A metric space $(X,d)$ is called an ultrametric space if it is a metric space and fulfills the stronger triangle inequality (see Wikipedia) $$ d(x,y) \le \max\{ d(x,z), d(z, y) \}. $$ Examples are the Cantor-Space $\{0,1\}^{\mathbb N}$ with the common-prefix-metric $d(\xi, \eta) = 1/k$ where $k$ is the smallest $k > 0$ such that $\xi_k \ne \eta_k$, or the $p$-adic numbers. Now is there anything known about spaces which fulfill the even stronger requirement $$ d(x,y) = \max\{ d(x,z), d(z, y) \}. $$ for pairwise different points $x,y,z$. The Cantor-Space with the common-prefix-metric is an example.
EDIT: Initially I wrote: $d(x,y) = \min\{ d(x,z), d(z, y) \}$, which always induces the discrete space, see the comment.
The Cantor space does not satisfy this stronger requirement: if $x = \langle 1 , 0 , 0 , \ldots \rangle$, $y = \langle 1, 1, 0 , 0 \ldots \rangle$ and $z = \langle 0, 0, 0, \ldots \rangle$, then $d (x,y) = 1/2$ and $d(x,z) = d(y,z) = 1/1$.
Note that your condition implies that all (non-degenerate) triangles are "equilateral": given distinct points $x,y,z$, we know that $d(x,y)$ must equal the greater of $d(x,z)$ and $d(y,z)$, and without loss of generality we may assume that it $d(x,y) = d(x,z)$. But now that same condition implies that $d(y,z) = \max \{ d(x,y) , d (x,z) \} = d(x,y) = d(x,z)$.
This in turn shows that in such a metric space there is a $\delta > 0$ such that $d(x,y) = \delta$ for all distinct $x,y$, resulting in a discrete topological space.