$S:=\{3\ln n-\ln(m^2+2m+4n^3): m,n\in\mathbb N\}\;\sup S,\inf S=?$ 
If exist, find $\sup S\;\&\;\inf S$
$$S:=\{3\ln n-\ln(m^2+2m+4n^3): m,n\in\mathbb N\}$$

My attempt:
$\inf (-S)=\sup S$
$$S\Bigg\{-\ln{\frac{m^2+2m+4n^3}{n^3}}:m,n\in\mathbb N\Bigg\}$$
I didn't manage to either separate $S$ as a combination of two possibly disjoint sets nor use arithmetic means because I thought the term under the logarithm should be symmetric, so I examined some concrete cases:
$n=1$: $$\lim_{m\to \infty}-\ln{(m^2+2m+4)}=-\infty$$
$m=1$: $$\lim_{n\to\infty}=-\ln4$$
$n,m=1$ $$s\in S=-7$$
However, I can't prove if any of these are $\sup S,\inf S.$
What would be a better approach?
 A: So
\begin{align}
3\ln n-\ln\left(m^2+2m+4n^3\right)&=-\ln\left(\frac{m^2+2m+4n^3}{n^3}\right)\\
&=-\ln\left(\frac{m^2+2m}{n^3}+4 \right).
\end{align}
The function $\ln$ is monotone, so it suffices to just consider $\frac{m^2+2m}{n^3}+4$. Its easy to see that there is no maximum by taking $n=1$ and $m$ to be arbitrarily large. Thus the set $S$ is unbounded below. On the other hand, since $m,n\in\Bbb N$, the number $\frac{m^2+2m}{n^3}$ is always positive. But taking $m=1$ and $n$ to be arbitrarily large, this term tends to $0$. Thus the infimum of $\frac{m^2+2m}{n^3}+4$ is $4$, and the supremum of $S$ is thus $-\ln 4$.
A: Using that the logarithm is a monotonous function, and that $3\ln n - \ln(m^2+2m+4n^3) = \ln(\frac{n^3}{m^2+2m+4n^3})$ we just need to find the $\sup$ and $\inf$ from that fraction.
Now, for $n=1$ and $m\rightarrow \infty$ the function diverges to $-\infty$, so there is no $\inf$.
For the $\sup$, we want to maximize the fraction $f(n,m)=\frac{n^3}{m^2+2m+2n^3}$, this implies that we should minimize $m$, because it just makes our denominator higher, we can prove this way that $f(n,m) \leq f(n,1)$. We set $m=1$ as we want to maximize this value of $f(n,m)$. Now, we have the function:
$$f(n,1) = \frac{n^3}{3+4n^3} = \frac{1}{\frac{3}{n^3} + 4}$$
As $\frac{3}{n^3} + 4$ is monotonous and decreasing sequence, we minimize the denominator with higher $n$. So we take the limit and we obtain $\sup(S) = \ln(\frac{1}{4}) = -\ln(4)$, which was what you thought it would be. 
