<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] Below are two geometric ways to get better approximations than the one requested.  Neither of the two involve the expression $\sqrt{5}+1$ directly so it would still take some work to show that these results are in fact stronger.  However, a trick combining both results will show that $\pi < \frac{16}{5}$ which is quickly seen to be sufficient.
1. Let $0 < x \leq \frac{\pi}{2}$.  Draw two lines through $(1, \frac{x}{2})$ that are tangent to the unit circle.  One line touches the unit circle at $(1,0)$ the other one at $$p = \left(\frac{4-x^2}{4+x^2}, \frac{4x}{4+x^2}\right).$$ 
Since the distance over the unit circle from $(1,0)$ to $p$ is less than $x$ we get the inequality
$$\sin(x) > \frac{4x}{4+x^2}.$$
In particular $\sin(4-2\sqrt{3}) > \frac{1}{2}$ and therefore $\frac{\pi}{6} < 4-2\sqrt{3}$.
2. Let $$\begin{eqnarray}x(t) &=& 1 - \frac{t^2}{2}\\ y(t) &=& t\end{eqnarray}$$ Then $x^2+y^2 \geq 1$. Let $\alpha = \sqrt{3} - 1$ then $x(\alpha) = y(\alpha) = \alpha$. So for $t \in [0, \alpha]$ the curve $(x,y)$ travels from $(1,0)$ to $(\alpha, \alpha)$ in counter clockwise direction.  The area of the sector bounded by this piece of curve is
$$
\frac{1}{2}\int_0^{\alpha} \begin{vmatrix} x & \dot{x} \\ y & \dot{y} \end{vmatrix} dt = \frac{1}{2}\int_0^{\alpha} \left(1+\frac{t^2}{2}\right) dt = \sqrt{3} - \frac{4}{3}.
$$
Since the curve is at least at a distance $1$ from the origin we get the inequality $ \frac{\pi}{8} < \sqrt{3} - \frac{4}{3}$.
Trick. Take a convex combination of the inequalities
$$
\begin{eqnarray}
\pi &<& 24 - 12 \sqrt{3} \\
\pi &<& -\frac{32}{3} + 8 \sqrt{3}
\end{eqnarray}
$$
with weights $\frac{2}{5}$ and $\frac{3}{5}$ to get $\pi < \frac{16}{5}$.
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