# 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.

• I remember $\pi^2\approx10$ and $e^3\approx20$; thus $\pi^6\approx1000$, whereas $(e\cdot2^{1/3})^6\approx1600$, which implies $\pi<e\cdot2^{1/3}$ Aug 26, 2019 at 2:40
• In fact, we can establish the inequality without relying on the approximations being good enough if we know $\pi^2 < 10$ and $e^3 = 20$. (This inequality turns out to be the unhelpful one, though, since the third quantity is the one in between these two.) Aug 26, 2019 at 2:42
• Notice that $e=2.7128...>2.71$ and ${1.25}^3=\frac{125}{64}<2 \rightarrow 2^{1/3}>1.25$, so $e \times 2^{1/3}> 2.71 \times 1.25 >3.2 >\pi$ Aug 26, 2019 at 2:50
• @Travis: Good point. Of course you meant $e^3>20$ Aug 26, 2019 at 2:54
• @J.W.Tanner Oops, yes, that's what I meant. Aug 26, 2019 at 2:55

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 $$\left(\frac{1 + \sqrt 2}{\sqrt 3 - 1}\right)^3 < \left(\frac{10}{3}\right)^3 < 40 = 2 \cdot 20 < (\sqrt{2} e)^3,$$ establishing

$$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt{2} e}} .$$

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 $$(\sqrt{2} e)^3 > 40$$, but $$\left(\frac{17}{5}\right)^3 < 40,$$ giving the order $$\color{#bf0000}{\boxed{\pi < \frac{1 + \sqrt{2}}{\sqrt{3} - 1} < \sqrt{2} e}} .$$

See this this follow-up question that discusses methods for deriving the inequality $$e^3 > 20$$ by hand.

• Just do 2.718 cubed. It’s not that hard by hand.
– Gabe
Aug 26, 2019 at 14:56
• The bound is very tight---the relative difference is $< 0.5\%$---so I'm not sure how efficient one can be. Naively, $e = \sum_{k=0}^\infty \frac{1}{k!} > \sum_{k=0}^5 = \frac{163}{60}$, and so we need only show $20 < \left(\frac{163}{60}\right)$, which is equivalent to $4\,320\,000 < 4\,330\,747$. Aug 26, 2019 at 15:15
• @TheSimpliFire I was surprised that hadn't been asked on this site before, so I wrote it up as a question, giving two elementary (but not terribly enjoyable) methods, including the one in my previous comment: math.stackexchange.com/questions/3335059/show-that-e3-20 . Aug 26, 2019 at 17:40
• (The quantity in parentheses in my previous answer should be raised to the third power.) Sep 21, 2019 at 5:09

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.