<s>[INST] Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1.$ 
Let $n$ be a positive integer and let $z$ be a root of $x^n-x-1.$ Show that $2 \operatorname{Re}(z-1/z) > 1/|z|^2 - 1$.

Solution. Write $z = re^{i\theta}, r > 0$ as $0$ is obviously not a root. Then $$1+2r\cos \theta = |z+1|^2-r^2 = r^{2n} - r^2.$$ We need to show that $$2(r\cos\theta - 1/r \cos \theta) =  2 \operatorname{Re}(re^{i\theta} - 1/r e^{-i\theta}) > 1/r^2 - 1.$$ We know that $2r\cos \theta = r^{2n}-r^2 - 1\Rightarrow \cos \theta = \dfrac{r^{2n}-r^2-1}{2r}$ from above. So we need to show that $$r^{2n}-r^2 - 1 - \dfrac{r^{2n}-r^2-1}{r^2} > 1/r^2 - 1\Leftrightarrow r^{2n+2} - r^4 - r^2 - r^{2n} + r^2 + 1 - 1 + r^2>0 \\ \Longleftrightarrow r^{2n+2} - r^{2n} - r^4 + r^2 = r^2(r^{2n} - r^{2n-2} -r^2 + 1).$$
I don't think I've made any algebraic mistakes, so I was wondering how I can show the last quantity is positive?
 [/INST] $n>1$ and your last parenthesis is $(r^2-1)(r^{2n-2}-1),$ so you just have to prove that $r\ne1.$
Let us prove it by contradiction. If $r=1$ then (by what you already proved) $\cos\theta=-\frac12$ hence $z=e^{\pm i2\pi/3},$ but such a $z$ satisfies $z+1=-z^2\ne z^n.$
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