An exercise concerning complex numbers Assume that $| z + 1 | > 2$. Show that $|z^3 + 1| > 1$.
My try was:
$$|z^3 + 1| = |z + 1| |z^2 - z + 1| > 2 |z^2 - z + 1| $$
but I'm stuck proving that $|z^2 - z + 1| > \frac 1 2$

Write $z = -1+ w$, so $|z+1|>2$ means $|w| > 2$.  Then 
$z^3 + 1 = w^3 - 3 w^2 + 3 w = w (w^2 - 3 w + 3)$.  The claim is that 
if $|w| > 2$, $|w^2 - 3 w + 3| > 1/2$. If $w = r \exp(i\theta)$, 
$$|w^2 - 3 w + 3|^2 = r^4 - 6 r^3 \cos(\theta) + (3 + 12 \cos(\theta)^2) r^2 - 18 r \cos(\theta)+ 9$$
Call that $F(r,\theta)$.  We have 
$$ \dfrac{\partial F}{\partial \theta} =  6 r \sin(\theta) (r^2 - 4 r \cos(\theta) + 3) $$
Thus for any $r > 2$ the minimum of $F(r,\theta)$ must occur at 
one of the values $\theta = 0, \pi$, or $\pm \arccos((r^2+3)/(4r))$
(note that $0 < (r^2+3)/(4r) \le 1$ if $2 < r \le 3$).
For $r>2$ we  have $$\eqalign{F(r,0) &= (r^2 - 3 r + 3)^2 > 1\cr
F(r,\pi) &= (r^2+3r+3)^2 > 169\cr F(r,\pm \arccos((r^2+3)/(4r))) &= 
\dfrac{(r^2-3)^2}{4} > \dfrac{1}{4}\cr}$$
Thus the minimum is greater than $1/4$, which establishes the claim.
