Show that $d_2$ defined by $d_2(x,y)=\frac{|x-y|}{1+|x-y|}$ is a metric
Solution 1:
$d_2 (x,y)+d_2 (y,z)=\frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)}$
$ \geq \frac{d(x,y)}{1+d(x,y)+d(y,z)}+ \frac{d(y,z)}{1+d(x,y)+d(y,z)} = \frac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}$
$= 1-\frac{1}{1+d(x,y)+d(y,z)} \geq 1-\frac{1}{1+d(x,z)}$
$= \frac{d(x,z)}{1+d(x,z)}=d_2 (x,z)$