# 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?

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Tip: Use \displaystyle in this manner: \displaystyle \frac{1}{sqrt(2)} - \frac{1}{sqrt(6)} –  Git Gud Feb 3 '13 at 9:15
how do you get surds? –  fosho Feb 3 '13 at 9:18
\sqrt {x+y} to get $\sqrt {x+y}$ –  Git Gud Feb 3 '13 at 9:19
There were a few edits that seem to have changed/corrected mathematical content in the OP. I don't think it's a good idea to do that and reviewers should look more closely... –  t.b. Feb 3 '13 at 9:30

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

$$\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}} < \frac{3}{10}$$

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

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But $1/\sqrt{2}-1/\sqrt{6}=0.29885...$ so something is off. –  coffeemath Feb 3 '13 at 22:21
Got right, there was a "9" in place of a "6". Mistake corrected. –  Jack D'Aurizio Feb 4 '13 at 13:46