An Angle Inequality in Simple Polygons

Dan Sitaru posted an elegant inequality and its proof at the CutTheKnotMath facebook page.

Let $A_1A_2\ldots A_n,$ $n\ge 3,$ be a simple (without self intersections) polygon, with the inner angles $\alpha_1,$ $\alpha_2,\ldots$ Then

$\displaystyle\prod_{k=1}^{n}(1+\alpha_k)\le \sum_{k=1}^{n}\frac{\pi^{k}(n-2)^{k}}{k!}.$


In a triangle with angles $\alpha,$ $\beta,$ $\gamma,$

$\displaystyle (1+\alpha )(1+\beta )(1+\gamma )\le 1+\pi+\frac{\pi^{2}}{2}+\frac{\pi^{3}}{6}.$

In a simple quadrilateral with angles $\alpha,$ $\beta,$ $\gamma,$ $\delta$

$\displaystyle (1+\alpha )(1+\beta )(1+\gamma )(1+\delta)\le 1+2\pi+2\pi^{2}+\frac{4\pi^{3}}{3}+\frac{2\pi^{4}}{3}.$


Define $\displaystyle S=\sum_{k=1}^{n}A_{k}$ (which is known to be $(n-2)\pi ).$ By the Geometric Mean - Arithmetic Mean inequality,

$\displaystyle\prod_{k=1}^{n}(1+\alpha_k)\le \left(1+\frac{S}{n}\right)^{n}.$

Implying the Binomial expansion,

$\displaystyle\begin{align} \left(1+\frac{S}{n}\right)^{n}&=\sum_{k=1}^{n}\frac{S^{k}}{n^{k}}{n \choose k}\\ &\le\sum_{k=1}^{n}\frac{S^{k}}{k!}\\ &=\sum_{k=1}^{n}\frac{\left[\pi(n-2)\right]^{k}}{k!}. \end{align}$

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