<s>[INST] How to compare $\pi, e\cdot 2^{1/3}, \frac{1+\sqrt{2}}{\sqrt{3}-1}$ This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and compute the nominator with the value of square root 2 and 3 memorized. And as to $e\cdot 2^{1/3}$, I just don't see how to compare it to other two items without take cubic and compute. This whole process is very time consuming. 
I have seen some tricks to compare say $2^\pi,\pi^2$. But the technique does not seem to apply here.
 [/INST] If you raise them to the third power you get
$\pi^3 \approx (3+\frac 17)^3 \approx 3^3 + 3*3^2*\frac 17 \approx 27*(1\frac 17)$
$e^3*2 \approx 2(3-0.29)^3 \approx 2(3^3 - 3*3^2*0.29)\approx 27*2*(0.71)\approx 27*1.42$ so $e*2^{\frac 13} > \pi$.  (And as a weird unexpected bonus I get that $e^3*2 \approx 27*\sqrt 2$.)
$(\frac {1 +\sqrt 2}{\sqrt 3-1})^2 =\frac {3+2\sqrt 2}{4- 2\sqrt 3}$ and
$(\frac {1 +\sqrt 2}{\sqrt 3-1})^3=\frac {3+2\sqrt 2}{4- 2\sqrt 3}\frac {1 +\sqrt 2}{\sqrt 3-1}=\frac {3+4+5\sqrt 2}{-6-4+6\sqrt 3}$
$=\frac {7+5\sqrt 2}{6\sqrt 3 - 10}=\frac {7+5\sqrt 2}{6\sqrt 3 - 10}\frac {6\sqrt 3 + 10}{6\sqrt 3 + 10}$
$=\frac{30\sqrt 6 +42*\sqrt 3 + 50\sqrt 2 +70}{36*3-100}$
$\approx \frac 18(30*1.7*1.4 + 42*1.7 + 50*1.4 + 70)\approx$
$\frac 18(51*1.4 + 7*10.2 + 70 + 70)\approx$
$\frac 18(70*4+1.4+1.4) \approx \frac {70.7}2 \approx 35.35$.
$\approx 27 + 8 \approx 27(1\frac 14)$
And $1\frac 17 < 1\frac 14 < 1.42$ so
So $\pi < \frac {1 +\sqrt 2}{\sqrt 3-1} < e*2^{\frac 13}$
.... which.... was really way too much work for too little result but.... well, it's one way to do it.
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