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.

In my functional analysis class, we were allowed to prove metric equivalence as follows. Plot the function $f(x)=\dfrac{1}{1+x}$ to obtain a graph that is decreasing from $y=1$ to $y=0.$ Then note that inside any open ball on the $x$-axis, you can pull back those values to the $y$ axis and fit an open ball inside the result. Conversely, once you have an open ball on the $y$ axis, you can trace those back down to the $x$ axis and fit an open ball inside it. Therefore, the two metrics are equivalent.
