Proof of inequality involving surds Determine whether $ \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$ is greater or less than $\frac{3}{10}$
So what I did was:
Add the fraction with surds to get $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{12}}$
Then squaring both sides results in $\frac{8-4\sqrt{3}}{12} = \frac{2-\sqrt{3}}{3}$
Therefore we have $\frac{2-\sqrt{3}}{3} ? \frac{9}{100}$ (using question mark in place of < or > therefore $200-100\sqrt{3} = 27$, but $1\times1 = 1$ and $2\times2 = 4$ thefore $1<\sqrt{3}<2$ and since $100-100\times1 = 0 <27$, $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}} < \frac{3}{10}$
I checked the answer and they had a method but same result, however would this suffice?

$$ \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = \frac{\sqrt{2}}{\sqrt{3}(1+\sqrt{3})}=\frac{\sqrt{2}}{3+\sqrt{3}}=\frac{\sqrt{2}(3-\sqrt{3})}{6}<\frac{7\sqrt{2}}{33},$$
where we have used $\sqrt{3}>\frac{19}{11}$. Now $\frac{7\sqrt{2}}{33}<\frac{3}{10}$, since $9800=99^2-1<99^2$.
