Distance triples in metric space

Question

Given is a metric space $(X,d)$ over a finite point-set $X$ with distance function $d$.

Let $x \in X$ be an arbitrary point and $y \in X$ a point of maximum distance from $x$.

Given three arbitrary points $a,b,c$ of $X$, I would like to show that $$\max \{d(a,y), d(b,y), d(c,y)\} \geq \min \{d(a,b), d(a,c), d(b,c)\}. $$

While intuitively the claim looks clear to me - if all $d(a,b), d(a,c), d(b,c)$ would be greater than the furthest distance from $y$ to $a,b,c$, then one of $a,b,c$ would be furthest away from $x$, so $y$ would not be maximal - I do not see the appropriate tools to prove it formally. I assume that using only the triangle inequality should be enough, but I would appreciate any hints.

EDIT: For the setting above, there exists a counterexample as shown below. But what if the metric space if path-connected?

Answer 1

For a counterexample, define the following metric on the set ${a,b,c,x,y}$. First, $d(a,y)=d(b,y)=d(c,y)=1$ and $d(a,b)=d(b,c)=d(a,c)=2$; this is just the path metric on a triod such as the letter Y. Next, define the distances from $x$ to each of $a,b,c,y$ to be $1000000$.

Added to address the edited version: To make this example path connected, start with the triod, then define the cone on that triod using cone point $x$. The cone from $x$ to each segment of the triod is given the metric of a Euclidean isosceles triangle, with base length $1$ and two side lengths $1000000$. So the space is basically three such isosceles triangles, with a long side of each such triangle identified to a single arc of length $1000000$. The metric on the space as a whole is given by the length of the shortest path between two points. The distances $d(a,b)$, $d(a,c)$, $d(b,c)$ are no longer equal to $2$ on the nose, but they are only very, very, very slightly smaller than $2$.