Is this function a metric? Let $X$,$d$ be a metric space.
Define $d'$ as the minimum of $1$ and $d$:
$$ d':\ X^2 \rightarrow \mathbb{R}:\ d'(x,y) = \min\{1,d(x,y)\} $$
The question is whether $d'$ is a metric.
I've managed to prove that the first two properties of a metric hold for this function, but I can't seem to figure out how to prove the triangle inequality. Any hints?

By definition $d(x,y) \ge 0 $ and $d(x,y)+d(y,z) \ge d(x,z) $
giving:
$$d'(x,y)+d'(y,z)$$ $$ =  \min\{1,d(x,y)\} +  \min\{1,d(y,z)\} $$ $$= \min\{1+1,1+d(x,y),d(y,z)+1,d(x,y)+d(y,z)\}$$ $$ \ge \min\{1,d(x,y)+d(y,z)\} $$ $$\ge  \min\{1,d(x,z)\}$$ $$ = d'(x,z)$$
