How to compare $\pi, e\cdot 2^{1/3}, \frac{1+\sqrt{2}}{\sqrt{3}-1}$

Question

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.

Answer 1

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.

Answer 2

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.