How to sort out different cases in a proof related to a metric space Problem:
Let $M = (X, d)$ be a metric space.
Show that 
$e(x, y) = min(1, d(x, y))$ is a metric.
To prove the triangle inequality for metric $e$,
three cases are considered.
Let $x, y, z \in X$.
(A) $d(x, y) \le 1$ and  $d(y, z) \le 1$
(B) $d(x, y) > 1$
(C) $d(y, z) > 1$
Why the above three cases are exhaustive?
Is there a systematic method to find the different cases? 

This is specific to this exercise: the case (B) $\ d(x,y) >1$ [or (C) $\ d(y, z) >1$] means $e(x, y) =1$ [or $e(y, z) =1$], and if neither holds, that just means $d(x, y)\le1$ and $d(y, z)\le1$, i.e. case (A), when $e(x, y) =d(x, y)$ and $e(y, z) =d(y, z) $. 
So these 3 scenarios are indeed exhaustive.
Nevertheless, to provide full details, one also has to consider subcases $d(x, z) \le1$ and $d(x, z) >1 $. 
