# Help proving $\forall x \geq 2: \cos(2 \pi/x) \leq 1 - 1/x^2$

I want to prove the following inequality: $$\forall x \geq 2: \cos(2 \pi/x) \leq 1 - 1/x^2$$ I believe it holds: numerical evidence.

I tried to prove it using taylor series, but it didn't work so well. Any ideas?

Note that by Taylor's series $\forall x$

$$\cos x\le 1-\frac{x^2}2+\frac{x^4}{24}$$

thus

$$\cos \frac{2\pi}{x}\le 1-\frac{2\pi^2}{x^2}+\frac{16\pi^4}{24x^4}\le 1-\frac{1}{x^2}$$

indeed

$$1-\frac{2\pi^2}{x^2}+\frac{2\pi^4}{3x^4}\le 1-\frac{1}{x^2}\iff\frac{1}{x^2}-\frac{2\pi^2}{x^2}+\frac{2\pi^4}{3x^4}\le0$$

$$\iff 3x^2-6\pi^2x^2+2\pi^4\le0\iff x^2(6\pi^2-3)\ge2\pi^4\iff x^2\ge \frac{2\pi^4}{6\pi^2-3}\approx3.465$$

You do not need Taylor series. The given inequality is equivalent to $\cos(2\pi z)\leq 1-z^2$ for any $z\in\left(0,\frac{1}{2}\right]$, or to $\cos^2(\pi z)\leq 1-\frac{z^2}{2}$, or to $\frac{z^2}{2}\leq \sin^2(\pi z)$, or to $\sin(\pi z)\geq \frac{z}{\sqrt{2}}$ over the same range, which is trivial by the concavity of the sine function.

An improved inequality is $\sin(\pi z)\geq 2z$, leading to $\cos(2\pi z)\leq 1-8z^2$ and $$\forall x\geq 2,\quad \cos\left(\frac{2\pi }{x}\right)\leq 1-\frac{\color{red}{8}}{x^2}.$$