Why does $f$ strictly increasing implies the triangular inequality for this metric? Assume $(X,d)$ is a metric space, and define a new metric $\tilde{d}$ on $X$.
Set $\tilde{d} = \frac{d(x,y)}{1+d(x,y)}$. Now with manipulation and since $d$ is a metric, I manage to show that $\tilde{d}$ satiesfies the triangular ineqality. But I read that one can conside the function $f(t) = \frac{t}{1+t}$ and show that this is a strictly increasing function (no problem). But they argue that the triangular inequality follows from $f$ is stricly increasing, why?
I guess I could pick three values $f(t_1), f(t_2), f(t_3)$ which all lies in $\mathbb{R}$ then on $\mathbb{R}$ we have $p(x,y) = |x-y|$ and thus $p(f(t_1),f(t_2)) \leq p(f(t_1), f(t_3)) + p(f(t_3), f(t_2))$ is that it?

Let $u=d(x,z),v=u=d(x,y),w=d(y,z)$. It is enough to check the condition $$u\le v+w\implies f(u)\le f(v)+f(w).$$ By increasingness we have $f(u)\le f(v+w)$. Elementary computation shows that $f(v+w)\le f(v)+f(w)$, which finishes this argument.
