Finding supremum and infimum of a set of rational numbers 
Let $f(n,m) = \dfrac{mn}{1+m+n} $. Put $S = \{ f(n,m) : n, m \in
 \mathbb{N} \} $. Find $\inf S $ and $\sup S$

Attempt:
First, it is clear that $f(n,m) \geq 0$ for all $n,m$. So $0$ is a lower bound. We need show that if $ f(n,m) \geq l$, then $0 \geq l$. If $l > 0$, then by the archimidean property we find and integer $N$ so that $\dfrac{1}{N} < l $ but observe that
$$ \dfrac{1}{N+2} < \dfrac{1}{N} < l $$
This does not seem to help because I can't find a $f(N,M)$ that is less than $l$. Any ideas?
As for the supremum, clearly, $\sup S = \infty$. To prove this: We can find some $f(n_0,m_0)$ greater than $f(n,m)$. For example
$$ \dfrac{n^2 }{1+2n} \geq \dfrac{mn}{1+m+n} $$
does this work?

Actually, $\inf S=\frac13=\frac{1\times1}{1+1+1}$. If $m,n\in\Bbb N$, then $m\geqslant n$ or $n\geqslant m$. If $m\geqslant n$, then$$3mn\geqslant3m=m+m+m\geqslant m+n+1$$which means that$$\frac{mn}{m+n+1}\geqslant\frac13,$$and, of course, if $n\geqslant m$, then you reach the same conclusion.
And, yes, $S$ has no upper bounds and so $\sup S=\infty$ (or it doesn't exist, depending upon the definition).
