Prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I'm trying to prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. However, I came up with this:
Since $\sqrt{4}=2$ and $\sqrt{9}=3$ we have a following inequality: $2 < \sqrt{6} < 3$. 
On the other hand we have $\sqrt{1}=1$ and $\sqrt{4}=2$ so we have $1 < \sqrt{2} < 2$.
Let $n = \sqrt{6}-\sqrt{2}$,
Therefore $ 1<n<3$, which implies that $n>1$.
Is this a correct proof?
Thank you.

$\sqrt{6}-\sqrt{2}$ is positive because $\sqrt{x}$ is minimal at $0$ and always increasing at $x>0$.
Because $x^2$ is $1$ at $x=1$ and always increasing at $x>0$, numbers greater than 1 remain greater than $1$ when squared and $8-\sqrt{48}$ results from squaring $\sqrt{6}-\sqrt{2}\,$.
Since $8-\sqrt{49}=1$ and $\sqrt{49}$ exceeds $\sqrt{48}$ given properties of $\sqrt{x}$, it then follows that $\sqrt{6}-\sqrt{2} > 1$.
