How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$

\begin{align}
\frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6}
\iff&3+3\sqrt{5}\lt\pi^2\\
\iff&3\sqrt{5}\lt\pi^2-3\\
\iff&(3\sqrt{5})^2\lt(\pi^2-3)^2\quad(\text{both sides are positive})\\
\iff&45\lt\pi^4-6\pi^2+9\\
\iff&\pi^4-6\pi^2-36\gt0\\
\end{align}
Using, for example, $3.14\lt\pi\lt3.15$ we have that
$$\pi^4-6\pi^2-36\gt3.14^4-6\cdot3.15^2-36\gt0$$
which implies that
$$\frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6}$$
