# Inequality derived from a famous problem

Let's have the following inequality: $$\frac{2}{3}(\sqrt 5-1)^3\lessgtr\sqrt[3]{2}$$.

Which part is greater, the left or the right? No calculator solutions are accepted.

• The doubling of the cube problem remained unsolved for two thousand three hundred years. One hundred fifty years ago someone proved that the doubling of the cube is impossible with numbers constructible with compass and straight edge. But when we allow other regular solids, such as cones and spheres, then the doubling of the volume by compass and straight edge is achievable. Let's have a cone with $r =(\sqrt{5}-1)$ and $h=2(\sqrt{5}-1)$, the $V_c=\frac{2}{3}(\sqrt{5}-1)^3\pi$. A sphere with the same radius has $V_s=\frac{4}{3}(\sqrt{5}-1)^3\pi$ which is exactly twice the volume of the cone. Commented May 24, 2021 at 0:43

The LHS is

smaller

than the RHS. Here's how we're going to prove it.

First, we'll turn the requested comparison into one involving the famous number $$\frac{1+\sqrt5}2$$. Call this $$\phi$$ and write $$\bar\phi=\frac{1-\sqrt5}2$$; we have the following key facts. (1) $$\phi\cdot\bar\phi=-1$$ (which is how we turn the original comparison into one involving $$\phi$$). (2) $$\phi,\bar\phi$$ are what are called conjugates, which has the consequence that various things built from them are integers. (3) Specifically, the numbers $$\phi^n+\overline\phi^n$$ are integers when $$n$$ is an integer, and there's a nice way to calculate them. (4) $$\bar\phi$$ is small, which means that the values of $$\phi^n$$ are well approximated by these integers. Putting these facts together will lead us where we need to go.

OK, let's get started. We want to know whether $$\frac23\left(\sqrt5-1\right)^3<\sqrt[3]2$$. So,

get it in terms of $$-\bar\phi=\frac{\sqrt{5}-1}2$$ by pushing some factors of 2 around: $$\frac{16}3\left(\frac{\sqrt5-1}2\right)^3<\sqrt[3]2$$;
exploit the relationship between $$\phi,\bar\phi$$: $$\frac{16}3\left(\frac{\sqrt5+1}2\right)^{-3}<\sqrt[3]2$$;
multiply both sides by that thing involving $$\phi$$: $$\frac{16}3<\sqrt[3]2\left(\frac{\sqrt5+1}2\right)^3$$;
get rid of the cube root by cubing both sides: $$\frac{4096}{27}<2\left(\frac{\sqrt5+1}2\right)^9$$;
cancel a factor of 2 from both sides: $$\frac{2048}{27}<\left(\frac{\sqrt5+1}2\right)^9$$.

That's the first step completed. Now

let's look at the thing we now have on the RHS. It's $$\phi^9$$. So, continuing with our plan, write $$u_n=\phi^n+\bar\phi^n$$. If you replace $$\phi,\bar\phi$$ with their explicit expressions in terms of $$\sqrt5$$ and expand out the n'th powers using the binomial theorem, you'll see that all the terms with an odd number of factors of $$\sqrt5$$ cancel out, so $$u_n$$ is rational. In fact, we can do better than that; a quick calculation shows that $$u_n$$ obeys the same recurrence relation as the Fibonacci numbers do ($$u_{n+2}=u_{n+1}+u_n$$), and that $$u_0=2,u_1=1$$ -- so in fact all the $$u_n$$ are integers, and we can easily calculate them with that recurrence relation. For obvious reasons we are interested in $$u_9$$ which comes out to be 76.

Nearly there. Finally,

let's see how far away from 76 $$u_9$$ actually is. The difference is $$\bar\phi^9$$. Remember that $$\left|\bar\phi\right|<1$$; in fact $$\bar\phi\cong-0.618$$. So its 9th power should be pretty small. But we can be more precise than that; remember that $$\phi\cdot\bar\phi=-1$$, so $$\left|\bar\phi^9\right|$$ is the reciprocal of $$\phi^9$$ -- which, since certainly $$\left|\bar\phi^9\right|<1$$, is bigger than $$u_9-1=75$$. Hence in fact $$\left|\bar\phi^9\right|<1/75$$, which means that $$|\phi^9-76|<1/75$$. And now we really are done, because 2048/27=76-4/27 which is certainly less than 76 by more than 1/75. So the LHS is smaller than the RHS.

(The above is fairly long, but only because I've gone into quite a lot of detail. The actual calculation is rather quick, and for those familiar with these ideas each step is more or less the "obvious" one.)

• That conjugate trick is actually quite pretty. Commented May 24, 2021 at 0:34
• Glad you like it! :-) Commented May 24, 2021 at 0:42
• Could you elaborate on the steps a bit more? I'm sure this solution is correct but the explanation is hard to follow. Commented May 24, 2021 at 1:23
• How about now? I've been more explicit about several of the steps, and put a sort of outline of the strategy and key ideas at the start. Commented May 24, 2021 at 2:03
• Thank you, the explanation is very educational and is much easier to follow! Commented May 24, 2021 at 2:17