I have found this weird inequality in a mathematical magazine. $$\sin \left(\frac{\pi}{4n}\right) \ge \frac{\sqrt{2}}{2n}, n\in \mathbb{N}$$
I tried using induction, but couldn't continue. Then, using the double angle formula, I managed to get an equivalent inequality: $\cos\left(\frac{\pi}{2n}\right) \le 1-\frac{1}{n^2}$ and tried proving it with induction, but, again, it couldn't do it.
Could you help me? I am not very good with trigonometric inequalities…
Best Answer
$\sin$ is a concave function on $[0,\frac{\pi}{4}]$, because $\forall x \in [0,\frac{\pi}{4}], sin’’(x)=-sin(x) \leq 0$. Therefore, the function is above all its chords, and, in particular, above the one that goes through the points $(0,0)$ and $(\frac{\pi}{4}, \frac{\sqrt{2}}{2})$. Hence:
$$\forall x \in [0, \frac{\pi}{4}], \sin x \geq \frac{2 \sqrt{2}}{\pi} x$$
The conclusion follows.