Kreyszig 1.6.7: If $(X,d)$ is complete, show that $(X,\tilde{d})$, where $\tilde{d}=d/(1+d)$, is complete. The problem in Kreyzig states:

If $(X,d)$ is complete, show that $(X,\tilde{d})$, where $\tilde{d}=d/(1+d)$, is complete.

My attempt to solve it:
1) $(X,d)$ is complete, so any Cauchy sequence $\{x\}_{n=0}^\infty$ converges and $\forall \epsilon > 0,\; \exists N\in \mathbb{N}$ such that $\forall n,\forall m > N,\; m,n\in \mathbb{N}$ we have $d(x_n,x_m)<\epsilon$.
2) By the property of a metric $d(x,y)\geq0, \; \forall x,y \in X$.
3) $\tilde{d}(x_n,x_m) = d(x_n,x_m)/(1+d(x_n,x_m)) \leq d(x_n,x_m)<\epsilon$ so the arbitrary Cauchy sequence in $X$ converges with the metric $\tilde{d}$.
Is the proof correct or there are some issues with it?
Is there any other way to prove it?
Thank you for your help in advance.
 A: You want to show that: if $\{x_n\}$ is a Cauchy sequence in $(X, \tilde d)$, then $x_n \to x$ (in the metric $\tilde d$) for some $x\in X$. 
So choose such a sequence $\{x_n\}$. Then for all $\epsilon >0$ and $\epsilon<1/2$, there is $N$ so that 
$$ \tilde d (x_n, x_m) <\epsilon$$
for all $n, m\ge N$. Then 
$$ \frac{d(x_n, x_m)}{1 + d(x_n, x_m)} <\epsilon.$$
Moving terms around, we have 
$$ d(x_n, x_m) < \frac{\epsilon}{1-\epsilon}< 2\epsilon.$$
(we used $\epsilon<1/2$ in the last inequality). Thus $\{x_n\}$ is also a Cauchy sequence in $(X, d)$ and so $x_n\to x $ in the metric $d$. That is, for all $\epsilon >0$, there is $N$ so that 
$$ d(x_n, x)<\epsilon$$
for all $n\ge N$. But since 
$$ \tilde d(x_n , x) \le d(x_n, x) <\epsilon,$$
this shows that $\{x_n\}$ also converge to $x$ in the metric $\tilde d$. 
A: From hypothesis the space $(X,d)$ is complete with respect to $d$,but you have to prove that $(X,\bar{d})$ is complete with respect to $\bar{d}$
Lets take a Cauchy sequence $x_n \in (X,\bar{d})$ and  prove that $x_n$ is Cauchy with respect to $d$.
We have that exists $n_0 \in \Bbb{N}$ such that $\frac{d(x_n,x_m)}{1+d(x_n,x_m)}<\epsilon,\forall n,m \geq n_0$
We have that $d(x_n,x_m) \leq (1 +d(x_n,x_m))\epsilon \Rightarrow d(x_n,x_m) \leq \frac{\epsilon}{1-\epsilon},\forall n,m \geq n_0$
Thus $\limsup_{\epsilon \to 0}d(x_n,x_m) \leq 0 $
Using this you can prove that $\lim_{m,n} d(x_n,x_m)=0$
Now you have to prove that $x_n$ converges with respect to  the new metric.
From the fact than  $(X,d)$ is complete exists $x \in X$ such that $d(x_n,x) \to 0$.
But $0 \leq \bar{d}(x_n,x) \leq d(x_n,x) \to 0$
Thus $x_n \to^{\bar{d}} x$ , so  $(X,\bar{d})$ is complete.
A: Your third line is a bit off. You shouldn't have $\tilde{d}(x_n,y_m)$. You want to look at elements of the same sequence: so, $\tilde{d}(x_n,x_m)$. The other thing you have that is suspect is $\tilde{d}(x_n,y_m)$ (sic) $\le d(x_n,y_n)$. The proper conclusion would be this: $\tilde{d}(x_n,x_m) \le d(x_n, x_m)$. 
Now since $(X,d)$ is complete, we have $x_n \to x$ for some $x\in X$ (w.r.t the $d$ metric). But by what you've shown, this implies $\tilde{d}(x_n,x) \to 0$.
