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 $n=2m+1$. Observe that $$z^n+1=(z+1)(z^{2m}-z^{2m-1}+\cdots+z^2-z+1)$$ has simple roots (consisting of roots of unity). Let $f(z)$ be the second factor of the right hand side. Then $$f(z)=(z-1)(z^{2m-1}+z^{2m-3}+\cdots+z)+1=0$$ if $z$ is a root of $z^n+1$ and $z\neq -1$.
Applying similar argument as yours, one has $$z-1=\frac{-1}{z^{2m-1}+z^{2m-3}+\cdots+z}$$ $$\Rightarrow |z-1|\geq \frac 1m=\frac 2{n-1},$$ as required. QED
