Prove that $\frac{\pi}{\phi^2}<\frac65 $

Question

How to prove that$$\frac{\pi}{\phi^2}<\frac{6}5 $$ from the definition of $\pi$ and $\phi$? ($\pi=4\int_{0}^1\sqrt{1-t^2}dt,\phi=\dfrac{\sqrt5+1}2.$)

Answer 1

It we express $\pi$ and $\phi$ as simple continued fractions, it is known: $$\pi = [3;7,15,1,292,1,1,,1,2,\ldots] \quad\text{ and }\quad \phi = [1;1,1,1\ldots].$$ The first few convergents of $\pi$ and $\phi$ are

$$3, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \ldots \quad\text{ and }\quad 1, \frac{2}{1}, \frac{3}{2}, \frac{5}{3} \ldots, \frac{F_k}{F_{k-1}} \ldots$$ where $F_k$ are the Fibonacci numbers. It is also known these convergents sandwich the values of $\pi$ and $\phi$ in alternate fashion.

$$3 < \frac{333}{106} < \pi < \frac{355}{113} < \frac{22}{7}$$ $$1 < \frac{3}{2} < \cdots < \frac{F_{2k}}{F_{2k-1}} < \phi < \frac{F_{2k+1}}{F_{2k}} < \cdots < \frac{5}{3} < \frac{2}{1}$$

In particular, $\pi < \frac{355}{113}$ and $ \frac{377}{233} = \frac{F_{14}}{F_{13}} = \frac{\phi^{14} - \phi^{-14}}{\phi^{13}+\phi^{-13}} < \phi $, this implies

$$\frac{5\pi}{6\phi^2} < \frac{5\left(\frac{355}{113}\right)}{6\left(1+\left(\frac{377}{233}\right)\right)} = \frac{82715}{82716} < 1$$