Inequalities involving the Roots of unity Let $\epsilon \ne 1$ be a nth root of unity. Then prove that,
$\bullet \ |1-\epsilon|\ge\frac{2}{n-1}$
$\bullet |\sin\frac{k}{n}| \ge \frac{1}{n-1}$
Here are my solutions:

$\bullet$ To prove $\implies \ |1-\epsilon| \ge\frac{2}{n-1}$

We have $$\sum_{i=1}^{n-1} {\epsilon}^i = -1 $$ 
$$\implies \sum_{i=1}^{n-1} ({\epsilon}^i -1 ) = -n $$
$$\implies (\epsilon - 1) (\epsilon^{n-2} + 2\epsilon^{n-3} + \cdots + (n-2)\epsilon + n-1) = -n $$
Taking absolute value on both sides
$$|\epsilon -1||\epsilon^{n-2} + 2\epsilon^{n-3} + \cdots + (n-2)\epsilon + n-1| = n$$
$$\implies |1-\epsilon| (|\epsilon^{n-2}| + 2 |\epsilon^{n-3}| + \cdots (n-2)|\epsilon| + n-1) \ge n$$
$$\implies |1-\epsilon| \cdot \frac{n(n-1)}{2} \ge n $$
$$\implies \boxed{|1-\epsilon| \ge \frac{2}{n-1}}$$

$\bullet$ To prove $\implies |\sin\frac{k\pi}{n}| \ge \frac{1}{n-1}$

Using $|1-\epsilon| \ge \frac{2}{n-1}$
we have,
$$| 1- \cos\frac{2k\pi}{n} -i\sin\frac{2k\pi}{n}| \ge \frac{2}{n-1}$$
$$\implies |2\sin^2\frac{k\pi}{n} - 2i\sin\frac{k\pi}{n} \cdot \cos\frac{k\pi}{n}| \ge \frac{2}{n-1}$$
$$\implies |\sin\frac{k\pi}{n}||\sin\frac{k\pi}{n} -i\cos\frac{k\pi}{n}| \ge \frac{1}{n-1}$$
$$\boxed{|\sin\frac{k\pi}{n}| \ge \frac{1}{n-1}}$$
Do you have any other solution?!! 

HINT:
Let $f(x)=\sin\left(\frac{\pi}{x}\right)-\frac{1}{x-1}$ for $x\ge 2$. Note that $f(0)=0$ and $\lim_{x\to \infty}f(x)=0$.  The second derivative is given by
$$f''(x)=-\frac{2}{(x-1)^3}+\frac{2\pi\cos\left(\frac{\pi}{x}\right)}{x^3}-\frac{\pi^2\sin\left(\frac{\pi}{x}\right)}{x^4}$$
Show that $f''(x)>0$ for $x\ge 2$ and conclude from the concavity that $f(x)> 0$ for $x>2$.
