Find the infimum of the set $\{x+\frac 1x\}$

Question

I am pretty new to the website, so I apologize in advance if I ask my question incorrectly.

I am trying to find the infimum of the Set $$A=\left\{x+\frac{1}{x} \text{ for all }x \in \mathbb N\right\}.$$

Here's how I started: Using the definition of infimum, there is a $u$ in the set $A$ such that $u<x+\frac{1}{x}$.

From the Archimedean property, $\epsilon>0$ implies $\frac{1}{\epsilon}>0$.

I am not sure how to continue. Any advice?

Answer 1

We can use AM-GM: $$n + \frac{1}{n} \ge 2\sqrt{1} = 2.$$

Answer 2

There's no need for any epsilon arguments or fancy inequalities here. One can simply observe that for all $n \in \mathbb N$, $$n + \frac{1}{n} > n.$$ Therefore for $n \geq 2$, we have $$n + \frac{1}{n} > 2.$$ On the other hand, if $n=1$, we have $$n + \frac{1}{n} = 2.$$ Therefore the sequence has a minimum of $2$ which occurs at $n=1$.

Answer 3

Observe that the sequence $(a)_n = \{n + \frac{1}{n}\}_{n \in \mathbb{N}}$ is strictly increasing, and that $2 = 1 + \frac{1}{1} \in (a)_n$ is the first element in the sequence, thus a lower bound for it, by our previous comment. Indeed, $2$ is the greatest lower bound for $(a)_n -$ if $L$ is a greater lower bound, then $L \not< 2$.