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. =)

Start with an unit circle inscribed inside a $2 \times 2$ square. One can chop off 4 right angled isosceles triangle whose shorter side has length $2 - \sqrt{2}$ from the four corners. This will turn the square into a octagon with the unit circle inscribed inside it. By comparing the area of the circle and the octagon, we have
$$\pi = \text{Area(circle)} < \text{Area(octagon)} = 4 - 2 (2 - \sqrt{2})^2 = 8(\sqrt{2}-1)$$
Since $2 \cdot 12^2 = 288 < 289 = 17^2$, we have
$$12 \sqrt{2} < 17 
\implies 8 (\sqrt{2}-1) < \frac{10}{3}
\implies \frac{3\pi}{10} < 1
$$
For $x \in (0,1)$, if one look at the Taylor expansion of $\sin x$ at $x = 0$, we have
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
One will notice the magnitude of each term is monotonic decreasing while the sign is alternating. This implies 
$$\sin x > x ( 1 - \frac{x^2}{6} ) > \frac{5x}{6}\quad\text{ for } x \in (0,1)$$
In particular, this is true for $ \displaystyle x = \frac{3\pi}{10}$. As a result,
$$ 1 + \sqrt{5} = 4 \cos\left(\frac{\pi}{5}\right) = 4 \sin\left(\frac{3\pi}{10}\right)
> 4 \left(\frac{5}{6}\right)\left(\frac{3\pi}{10}\right) = \pi$$
Update
An alternate proof without calculus.

Start with a circle centered at $O$ with radius $2$. Inscribe a regular pentagon inside it.
Let $AB$ be an edge of the pentagon and $C$ be its mid-point. Extend $OC$ until it hit the circle at $D$. Construct a line through $O$ perpendicular to $OC$ and let it hit the circle at $E$. Let $F$ be the point so that $OCFE$ forms a rectangle.
From elementary geometry, we know the area of rectangle $OCFE$ is $1 + \sqrt{5}$ and the area of the quarter circle (the red one as shown) $ODE$ is $\pi$. To prove the inequality
$1 + \sqrt{5} > \pi$, one just suffices to show the area of the green shape $BFE$ above
is larger than that of the shape $BCD$.
Extend $OB$ to the point $G$ where $BG = OB = 2$. Construct a line through $G$ parallel to $OD$. Let $C'$ be its intersection with the line $CF$. Let $D'$ be a point on the line at a distance $2$ from $G$. By symmetry, the shape $BC'D'$ enclosed by the red dashed lines
has the same area as the shape $BCD$. 
Construct a line through $D'$ parallel to $OE$ and let it hit the line $EF$ at $H$.
The area of that portion of the shape $BC'D'$ outside the rectangle $OCFE$ is smaller than
that of the rectangle $FC'D'H$. Extend $D'H$ to the point $D''$ where $HD'' = D'H$. Extend $FH$ to the point $C''$ where $HC'' = 2FH$. The triangle $HC''D''$ has the same area as the rectangle $FC'D'H$'. It is clear this triangle lies completely in the green shape $BEF$ and 
disjoint from the shape $BC'D'$. As a result, we have:
$$\text{Area}(BFE) > \text{Area}(BC'D') = \text{Area}(BCD)$$
and hence
$$1 + \sqrt{5} = \text{Area}(OCFE) > \text{Area}(ODE) = \pi$$
