Question Number 178 by 123456 last updated on 14/Dec/14
$$\mathrm{proof}\:\mathrm{that} \\ $$$$\mid\mathrm{sin}\:\mathrm{1sin}\:\mathrm{2}...\mathrm{sin}\:{n}\mid\leqslant\mathrm{sin}\:\frac{\pi}{{n}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{{n}}...\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}} \\ $$$$\mathrm{for}\:\forall{n}\in\mathbb{N}\backslash\left\{\mathrm{0},\mathrm{1}\right\} \\ $$