Prove $\forall n \in \mathbb{N}$ where $ n \neq 1 , n + \frac{1}{n} > 2$ using completing the square.

Question

I have got this far; I am only unable to understand how to finish the proof.

$n>0 \implies n + 1/n > 0 \implies n + 1/n + 2 - 2 > 0 \implies {\big(\sqrt{n}+\frac{1}{\sqrt{n}}\big)}^2 - 2 > 0 \implies {\big(\sqrt{n}+\frac{1}{\sqrt{n}}\big)}^2 > 2$

How can I use this to prove what's being asked in the question?

Answer 1

$${ \left( \sqrt { n } -\frac { 1 }{ \sqrt { n } } \right) }^{ 2 }>0\Rightarrow n-2+\frac { 1 }{ n } >0\Rightarrow n+\frac { 1 }{ n } >2$$

Answer 2

A simple solution, but without using completing the square:

As $n\in \mathbb{N}$ and $n\neq 1$, so $n\geq 2$, and hence $n + 1/n>2$.

Answer 3

Completing the square a different way: $$n+\frac1n=\frac{n^2+1}n = \frac{n^2-2n+1+2n}n = \frac{(n-1)^2}n + 2>2$$ The final inequality follows from the assumption that $n$ is not 1, hence the quantity $\frac{(n-1)^2}n$ is strictly positive.