Solution: We recall that two metrics $d_1$ and $d_2$ on a set $X$ are equivalent, if for any $x\in X$ and $r > 0$, there exist $s_1,s_2 >0$ such that \[ B_{d_1}(x,s_1) \subseteq B_{d_2}(x,r) \text{ and } B_{d_2}(x,s_2) \subseteq B_{d_1}(x,r). \] To prove that the metrics $d$ and $\rho $ are equivalent, we will show that for any $x\in X$ and $r>0$ there exist $s_1, s_2 >0$ such that \[ B_\rho (x,s_1) \subseteq B_d(x,r) \subseteq B_\rho (x,s_2). \] Take \[ s_1 = \frac{r}{1+r}. \] Then, for any $y \in B_\rho (x,s_1)$ \[ \rho (x,y) \lt s_1 = \frac{r}{1 + r} \implies \frac{d(x,y)}{1 + d(x,y)} \lt \frac{r}{1+r} \implies d(x,y) \lt r, \] and hence, $y\in B_d(x,r)$. This proves that $B_\rho (x,s_1) \subseteq B_d(x,r)$.

Now for the other inequality, we take \[ s_2 = \frac{r}{1 + r}. \] Then, for any $y\in B_d(x,r)$, \begin{align*} d(x,y) \lt r & \implies \frac{1}{d(x,y)} \gt \frac{1}{r} \\ & \implies 1 + \frac{1}{d(x,y)} \gt 1 + \frac{1}{r} \\ & \implies \frac{1 + d(x,y)}{d(x,y)} \gt \frac{1 + r}{r} \\ & \implies \frac{d(x,y)}{1 + d(x,y)} \lt \frac{r}{1 + r} \\ & \implies \rho (x,y) \lt s_2. \end{align*} Thus, we obtained the other inequality. Hence, the metrics $d$ and $\rho $ are equivalent.