<s>[INST] 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}} $
 [/INST] 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.)
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