Show that two metrics known not to be strongly equivalent actually induce the same topology. Suppose on $\mathbb{R}$, we have the usual Euclidean metric, $\rho_{1}(x,y) = \Vert x-y \Vert$, and also the metric $\rho_{2}(x,y) = \displaystyle \frac{\rho_{1}(x,y)}{1+\rho_{1}(x,y)}$. 
I need to show that the topologies that $\rho_{1}$ and $\rho_{2}$ respectively induce are equal to each other (i.e., if $\rho_{1}$ induces $\tau_{1}$ and $\rho_{2}$ induces $\tau_{2}$, then $\tau_{1} = \tau_{2}$).
I understand that $\rho_{1}$ and $\rho_{2}$ are not strongly equivalent metrics on $\mathbb{R}$, so I am looking to use them as an example of two metrics that induce the same topology but are not strongly equivalent.
The only problem is I don't know how to show that $\rho_{1}$ and $\rho_{2}$ induce the same topology...Could someone please let me know how to do this?  Thanks. I would prefer something very elementary rather than something very technical.
I am putting a bounty on this question for a fully worked solution.
 A: In fact something more general is true: if $\langle X,\rho\rangle$ is any metric space, and we define 
$$\rho_1:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\frac{\rho(x,y)}{1+\rho(x,y)}\;,$$
then $\rho_1$ is a metric on $X$ and is equivalent to $\rho$ (i.e., generates the same topology).
HINT: To show the equivalence, it suffices to show that for each $\epsilon>0$ and $x\in X$,


*

*if $y\in B_\rho(x,\epsilon)$, then there is $\delta>0$ such that $B_{\rho_1}(x,\delta)\subseteq B_\rho(x,\epsilon)$, and  

*if $y\in B_{\rho_1}(x,\epsilon)$, then there is $\delta>0$ such that $B_\rho(x,\delta)\subseteq B_{\rho_1}(x,\epsilon)$.


If $\tau$ and $\tau_1$ are the topologies generated by $\rho$ and $\rho_1$, respectively, the first of these points shows that $\tau\subseteq\tau_1$, and the second shows that $\tau_1\subseteq\tau$.
For the first, you can begin by using the triangle inequality to show that if $\alpha=\epsilon-\rho(x,y)$, then $B_\rho(y,\alpha)\subseteq B_\rho(x,\epsilon)$. Then you just need to find a $\delta>0$ such that $B_{\rho_1}(y,\delta)\subseteq B_\rho(y,\alpha)$. The second point can be approached similarly.
It may be helpful to notice that if
$$b=\frac{a}{1+a}\;,$$
then
$$a=\frac{b}{1-b}\;.$$
A: Theorem:
Let $(X,\rho)$ be a metric space and consider the function $d:X\times X\to \mathbb{R}$, defined by $d(x,y):=\rho(x,y)/(1+\rho(x,y))$. Then $d$ is a metric and induces on $X$ the same topology than $\rho$. 
Proof: Consider the function $f:[0,\infty)\to\mathbb{R}$, $f(x)=x/(1+x)$. Then $d(x,y)=f(\rho(x,y))$. Notice that $f$ is an increasing and continuous function (you can verify this by examining the first derivative of $f$). Also notice that $f$ is a subadditive function. In fact, if $a,b\geq0$ then
$$f(a+b)\leq f(ab+a+b)=\frac{ab+a+b}{1+ab+a+b}\leq\frac{2ab+a+b}{1+ab+a+b}=f(a)+f(b),$$
where the first inequality is due to the fact that $f$ is increasing.
Claim 1: $d$ is a metric on $X$.
It is enough to show that $d$ satisfies the triangle inequality. 
Since $f$ is increasing and subadditive we have that:
$$d(x,y)=f(\rho(x,y))\leq f(\rho(x,z)+\rho(z,y))\leq f(\rho(x,z))+f(\rho(z,y))=d(x,z)+d(z,y).$$
Claim 2: For every $x\in X$ and every $\varepsilon>0$ there exists $\delta>0$ such that $B_\rho(x,\delta)\subset B_d(x,\varepsilon)$.
Since $f$ is continuos at $0$, for any $\varepsilon>0$ given, there exist $\delta>0$ such that: $$0\leq a<\delta\Longrightarrow f(a)<\varepsilon.$$
Thus, for any $x,y\in X$: $$\rho(x,y)<\delta\Longrightarrow d(x,y)<\varepsilon.$$
Claim 3: For any $a,b\geq0$: $f(b)<\frac{f(a)}{2}\Longrightarrow b<\frac{a}{2}$.
It is enough to prove its (equivalent) contrapositive:
for any $a,b\geq0$: $a\leq2b\Longrightarrow f(a)\leq2 f(b)$.
Let $a,b\geq0$ such that $a\leq2b$. Since $f$ is increasing and subadditive, we have that:
$$f(a)\leq f(b+b)\leq f(b)+f(b).$$
Claim 4: For every $x\in X$ and every $\varepsilon>0$ there exists $\delta>0$ such that $B_d(x,\delta)\subset B_\rho(x,\varepsilon)$.
By claim 3 we have that:
$$d(x,y)=f(\rho(x,y))<\frac{f(\varepsilon)}{2}\Longrightarrow \rho(x,y)<\frac{\varepsilon}{2}<\varepsilon.$$
Then for any given $x\in X$ and  $\varepsilon>0$, choose $\delta=\frac{f(\varepsilon)}{2}$ to satisfy the claim.
Thus we have proved that $d$ and $\rho$ induce  the same topology on $X$. $\blacksquare$

Bonus Track: Actually the same proof works to prove this more general result:
Let $(X,\rho)$ be a metric space and consider the function $d:X\times X\to \mathbb{R}$, defined by $d(x,y):=f(\rho(x,y))$, where $f:[0,\infty)\to\mathbb{R}$ is an increasing, continuous and subadditive function such that $f^{-1}(\{0\})=\{0\}$. Then $d$ is a metric and induces the same topology than $\rho$. 

Remark: Notice that $\rho/(1+\rho)$ is a bounded metric. If $\rho$ is not a bounded metric then it cannot be strongly equivalent to $\rho/(1+\rho)$.
