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?

Assume the LHS is $<$ the RHS.  Square the LHS
$$\frac{1}{2} + \frac{1}{6} - \frac{2}{\sqrt{12}} = \frac{2}{3} - \frac{1}{\sqrt{3}} < \frac{9}{100}$$
Subtract $2/3$ from the LHS and the RHS and we are assuming that
$$ \frac{1}{\sqrt{3}} > \frac{2}{3} - \frac{9}{100} = \frac{173}{300}$$
Square both sides again:
$$\frac{1}{3} > \frac{29929}{90000}$$
which is true, and therefore the assumption is true.  Therefore
$$\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}} < \frac{3}{10}$$
