satisfied for all $w,x,y,z\in X$ and for some $\delta>0$ $$(x,z)_w\geq \min((x,y)_w,(y,z)_w)-\delta$$ where $$(y,z)_x=\frac{1}{2}(d(x,y)+d(x,z)-d(y,z))$$. If $\Delta(x,y,z)$ is a triangle in X and $p$, $q$ are points on segments, say $[x,y]$ and $[z,y]$, I don't know how to relate $d(p,q)$ to $d(\bar{p},\bar{q})$ in a comparison triangle, especially when $X$ is not geodesic. I suppose I have to get the CAT inequality from the aforementioned four point property for $\delta-$hyperbolic space, but I don't know how to start.

Answer 1

Many hyperbolic spaces are not CAT(0). For instance:

Let $X$ be any bounded graph. Then $X$ is $\delta$-hyperbolic when $\delta>\mathrm{diam}(X)$. However, it is CAT(0) if and only it is a tree. (Indeed, CAT(0) spaces are contractible.)

Typically, you do not have any control on the local geometry of a hyperbolic space. On the other hand, being CAT(0) is quite rigid. For instance, take any CAT(-1) space, and glue small loops at each point. Your space will be still hyperbolic, but it will be no longer CAT(0).

In the opposite direction, it is interesting to note that we do not know if all hyperbolic groups act geometrically on CAT(0) spaces, it is an open problem.