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?
Best Answer
I'm going to assume, as much as it pains me, that $0\not\in\mathbb{N}$. Otherwise, $f(0,0)=0\le f(n,m)$ for all $n,m$, and we're done.
Note that $f(1,1)=\frac{1}{3}$, so $\frac{1}{3}\ge \inf S$.
Given $n,m\in\mathbb{N}$, $$ f(n,m)=\frac{1}{\frac{1}{nm}+\frac{1}{n}+\frac{1}{m}}\ge \frac{1}{1+1+1}=\frac{1}{3} $$