Triangle inequality for Bergman metric

Question

Please help to prove triangle inequality for d(z,w) Where

Answer 1

If you require a direct proof using the formula above, it does require some computation and approximating some inequalities. If you only need to show that it satisfies the triangle inequality this follows directly from the definition of the Bergman metric as follows: The metric $d(x,z)$ is defined as the $\inf$ over the lengths of all rectifiable paths between the points $x$ and $z$. If there were some point $y$ such that $d(x,y) + d(y,z) < d(x,z)$ (i.e., the triangle inequality would be violated for those points), then from the definition you know that there are (for any $\epsilon >0$) rectifiable paths $\gamma_{xy}$ and and $\gamma_{yz}$ between $x,y$ and $y,z$ respectively with lengths $l(\gamma_{xy}) \leq d(x,y) + \epsilon$ and $l(\gamma_{yz}) \leq d(y,z) + \epsilon$. But then clearly the path $\gamma_{xz}$ which just follows $\gamma_{xy}$ first and then $\gamma_{yz}$ is a path of length $l(\gamma_{xz}) \leq d(x,y) + d(y,z) + 2 \epsilon$. And because epsilon is free, in the $\inf$ we have $d(x,z) \leq d(x,y)+d(y,z)$.