When defining the Kobayashi metric on a connected complex analytic space $X$, one makes the following auxiliary definition:

A holomorphic chain from $x\in X$ to $y\in X$ is a finite sequence of holomorphic maps $f_1,\ldots ,f_n\colon\Delta\to X$ (where $\Delta$ is the unit disk in $\mathbb{C}$) together with points $z_1,\ldots ,z_n,w_1,\ldots ,w_n\in\Delta$ such that $f_1(z_1)=x$, $f_i(z_i)=f_{i+1}(w_{i+1})$ for $1\le i< n$ and $f(w_n)=y$.

The length of a holomorphic chain is, with this notation, is $\sum_{i=1}^nd(z_i,w_i)$ (Poincaré metric on $\Delta$).

Finally the Kobayashi pseudo-distance on $X$ is obtained by setting $d(x,y)=$ infimum of lengths of all holomorphic chains from $x$ to $y$.

This "pseudo-distance" is obviously symmetric, satisfies $d(x,x)=0$ and the triangle inequality. The space is called Kobayashi hyperbolic if $d$ is in addition non-degenerate, i.e. if $d$ is a metric.

Now one could as well begin with the following much simpler construction:

Consider the function $\delta :X\times X\to [0,\infty ]$ with $\delta (x,y)=\inf d(z,w)$, the infimum running over all triples $(f,z,w)$ with $f:\Delta\to X$ holomorphic, $z,w\in\Delta$ and $f(z)=x$, $f(w)=y$.

This is still symmetric and satisfies $\delta (x,x)=0$, but now it is unclear whether

a) $\delta (x,y)$ is finite, i.e. the set of triples $(f,z,w)$ is non-empty;

b) $\delta$ satisfies the triangle inequality.

Clearly, a) and b) together are equivalent to $d=\delta$, and $d$ can be obtained from $\delta$ by an easy construction. Finally, my questions:

Under which circumstances is $d=\delta$?

Is there a simple example where $d\neq\delta$?

What is the logical relation between a) and b)?