Prove that this set $A=\{ \frac{m}{n} + \frac {4n}{m} : m,n \in N\}$ unbounded Prove that this set $A=\{ \frac{m}{n} + \frac {4n}{m} : m,n \in N\}$ unbounded.
Could anyone show me how to prove it please?
my only argument for the proof is that if you plug m=1 in the above expression you will see that it is unbounded for $n \in N$
 A: You can prove that a set is unbounded by proving that it has the following property:

For every $M\in\mathbb R$, there exists some $x\in A$ so that $x>M$.

The proof of a "for every $M\in\mathbb R$" statement usually starts with the words "let $M\in\mathbb R$", so we should start with that. Here's the template, now try to fill in the details:

Let $M\in\mathbb R$.
something should go in here
Therefore, $x>M$, and since $M\in\mathbb R$ was arbitrary and $x\in A$, we conclude that $A$ is not bounded. 

Hint
To discover the correct $x$ to prove your statement, use the following facts:


*

*$\frac{m}{n} + \frac{4n}{m} > \frac mn$

*You can always find some natural number $k$ so that $k>M$.

A: $m=1$ is a nice idea, but it would be simpler to take $m=4$: then, ${4\over n}+n \in A$ for any $n \in \mathbb N$. Now, for any $n \in \mathbb N$, we have an element of $A$ which is bigger than $n$ (namely ${4 \over n} + n$), which is the definition of unboundedness.
A: It is :
$$\frac{m}{n}+\frac{4n}{m} = \frac{m^2 + 4n^2}{mn}$$
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 $\inf K = 4$.
Also, 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 
$$
f(x)=x+4\cdot \frac{1}{x}
$$
