I read an article that claimed that among two regular polygons with equal perimeters, the one with more sides has a larger area, in other words $\frac{ \cot(\frac{\pi}{n+1}) }{ \cot(\frac{\pi}{n}) }\cdot\frac{n}{n+1} >1$ for $n \ge 3$.
I became curious about $f(x):=\frac{ \cot(\frac{\pi}{x+1}) }{ \cot(\frac{\pi}{x}) }\cdot\frac{x}{x+1}$ and decided to graph it. It seemed to be strictly decreasing on $(2,\infty)$
$$f'(x)=\frac{\cot(\frac{\pi}{x}) \cdot \csc(\frac{\pi}{x+1})\cdot \frac{\pi}{(x+1)^2}- \cot(\frac{\pi}{x+1}) \cdot \csc(\frac{\pi}{x})\cdot \frac{\pi}{x^2}}{\cot^2(\frac{\pi}{x}) }\bigg(1 -\frac{1}{x+1} \bigg) +\frac{ \cot(\frac{\pi}{x+1}) }{ \cot(\frac{\pi}{x}) }\cdot\frac{1}{(x+1)^2}
$$
I tried to prove this observation, but I failed. Finding the derivative of the function and then showing that $f′(x)<0$ for $x>2$ seemed impossible and dreadful, as the derivative involved trigonometric and rational functions.
I also want to know how to prove that $\frac{\cot(\frac{\pi}{x})}{x}$ is strictly increasing for $x\ge 3$ or in other words among two regular polygons with equal perimeters, the one with more sides has a larger area
Best Answer
It suffices to prove that, for all $x > 2$, $$(\ln f(x))' < 0, $$ or $$\left(\ln \cot \frac{\pi}{x + 1} - \ln\cot \frac{\pi}{x} + \ln x - \ln(x + 1)\right)' < 0,$$ or $$\frac{2\pi}{(x+1)^2\sin\frac{2\pi}{x+1}} - \frac{2\pi}{x^2\sin\frac{2\pi}{x}} + \frac{1}{x} - \frac{1}{x + 1} < 0. \tag{1} $$
Note that $g(u) := \frac{u}{\sin u}$ is non-decreasing on $(0, \pi)$. We have $$\frac{2\pi}{(x + 1)\sin \frac{2\pi}{x+ 1}} \le \frac{2\pi}{x\sin \frac{2\pi}{x}}. \tag{2}$$ (Note: We have $g'(u) = \frac{\sin u - u\cos u}{\sin^2 u} \ge 0$ on $(0, \pi)$.)
From (1) and (2), it suffices to prove that, for all $x > 2$, $$\frac{1}{x + 1} \cdot \frac{2\pi}{x\sin \frac{2\pi}{x}} - \frac{2\pi}{x^2\sin\frac{2\pi}{x}} + \frac{1}{x} - \frac{1}{x + 1} < 0,$$ or $$\sin \frac{2\pi}{x} < \frac{2\pi}{x}$$ which is true.
We are done.