The question is let (X,d) be a metric space. Show that $ d' = \frac{d}{1+d} $
I would like to know if I am on the right path with my solution. Any help would be greatly appreciated!
Solution:
Let $x,y,z \in X $
Then $ d'(x,z) = \frac{d(x,z)} {(1+d(x,z))}\le \frac{d(x,y) + d(y,z)}{ 1+d(x,y)+d(y,z)}=\frac{d(x,y)}{1+d(x,y) + d(y,z)}+\frac{d(y,z)}{1+d(x,y) + d(y,z)}\le \frac{d(x,y)}{1+d(x,y)}+ \frac{d(y,z)}{1+d(y,z)}$ $$= d'(x,y) + d'(y,z).$$ Thus we have proved that $d'=\frac{d}{1+d}$ is also a metric on $X$.
You haven't really justified why the first inequality works though.
I would start by considering the function $F(t)=\frac{t}{1+t}$, which has positive derivative on $[0,\infty)$.
This means $F(d(x,z)) \le F(d(x,y)+d(y,z))$, since $d(x,y) \le d(x,y)+d(y,z)$.
Then carry on, as you did, and you get the correct answer.
Maybe add a comment on why separation and symmetry follow very easily.