Question about proving a sequence is Cauchy 
If $a_n$ is a Cauchy sequence, show $\dfrac{3a_n^3}{1+a_n^2}$ is also Cauchy. 

So far I have:
$$\left|\frac{3a_m^3}{1+a_m^2} - \frac{3a_n^3}{1+a_n^2}\right| = \left|\frac{3a_m^3 + 3a_m^3a_n^2-3a_n^3-3a_n^3a_m^2}{(1+a_m^2)(1+a_n^2)}\right|$$
I'm not really sure how to proceed from here.
Thanks in advance for the help.
 A: Hint. Let $f(x)=3x^3/(1+x^2)$ then
$$f(x)-f(y)=3\cdot \frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\cdot (x-y).$$
Show that
$$0\leq \frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\leq 2,$$
that is
$$0\leq x^2y^2+x^2+xy+y^2\leq 2x^2y^2+2x^2+2y^2+2.$$
Hence $|f(a_n)-f(a_m)|\leq 6 |a_n-a_m|$.
P.S. More generally if $(a_n)_n$ is a Cauchy sequence then it is bounded: there is $M>0$ such that $|a_n|\leq M$ for all $n\in\mathbb{N}$. Moreover 
if $f$ is a $C^1(\mathbb{R})$ function then $f'$ is bounded over the compact set $[-M,M]$. Hence, by the Mean Value Theorem, if $a_n\not= a_m$, then there exists $t\in [-M,M]$ (which depends on $a_n$ and $a_m$) such that $f(a_n)-f(a_m)=f'(t)(a_n-a_m).$
Hence 
$$|f(a_n)-f(a_m)|\leq C |a_n-a_m|$$ 
where $C:=\max_{x\in [-M,M]}|f'(x)|$.
A: If $\{a_n\}$ is Cauchy, then it converges. Say $a_n\to a\in\mathbb R$. Then clearly,
$$
\frac{3a_n^3}{a_n^2+1}\to\frac{3a^3}{a^2+1}
$$
and thus, since $\,\big\{\frac{3a_n^3}{a_n^2+1}\big\},\,$ is a convergent sequence, it is a Cauchy sequence, as well. 
