Proving that $\rho(x,y) = \frac{d(x,y)}{1+d(x,y)}$ is a metric and that $\rho(x,y)$ and $d(x,y)$ are equivalent metrics As you can see the proof is divided into 2, the first part consists on proving that $\rho(x,y)$ is a metric
My attempt
i) $\rho(x,y) \geq 0$, which is clear since $d(x,y)$ is a metric, and it is $0$ if and only if $x=y$
ii) Symmetry, which also comes from the fact that $d(x,y)$ is a metric
iii) The triangular inequality, for which I have already proven that $f(x)=\frac{x}{1+x}$ is non decreasing, however I am stuck proving f(x) is concave. A fact that I need in order to show that $f(d(x,y)) \leq f(d(x,z)) + f(d(y,z))$ which I need to show the triangular inequality. I know how to prove a function is concave, I am only getting a little bit stuck on the algebra. 
For the second part, I know that the $\rho(x,y) , d(x,y)$ metrics are equivalent if $\exists c_1, c_2 \in \mathbb{R}$ such that $c_1d(x,y)\leq \rho(x,y) \leq c_2d(x,y)$ 
I have proven that $\rho(x,y) \leq d(x,y) $ since $\frac{1}{1 + d(x,y)} \leq 1$ Hence, $c_2 = 1$. However I can't get the other inequality.

For any sequence $x_n$, $x_n\to 0^+ \iff \frac{x_n}{1+x_n} \to 0^+.$ Now, for any sequnece $x_n$ in the corresponding metric consider the sequence of real numbers $d(x_n,x),$ which gives us the equivalence in convergence between two metrices. This implies the equivalence between them. 
