Charming approximation of $\pi$: $2\left(\frac{1}{2}\right)^{\phi/2}+2< \pi$, where $\phi$ is the golden ratio

If we'll prove that $$2^{\sqrt5}>\frac{212}{45},$$ so it's enough to prove that: $$\frac{2}{\sqrt[4]2\cdot\sqrt[4]{\frac{212}{45}}}+2<\pi$$ or $$\sqrt[4]{\frac{90}{53}}<\pi-2$$ for which it's enough to prove that $$\sqrt[4]{\frac{90}{53}}<\frac{1613}{1413},$$ which is true because $$1413^4\cdot90-1613^4\cdot53=-1802797643<0.$$ I hope it will help.

To prove the main result

$$2\left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2<\pi$$

we shall show that $ \pi <\sqrt{10}$ and interestingly, we also deduce that $ 3<\pi<4$

Preliminaries

Consider the set $$S=\left\{X_n= \frac{1}{n^3(n+1)^3} : n\in\mathbb {N} \right\}$$ and here we show that set $S$ is bounded set with lower and bounds $0$ and $\frac{1}{8}$ respectively. Note that for $n\geq 1$ the $$ \begin{aligned} S_{n+1}-S_n & =\frac{1}{(n+1)^3}\left[\frac{1}{(n+2)^3}-\frac{1}{n^3}\right]\\ & =\frac{1}{(n+1)^3}\left[\frac{n^3-(n+2)^3}{n^3(n+2)^2}\right]\cdots(1)\end{aligned}$$ since for all $n>1$, $ n< n+2\implies n^3-(n+2)^3<0$ from $(1)$ it follows that $ S_{n+1}-S_n <0$ implies the sequence $X_n$ contained in set $S$ is a decreasing sequence and thus $$ \begin{aligned} \operatorname{sup}\left\{X_n : n\in \mathbb{N}\right\}&=\frac{1}{8}<1\\ \operatorname {inf}\left\{X_n:n\in\mathbb{N} \right\}&=0\end{aligned}$$Therefore, $ 0 < X_n \leq \frac{1}{8} <1$. Futher, $n^3(n+1)^3> n(n+1)=Y_n$ and hence $$0 <\sum_{n\geq 1} X_n < \sum_{n\geq 1} (Y_n)^{-1}=1\cdots(2)$$ since we have telescoping series as $\displaystyle \sum_{n\geq 1} (Y_n)^{-1} =\sum_{n\geq 1} \left(\frac{1}{n}-\frac{1}{n+1}\right)=1$ Now $$\begin{aligned} \sum_{n\geq 1} X_n & =\sum_{n\geq 1} \left(\frac{1}{Y_n}\right)^3\\&=\sum_{n \geq 1}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)\\&-\sum_{n\geq 1}\frac{3}{(Y_n)^2}=\zeta(3)-\zeta(3)+1\\& 1-3\sum_{n\geq 1} \left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)\\&-2\sum_{n\geq 1}\frac{1}{(Y_n)}= 1-6\zeta(2)+9 \\&= 10-\pi^2\end{aligned}$$ and thus from $(2)$ $$\sum_{n\geq 1}X_n >0\implies \pi <\sqrt{10}$$ since $ 9-\pi^2 < 10-\pi^2<16-\pi^2$ which implies $3< \pi < 4$.

Proof of the main result

If the left hand expression of main result has to be less than $\pi$ then it should also be less than $\sqrt {10}$. To prove the result we suppose that the inequality is true. That is; $$A= 2\left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2 < \pi ,\;\; A< \sqrt {10}$$ Squaring both sides we yield $$\begin{aligned}\left( \frac{1}{2}\right)^{\phi}+ \left(2\left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2\right)-1<\frac{5}{2}\end{aligned}\\ \left(\frac{1}{2}\right)^{\phi}+A<\frac{7}{2}$$ hence$$\left(\frac{1}{2}\right)^{\phi} <\frac{7-2\sqrt{10}}{2}=\left(1+\frac{5-2\sqrt{10}}{2}\right)^{\frac{1}{\phi}}=(1+y)^{\phi^{-1}}$$ since $ \phi>1\implies \frac{1}{\phi}<1$ and hence by Bernoulli inequality we have $$(1+y)^{\phi^{-1}}<1+ \frac{y}{\phi}=\\ 1-\frac{15}{(5+2\sqrt{10})(1+\sqrt{5})}<1-\frac{15}{11 \cdot 3} =1-\frac{15}{33}=\frac{18}{33}$$ Since $$\begin{aligned} \frac{7-2\sqrt {10}}{2}=(1+y)^{\phi{-1}} <\frac{18}{33}\end{aligned}$$. We claim that $A<\sqrt {10}$ which also says $\frac{1}{2} <(1+y)^{\phi^{-1}}$ also we have $(1+y)^{\phi^{-1}} <\frac{18}{33}$. Therefore we must have $\frac{1}{2} < \frac{18}{33}$ which is true since $$\frac{1}{2} -\frac{18}{33} =\frac{33-36}{66} =-\frac{1}{22}<0$$ As we claimed inequality to be true and hence we came up $-\frac{1}{22}<0$ to be true and thus,

$$2 \left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2<\pi$$ must be true.