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.

Here's a dirty decimal arithmetic method that presumes knowledge only of the bounds $1.41 < \sqrt{2} < 1.42$, $1.73 < \sqrt{3} < 1.74$---which you probably know if you're taking the GRE subject test---and the not-too-obscure fact $e^3 > 20$: Since $\sqrt 6 = \sqrt 2 \sqrt 3$ multiplying gives $2.43 < \sqrt{6} < 2.47$ Then, using the rationalization $$\frac{1 + \sqrt 2}{\sqrt 3 - 1} = \frac{1}{2} (1 + \sqrt 2 + \sqrt 3 + \sqrt 6) ,$$ and substituting the decimal values gives
$$\pi < 3.29 < \frac{1 + \sqrt 2}{\sqrt 3 - 1} < 3.32 .$$ Now, $3.32 < \frac{10}{3}$, so $$\left(\frac{1 + \sqrt 2}{\sqrt 3 - 1}\right)^3 < \left(\frac{10}{3}\right)^3 < 40 = 2 \cdot 20 < (\sqrt[3]{2} e)^3,$$
establishing
$$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt[3]{2} e}} .$$
Alternatively, here's a version that uses only estimates using fractions with small denominators (which themselves follow from the decimal bounds above): Since $$\frac{7}{5} < \sqrt{2} < \frac{10}{7} \qquad \textrm{and} \qquad \frac{12}{7} < \sqrt{3} < \frac{7}{4} ,$$ we have
$$\frac{1 + \sqrt{2}}{\sqrt{3} - 1} > \frac{1 + \frac{7}{5}}{\frac{7}{4} - 1} = \frac{16}{5}.$$ (Of course we can verify the bounds on $\sqrt{2}, \sqrt{3}$ without knowing anything about the numbers' decimal representations---just square all of the numbers, which reduces the problem to comparing rational numbers.) This is $3.2 > \pi$, but we can avoid decimal representations using $\frac{16}{5} > \frac{22}{7} > \pi$.
On the other hand,
$$\frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \frac{1 + \frac{10}{7}}{\frac{12}{7} - 1} = \frac{17}{5} .$$
Since $e^3 > 20$, we have $(\sqrt[3]{2} e)^3 > 40$, but
$$\left(\frac{17}{5}\right)^3 < 40,$$ giving the order
$$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt[3]{2} e}} .$$
See this this follow-up question that discusses methods for deriving the inequality $e^3 > 20$ by hand.
