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?
 A: 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
A: 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
$$|z-1|\ge\frac{2}{n-1}, \quad n\ge3.$$
A: If $z^n = -1$, we have
$$
z = \exp\left(\frac{(2k + 1)i \pi}{n}\right),
\qquad k \in \{0, \ldots,n - 1\}
$$
Denote for $k \in \{0, \ldots,n - 1\}$
$$
a_k := \frac{(2k + 1) \pi}{n}
$$
Then
\begin{align}
| z - 1 |^2
& = | \cos(a_k) + i \sin(a_k) - 1 |^2
= (\cos(a_k) - 1)^2 + \sin^2(a_k) \\
& = \cos^2(a_k) + \sin^2(a_k) - 2 \cos(a_k) + 1 \\
& = 2 - 2 \cos(a_k),
\end{align}
therefore
$$
| z - 1 |
= \sqrt{2} \sqrt{1 - \cos(a_k)}.
$$
As $\sqrt{2} \left| \sin\left(\frac{x}{2}\right)\right| = \sqrt{1 - \cos(x)}$, we have
$$
| z - 1| = 2 \left| \sin\left(\frac{(2k + 1)\pi}{2n}\right) \right|.
$$
Therefore it remains to show that
$$
\left| \sin\left(\frac{\left(k + \frac{1}{2}\right)\pi}{n}\right) \right|
\ge \frac{1}{n - 1}, \qquad
k \in \{0, \ldots,n - 1\}.
$$
Also, we can exclude one $k$, which is the $k$ for which $a_k = 0$.
Notice that $\left| \sin\left(\frac{\left(k + \frac{1}{2}\right)\pi}{n}\right) \right|$ is smallest (exercise!) for $k = 0$ and $k = n - 1$ (they yield the same result since $\frac{\pi}{2n} = \frac{(n - 1/2) \pi}{n}$), so it suffices to show it for $k = 0$, which is to say
$$
\left| \sin\left(\frac{\pi}{2n}\right) \right|
\ge \frac{1}{n - 1}.
$$
But this follows from the concavity (similar to here) of $x \mapsto \sin(x)$ on $\left[0, \frac{\pi}{6}\right]$ and $n \ge 3$.
A: 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 $\tan{t}\geq t,\ \forall\ t\in[0,\frac{\pi}{2}]$.
