Show that $|e^{-|x-y|}-1|$ is a metric

Show that $d(x,y)=|e^{-|x-y|}-1|$ is a metric, I already went throught the first 3 steps, but I'm having trouble with the triangular inequality, as far as i know is not possible to get $|e^{-|x-y|}-1|\leq |e^{-|x-z|}-1|+|e^{-|z-y|}-1|$ thanks for your help

Note that $e^x\le1$ for all $x\le0$.

$$\begin{aligned}&\quad |e^{-|x-z|}-1|+|e^{-|z-y|}-1| - (|e^{-|x-y|}-1|) \\ &= 1-e^{-|x-z|}+ 1 - e^{-|z-y|} - (1- e^{-|x-y|})\\ &= 1-e^{-|x-z|} - e^{-|z-y|} +e^{-|x-y|}\\ &\ge 1-e^{-|x-z|}- e^{-|z-y|} +e^{-(|x-z| + |z-y|)}\\ &=(1-e^{-|x-z|})(1 - e^{-|z-y|})\\ & \ge0 \end{aligned}$$ The first inequality holds since $|x-y|=|(x-z ) + (z-y)|\le |x-z| + |z-y|$ and the function $f(x)=e^x$ is an increasing function.