We have the following set: $\mathcal{A} = \{ \frac{m}{n} + \frac{4n}{m};\ \ m, n \in \mathbb{N} \} $
Attempting to prove that the set's minimum is 4 yields: $$\frac{m}{n}+\frac{4n}{m} = \frac{m^2 + 4 n ^2}{mn} \geq 4$$ $$m^2 + 4n^2 \geq 4mn$$ $$m^2 + 4n^2 - 4mn \geq 0$$ $$(2n-m)^2 \geq 0$$
I do not know how to proceed beyond this point, although I suspect induction may be required. Thank you for your time.
Edit: how can it be shown that the set has no upper-bound?