Simple Metric Space Question Let $(E,d)$ be a metric space and assume that the metric d satisfies $d(x,z) \leq \max(d(x,y),d(y,z))$ for all $x,y,z\in E$. 
Prove that if $d(x,y) \neq d(y,z)$ then $d(x,z)=\max(d(x,y),d(y,z))$.
I have spent literally hours on this and I feel like a moron because I can't figure it out. Please help! Thanks!
 A: Let $a=d(y,z)$, $b=d(x,z)$, $c=d(x,y)$.
Without loss of generality we may assume that $a\gt c$.
We have $a\le \max(b,c)$. Since $a\gt c$, we must have $a\le b$.
We have $b\le \max(a,c)$, so $b\le a$. 
It follows that $b=a$. 
A: We already know "$\leq$", so let's assume "$<$" and hope to reach a contradiction.
Without loss of generality, let's assume $d(x, y) < d(y, z)$, so that our new assumption is that $d(x, z) < d(y, z)$.
But we have $d(y, z) \leq max(d(y, x), d(x, z)) = max(d(x, y), d(x,z))$, a contradiction to (one of) the above two inequalities.
Further reading: The inequality that your metric satisfies is called the "ultrametric inequality" or the "non-archimedean triangle inequality". A metric space with a metric that satisfies the ultrametric inequality is called an ultrametric space. You can read more about them here: http://en.wikipedia.org/wiki/Ultrametric_space
A: Let $d(x,y)\leq d(y,z)\leq d(x,z)$ (WLOG)
Then, $d(x,z)\leq max(d(x,y),d(y,z))=d(y,z)\implies d(x,z)\leq d(y,z)$
Now, $d(y,z)\leq max(d(y,x),d(x,z))=d(x,z)\implies d(y,z)\leq d(x,z)$
Both of these $\implies d(x,z)=d(y,z)\implies d(x,z)= max(d(x,y),d(y,z))$
