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I proved the inequality shown in the title by using AM-GM and: $$\cos A + \cos B + \cos C \le \frac{3}{2}$$ Equality holds for an equilateral triangle.

For an $n$-sided convex polygon with internal angles $\alpha_1, \alpha_2, \dots, \alpha_n$, it appears to me that $\prod^n_{k=1}1 - \cos\alpha_k$ is maximum when $\alpha_1 = \alpha_2 = \dots = \alpha_n$, i.e. when $\alpha_i = \frac{(n-2)\pi}{n}$, thus $$\prod^n_{k=1}1 - \cos(\alpha_k) \le \cos^n \pi\left(1-\frac{2}{n}\right)$$ How can this inequality be proved or disproved?

Michael Rozenberg
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Ojas
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    See https://math.stackexchange.com/questions/779565/prove-sin-frac-alpha2-sin-frac-beta2-sin-frac-gamma2-leq-frac and https://math.stackexchange.com/questions/949530/in-triangle-abc-show-that-1-lt-cos-a-cos-b-cos-c-le-frac-32 – lab bhattacharjee Aug 26 '17 at 15:52

2 Answers2

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Let's take $f(x)=\log(1-cos(x))$

We have $f''(x)=\dfrac{1}{cos(x)-1}$ for $0\lt x\lt 2\pi$ (By Wolfram Alfa)

So $f(x)$ is a concave function in $(0,2\pi)$.

Because our polygon is convex, all $a_i$'s are in $(0,2\pi)$

By Using Jensen Inequality for $f(x)$

$$f(a_1)+f(a_2)+\cdots+f(a_n)\leq nf(\dfrac{a_1+a_2+\cdots+a_n}{n})=nf(\dfrac{(n-2)\pi}{n})$$

$$\Rightarrow \log(1-cos(a_1))+\cdots+\log(1-cos(a_n))\leq n\log(1-cos(\dfrac{(n-2)\pi}{n}))$$

$$\Rightarrow \log\left(\prod^n_{k=1}(1 - \cos(a_k)) \right)\leq \log\left [\left(1-cos(\dfrac{(n-2)\pi}{n})\right)^n\right ]$$

$$\Rightarrow \prod^n_{k=1}(1 - \cos(a_k)) \leq \left(1-cos(\dfrac{(n-2)\pi}{n})\right )^n$$

Merdanov
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For the case $n = 3$ which is of immediate need and fun to see, use the well-known identity: $1- \cos X = 2\sin^2(\frac{X}{2})$,and follows by AM-GM inequality: $xyz \le \dfrac{(x+y+z)^3}{27}$ for $x = \sin(\frac{A}{2}), y = \sin(\frac{B}{2}), z = \sin(\frac{C}{2})$,and finally use concavity of $\sin X$ to arrive at the result. This can be generalized to any $n \ge 3$.

DeepSea
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