TITLE: Elementary proof that $\pi < \sqrt{5} + 1$
QUESTION [11 upvotes]: I wanted to show that 
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove that $\dfrac{\pi}{\phi^2}<\dfrac{6}5 $.
By squaring my inequality (valid since both sides are positive), and dividing by $6$ I obtain. 
$$ \frac{\pi^2}{6} < 1 + \frac{\sqrt{5}}{3} $$
Where the left handside has a very neat series representation, alas the same does not hold for the right handside. However this is far from an elementary solution. Does someone have a relative simple proof for the equality? To be precise something that is not using advanced knowledge of series. =)

REPLY [2 votes]: Consider a dodecagon circumscribed around a circle of radius $1$.  The area of the dodecagon is
$$12\tan\left(\pi\over12\right)=12\tan\left({\pi\over3}-{\pi\over4} \right)=12{\tan{\pi\over3}-\tan{\pi\over4}\over1+\tan{\pi\over3}\tan{\pi\over4}}=12{\sqrt3-1\over1+\sqrt3}=24-12\sqrt3$$
Because the circle, which has area $\pi$, is inside the dodecagon, we have
$$\pi\lt24-12\sqrt3$$
so it remains to show that
$$24-12\sqrt3\lt\sqrt5+1$$
This can be verified by repeated squarings and simplifications:
$$\begin{align}
24-12\sqrt3\lt\sqrt5+1&\iff23\lt12\sqrt3+\sqrt5\\
&\iff529\lt144\cdot3+24\sqrt{15}+5\\
&\iff92\lt24\sqrt{15}\\
&\iff23\lt6\sqrt{15}\\
&\iff529\lt36\cdot15=540
\end{align}$$