What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
 A: I would multiply by the conjugate of denominator $$a=\frac{3 \sqrt{3}-4}{7-2 \sqrt{3}}=\frac{1}{37} \left(13 \sqrt{3}-10\right)$$ $$b=\frac{3 \sqrt{3}-8}{1-2 \sqrt{3}}=\frac{1}{11} \left(13 \sqrt{3}-10\right)$$ this makes $a<b$.
A: Using the rule $a\sqrt b=\sqrt{a^2b}$ for $a,b>0$ one checks the signs of numerator and denominator to be both positive for the first fraction and both negative for the second fraction. This is therefore a comparison of two positive numbers (so no verdict yet), and by flipping the signs of numerator and denominator on the right we can ensure that all individual factors are positive: compare $\frac{3\sqrt3-4}{7-2\sqrt3}$ and $\frac{8-3\sqrt3}{2\sqrt3-1}$. Now multiply by the (positive!) product of the denominators to compare $(3\sqrt3-4)(2\sqrt3-1)$ and $(8-3\sqrt3)(7-2\sqrt3)$ which amounts to comparing $22-11\sqrt3$ and $74-37\sqrt3$ or finally $26\sqrt3$ and $52$. Since by happy coincidence $26\times2=52$ and $\sqrt3<2$ the number on the left is smaller.
I'm not sure whether many people could do this easily by mental arithmetic (I certainly did not). Each of the terms in the final comparison are obtained by adding $4$ products from the terms in the original expression. One product, namely $(2\sqrt3)(3\sqrt3)$, contributes twice with opposite signs and could therefore have been cancelled beforehand (leaving $7\times8-4\times1=52$ for that term), but that's all I can see for simplifications. Personally my main problem with mental computation is that I very often get the sign of one term wrong in my head, but maybe with a lot of training one can limit this kind of errors.
A: $$\dfrac{3\sqrt3-8}{1-2\sqrt3}=\dfrac{8-3\sqrt3}{2\sqrt3-1}$$
Now $$\dfrac{3\sqrt3-4}{7-2\sqrt3}-\dfrac{8-3\sqrt3}{2\sqrt3-1}=\dfrac{(3\sqrt3-4)(2\sqrt3-1)-(7-2\sqrt3)(8-3\sqrt3)}{(7-2\sqrt3)(2\sqrt3-1)}$$
$$=\dfrac{26\sqrt3-52}{(7-2\sqrt3)(2\sqrt3-1)}$$
Now  the denominator $>0,$ and $26\sqrt3-52=26(\sqrt3-2)<0$
A: Using the approximation $\sqrt3\approx1.7$, one can do a bit of mental arithmetic and see that
$${3\sqrt3-4\over7-2\sqrt3}\approx{1.1\over3.6}\lt1\qquad\text{while}\qquad{3\sqrt3-8\over1-2\sqrt3}\approx{-2.9\over-2.4}={2.9\over2.4}\gt1$$
This only works, of course, because the two numbers are not at all close.  It's also not a rigorous proof, although it could be turned into one with a little more care.  For example, since $\sqrt3\lt2$, we have
$${3\sqrt3-4\over7-2\sqrt3}\lt{6-4\over7-4}={2\over3}\qquad\text{while}\qquad{3\sqrt3-8\over1-2\sqrt3}={8-3\sqrt3\over2\sqrt3-1}\gt{8-6\over4-1}={2\over3}$$
(Note the element of luck here, in that the two crude bounds happen to agree.)
A: Claim: 
$$\frac {3\sqrt {3}-4}{7-2\sqrt {3}}<\frac {3\sqrt {3}-8}{1-2\sqrt {3}}$$
Proof:
Let $u=\sqrt 3-1$. Notice $1>u>\frac12$.
If
$$\frac{3u-1}{5-2u}<\frac{3u-5}{-2u-1}$$
$$\Leftarrow-6u^2-u+1>-6u^2+25u-25$$
$$\Leftarrow26>26u$$
$$\Leftarrow1>u=\sqrt3-1$$
which is true.
