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?

Let $x,y,z\in X$; then you have two cases:


*

*if $d(x,z)\le 1$, then $d'(x,z)=d(x,z)\le d(x,y)+d(y,z)\le d'(x,y)+d'(y,z)$

*if $d(x,z)>1$, then $d'(x,z)=1$; since $1<d(x,z)\le d(x,y)+d(y,z)$, we must have either that one among $d(x,y)$ and $d(y,z)$ is at least $1$ (and we're done), or they are both less than $1$ and so $d'(x,y)+d'(y,z)=d(x,y)+d(y,z)\ge d(x,z)>d'(x,y)$.
