We denote by $d(.,.)$ a metric with the common properties.
The Four Point Condition for vertices $x,y,z,t$ in a tree is known as $$d(x,y)+d(z,t) \leq \max \left\lbrace d(x,z)+d(y,t), d(x,t)+d(y,z) \right\rbrace ~~(1)$$ (for the sake of this question, you could regard the vertices simply as some points in a metric space)
Now, I've read several times (e.g. in this question Four-point condition for tree distances: is there a detropicalization proof?) that the statement $(1)$ is equivalent to saying that
(2): the two larger sums of $d(x,y)+d(z,t)$, $d(x,z)+d(y,t)$ and $d(x,t)+d(y,z)$ are equal and not less than the third.
I wanted to proof this and the direction $(2) \Rightarrow (1)$ was fairly simply; however, I am having some difficulties showing the other direction. This is what I have so far:
By (1) we have $d(x,y)+d(z,t) \leq d(x,z)+d(y,t)$ and $d(x,y)+d(z,t) \leq d(x,t)+d(y,z)$. Now, I need to show that $d(x,z)+d(y,t)= d(x,t)+d(y,z)$. For this, assume that $d(x,z)+d(y,t) \neq d(x,t)+d(y,z)$, in particular $d(x,z)+d(y,t) < d(x,t)+d(y,z)$. Then, by using the triangle inequality and symmetry, we get
$$d(x,z)+d(y,t) < d(x,t)+d(y,z) \\ \Leftrightarrow d(x,z)+d(y,t) < d(x,y) + d(y,t)+d(y,z) \\ \Leftrightarrow d(x,z)+d(y,t) < d(x,y) + d(y,t)+d(y,t) + d(t,z) \\ \Leftrightarrow d(x,z)+d(y,t) < d(x,y) + d(y,t)+d(y,t) + d(t,z) \\ \Leftrightarrow d(x,z)+d(y,t) < d(x,y) + d(z,t)+2d(y,t) $$
My aim was to somehow show that I would obtain $d(x,z)+d(y,t) < d(x,y) + d(z,t)$ which would be wrong due to the Four Point Condition.
Any ideas or suggestion on how to proof this? Thank you!
