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?

Let $z=e^{ \frac{i\pi}{n}},$ for example. If we can show it for this point, it is true in general since this is and its conjugate are the closest to the point $z=1$.
$$|z-1|^2 = (\cos \frac{\pi}{n} -1)^2 + \sin^2 \frac{\pi}{n} = 2 \left( 1-\cos \frac{\pi}{n} \right).$$
$$|z-1|^2=2\left[ 1-\left(1-\frac{1}{2!}\left(\frac{\pi}{n} \right)^2 -\frac{1}{4!}\left( \frac{\pi}{n}\right )^4 + \frac{1}{6!}\left( \frac{\pi}{n}\right )^6 -\cdots\right) \right] $$
$$|z-1|^2=\left(\frac{\pi}{n} \right)^2 -\frac{2}{4!}\left( \frac{\pi}{n}\right)^4 + \frac{2}{6!}\left( \frac{\pi}{n}\right )^6 -\cdots$$
When $n=3, |z-1|=1.$
When $n>3$,
$$|z-1|^2 > \left(\frac{\pi}{n} \right)^2 -\frac{1}{12}\left( \frac{\pi}{n}\right)^4 > \frac{4}{(n-1)^2}$$
So
$$|z-1|\ge\frac{2}{n-1}, \quad n\ge3.$$
