Complex number inequality $|z-1| \ge \frac{2}{n-1}$ 
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$

I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq  -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n-1}+z^{n-2}+...+z^2+z+1|}$$
Also $|z|=1$, so if I use triangle inequality, I find:
$$|z-1|\ge \frac{2}{|z^{n-1}|+...+|z|+1}=\frac{2}{n}$$
but this is lower than $\frac{2}{n-1}$. Does this help in any way?

The $n$ roots of $z^n=-1$ are $n$ equidistant points on the unit circle. $|z-1|$ is simply the distance between one such root and $1$. Therefore, $|z-1|$ is minimized when $z=e^{\frac{\pi i}{n}}$. Therefore, $|z-1|\geq|e^{\frac{\pi i}{n}}-1|=2\sin{\frac{\pi}{2n}}$(Which can be found using geometry or algebra, as one wishes).
It only remains to show that $\sin{\frac{\pi}{2n}}\geq\frac{1}{n-1}$.
Define $f(x)=(x-1)\sin{\frac{\pi}{2x}}-1$. We will prove that $f(x)\geq0$ for $x\geq3$ and that would imply the claim. Notice that $f(3)=0$. If we now prove $f'(x)\geq 0$ for $x\geq3$, we would be done. This follows as $f'(x)=\sin{\frac{\pi}{2x}}-\frac{\pi(x-1)}{2x^2}\cos{\frac{\pi}{2x}}\geq\sin{\frac{\pi}{2x}}-\frac{\pi}{2x}\cos{\frac{\pi}{2x}}=\cos{\frac{\pi}{2x}}(\tan{\frac{\pi}{2x}}-\frac{\pi}{2x})\geq0$. Note the last inequality is true as $\tan{t}\geq t,\ \forall\ t\in[0,\frac{\pi}{2}]$.
