Let $\rho(x,y)= \frac{d(x,y)}{1+d(x,y)}$, show that $\rho$ is a metric equivalent to $d$ 
Let $(E,d)$ be a metric space and $\rho(x,y)= \frac{d(x,y)}{1+d(x,y)}$. Show that $(E,\rho)$ is a metric and $\rho$ and $d$ are equivalent.

I have proved the part about the metric space. However, I cant remember how to prove the equivalence. Also I'm stuck with a follow up which says:

Show that there are no constants $c,C >0$ such that for all $x,y \in E:$ $$cd(x,y) \leq \rho(x,y) \leq Cd(x,y)$$ 

 A: Hint: For the triangle inequality, use the following: if $0 \leq s \leq t$, then
\begin{align}
s+ts \leq t+ts.
\end{align}
Edit: Here's the detail. From the above inequality, it follows
\begin{align}
\frac{s}{1+s} \leq \frac{t}{1+t}
\end{align}
i.e. the function $x/(1+x)$ is monotone increasing. In particular, it follows
\begin{align}
\frac{d(x, y)}{1+d(x, y)} \leq \frac{d(x, z)+d(z, y)}{1+d(x, z)+d(z, y)}
\end{align}
since $d(x, y) \leq d(x, z)+d(z, y)$. 
A: T.P. that metric are equivalent
Observe that $B_{\rho}(a,r)=\lbrace x\in E: \rho(a,x)<r\rbrace$=$\lbrace x\in E: \dfrac{d(a,x)}{1+d(a,x)}<r\rbrace$=$\lbrace x\in E: d(a,x)(1-r)<r\rbrace$=$\lbrace x\in E: d(a,x)<\dfrac{r}{1-r}\rbrace$=$B_d(a,\dfrac{r}{1-r})$ let $r_1=\dfrac{r}{1-r}$. Then $B_{\rho}(a,r)\subseteq B_d(a,r_1)$ (you can even start with a ball in d with arbitrary radius $r_1$ and use the above to get the radius r such that $B_d(a,r_1)\subseteq B_{\rho}(a,r)$) And two metric are said to be equivalent if collection of open sets w.r.t the two matrices is same. So its enough T.P. that open ball w.r.t d is an open set in $\rho$ and vice versa. Refer https://en.wikipedia.org/wiki/Equivalence_of_metrics for the details.
Now the relation  cd(x,y)≤ρ(x,y)≤Cd(x,y) is actually strong equivalance condition.
