Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$. Problem :

Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.

My approach :
Since :
$(a^3+b^3)=(a+b)^3-3ab(a+b)$
$\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})$
Now I don't know further about this problem whether or not this will help here. Please guide to solve this problem will be of great help, thanks.

You have $(z+\frac{1}{z})^3-3(z+\frac{1}{z})=z^3+\frac{1}{z^3}$, and $\left|(z+\frac{1}{z})^3-3(z+\frac{1}{z})\right|=\left|z+\frac{1}{z}\right|\cdot\left|(z+\frac{1}{z})^2-3\right|$.
Hence if $\left|z+\frac{1}{z}\right|>2$, then $\left|z^3+\frac{1}{z^3}\right|>2\cdot(4-3)=2$. Contradiction.
