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.

$$\sqrt{6}-\sqrt{2} = \frac{6-2}{\sqrt{6}+\sqrt{2}}= \frac{2}{\frac{\sqrt{6}+\sqrt{2}}{2}}> \frac{2}{\sqrt{\frac{6+2}{2}}}=1$$
$\bf{Added:}$
$$\sqrt{6} - \sqrt{2} > \sqrt{6.25} - \sqrt{2.25} = 2.5 - 1.5 = 1$$
since the function $x\mapsto \sqrt{x}$ is concave.
