$\inf$ and $\sup$ of $K = \bigg\{ \frac{m}{n}+\frac{4n}{m} : m,n \in \mathbb N\bigg\}$ Exercise :

Consider the set :
  $$K = \bigg\{ \frac{m}{n}+\frac{4n}{m} : m,n \in \mathbb N\bigg\}$$
  Find $\inf K$ and $\sup K$.

Attempt :
$$\frac{m}{n}+\frac{4n}{m} = \frac{m^2 + 4n^2}{mn}$$
Now, it is :
$$\frac{m^2 + 4n^2}{mn} \geq 4 \Leftrightarrow (2n-m)^2 \geq 0$$
which holds $\forall \; n,m \in \mathbb N$. Also, it is :
$$\frac{m^2 + 4n^2}{mn} = 4 \Leftrightarrow m=2n$$
Thus $\exists \; n,m \in \mathbb N : \frac{m^2 + 4n^2}{mn} = 4$, thus $\inf K = 4$.
How should I proceed on finding $\sup K$ now ?
 A: Note that 
$$
K = \bigg\{ \frac{m}{n}+4\frac{1}{\frac{m}{n}} : m,n \in \mathbb N\bigg\}
$$
The rational numbers are dense in $\mathbb{R}\cup\{-\infty,+\infty\}$. Then the $\inf K$ and $\sup K$ coincide with the $\min f(x)$ and $\max f(x)$ for 
$$
f(x)=x+4\cdot \frac{1}{x}
$$
A: A note: if they gave you $$ r + \frac{3}{r}  $$
with rational $r > 0,$ there would still be an infimum by switching to $$ x + \frac{3}{x}$$ for positive real $x,$ taking the derivative $1 - \frac{3}{x^2},$ then plugging back $x = \sqrt 3$ giving $2 \sqrt 3.$
However, it is not possible to achieve this with rational $r,$ so there is an infimum that is not attained.  Officially, no minimum.
The original problem works because $4$ is a square
A: One can actually show that a supremum does not exists: if $\sup K$ does exists, call this $L$. Then the Archimedian property tells us that $L < m$ for some $m\in \mathbb N$. Then $L < m + \frac{4}{m}$. But $m + \frac{4}{m}$ is in $K$ (by choosing $n=1$). But this is impossible as $L$ is an upper bound of $K$. 
Thus $\sup K$ does not exists. 
