Find supremum and infimumm of a set with two variables $$A=  \left\{\frac{m}{n}+\frac{4n}{m}:m,n\in\mathbb{N}\right\}$$

Since $m,n\in \mathbb{N}$, infimum is zero because $m,n$ both are increasing to infinity. Then the supremum is $5$ when $m,n$ are equal to $1$. 
But I don't think my approach is right. Can someone give a hit or suggestion to get the right answer? Thanks. 
 A: Your answer isn't good. We see that $m/n+n/m=\frac{m^2+n^2}{mn}\geq 2$ by $MA\geq MG$. So $m/n+4n/m=m/n+n/m+3n/m\geq 2+3n/m$. Take $m=1$ the expression is ilimited. $\sup(A)=\infty$. Since $E=m/n+4n/m=\frac{m^2+4n^2}{nm}\geq 4$ by $MA\geq MG$ we have the infimumm is 4 and holds always that m=2n.
A: Here I assume $\mathbb{N} = \mathbb{Z}_{>0}$. If that's the case, then $\frac{m}{n}\in\mathbb{Q}_{>0}\subseteq\mathbb{R}_{>0}$ $\forall m,n\in\mathbb{N}$. So $A\subseteq B:=\{x + \frac{4}{x}|x\in\mathbb{R}_{>0}\}$ and so $\inf(B)\leq\inf(A)$ (this is a nice property of infimum we can make use of). The function $f(x) = x + \frac{4}{x}$ is differentiable on $\mathbb{R}_{>0}$, so we can consider $f^{\prime}(x) = 1 - \frac{4}{x^2}$ $\forall x\in\mathbb{R}_{>0}$. Notice that $f^{\prime}(x) = 0$ $\Leftrightarrow$ $1 = \frac{4}{x^2}$ and the only $x\in\mathbb{R}_{>0}$ for which this is possible is $x = 2$. Thus $f(2) = 2 + \frac{4}{2} = 4$ is the smallest value of $f$ on $\mathbb{R}_{>0}$. Since $2 = \frac{m}{n}$ for $m = 2$, $n = 1$, we have that $4\in A$, and thus that $\inf(A) \leq 4$. Combining everything, we have $4 = \inf(B)\leq\inf(A)\leq 4$, and so we can conclude $\inf(A) = 4$. 
For the supremum, we can take $m = 1$, and let $n \rightarrow \infty$ to see that $\sup(A) = \infty$.
