Finding bounds of the set: $ A = \{ \frac{m\cdot n}{m+n}: m,n \in \mathbb N \} $ I'm having some problems with this. 
I know that lower bound will be $\dfrac{1}{2}$. Should I just find $m,n$ for which  $\dfrac{m\cdot n}{m+n} \lt \dfrac{1}{2} + \epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.
Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.
Thank you for any hints.
 A: The number $\dfrac12$ is a lower bound because, for each $m,n\in\mathbb N$, you have$$\frac{mn}{m+n}\geqslant\frac12.$$That's so because\begin{align}\frac{mn}{m+n}\geqslant\frac12&\iff2mn\geqslant m+n\\&\iff2mn-m-n\geqslant0\\&\iff mn+mn-m-n+1\geqslant1\\&\iff mn+(m-1)(n-1)\geqslant1,\end{align}which is true. On the other hand $\dfrac{1\times1}{1+1}=\dfrac12$ and therefore $\dfrac12$ is the greatest lower bound.
On the other hand, $\lim_{n\to\infty}\dfrac{n^2}{2n}=\infty$, and therefore your set has no upper bound.
A: Consider $(m-\frac{1}{2})(n-\frac{1}{2}) \ge \frac{1}{4}$ if $m,n\ge1$.
A: Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.
Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $\dfrac {n}{n+1}$. Now, how does $\dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $\dfrac  12, \dfrac 23, \dfrac 34, ...$ Can you prove it using induction?
Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?
If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $\dfrac 12$.
