Elementary proof that $\pi < \sqrt{5} + 1$

Question

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. =)

Answer 1

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 $$1 + \sqrt{5} = \text{Area}(OCFE) > \text{Area}(ODE) = \pi$$

Answer 2

The function $$f(x):=\cos x-\sin x=\sqrt{2}\sin\left({\pi\over4}-x\right)\ .$$ is monotonically decreasing on $\bigl[0,{\pi\over2}\bigr]$ and has a zero at $x={\pi\over4}$. As the Taylor expansions of $\cos$ and $\sin$ are alternating we conclude that for $0<x<1$ one has the estimate $$f(x)<\left(1-{x^2\over2}+{x^4\over24}\right)-\left(x-{x^3\over 6}\right)\ .$$ In particular one finds that $$f\left({4\over5}\right)<-{11\over625}<0\ .$$ From this it follows that ${4\over5}>{\pi\over4}$, or $$\pi<{16\over5}\ .$$

On the other hand from $121<125$ it we obtain ${11\over5}<\sqrt{5}$, or $${16\over5}<1+\sqrt{5}\ .$$

Answer 3

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} $

Answer 4

What known results can you use? For example, can you use the fact that $\frac{\pi^4}{90} = \sum_{k=1}^\infty k^{-4}$? If so, you can show that $\pi^2<10$ as follows, then continue with clark's answer.

$$\frac{\pi^4}{90} = 1+\frac{1}{16} + \sum_{k=3}^\infty k^{-4}< 1+\frac{1}{16} + \int_2^\infty x^{-4} dx = 1+\frac{1}{16}+\frac{1}{24}\textrm{, which implies that }$$ $$\pi^2<\sqrt{99.375}<10\textrm.$$

Answer 5

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 $\pi < \frac{16}{5}$.

Answer 6

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:

$$\begin{align} 24-12\sqrt3\lt\sqrt5+1&\iff23\lt12\sqrt3+\sqrt5\\ &\iff529\lt144\cdot3+24\sqrt{15}+5\\ &\iff92\lt24\sqrt{15}\\ &\iff23\lt6\sqrt{15}\\ &\iff529\lt36\cdot15=540 \end{align}$$