<s>[INST] Elementary proof that $\pi < \sqrt{5} + 1$ 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. =)
 [/INST] Just use that $\pi ^2 \leq \frac{61}{6}$ which comes from the series representation of $\frac{\pi^2}{6}$. 
Then
$
\begin{align}
\pi^2 <5+1+2\cdot \sqrt5&\Leftarrow\\
\frac{61}{6} <6+2\cdot \sqrt5&\Leftrightarrow\\
\frac{5^2}{6}<2\sqrt 5&\Leftrightarrow\\
\frac{5^3}{36}<4\Leftrightarrow\\
125<144
\end{align}
$
$$$$
Indeed you have that  $\frac{\pi^2}{6}= \sum _{k\geq1}\frac{1}{k^2}$
then
$
\begin{align}
\frac{\pi^2}{6}-1-\frac{1}{4}-\frac{1}{9}= \sum _{k\geq 4}\frac{1}{k^2} &\leq \\
\sum _{k\geq 4}\frac{1}{(k-1)k} &=\\= \sum _{k\geq4}\left ( \frac{1}{k-1} - \frac{1}{k} \right)&=\frac{1}{3}\\
\end{align}
$
</s>