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.
 A: You are on the right path; you have shown that
$$\left(z+\tfrac{1}{z}\right)^3=z^3+\tfrac{1}{z^3}+3\left(z+\tfrac{1}{z}\right).$$
Taking norms it follows from the triangle inequality that
$$\left|\left(z+\tfrac{1}{z}\right)^3\right|=\left|z^3+\tfrac{1}{z^3}+3\left(z+\tfrac{1}{z}\right)\right|\leq\left|z^3+\tfrac{1}{z^3}\right|+\left|3\left(z+\tfrac{1}{z}\right)\right|,$$
and rewriting and rearranging the terms a bit yields
$$\left|z+\tfrac{1}{z}\right|\cdot\left(\left|z+\tfrac{1}{z}\right|^2-3\right)\leq\left|z^3+\tfrac{1}{z^3}\right|\leq2.$$
Let $x:=\left|z+\tfrac{1}{z}\right|$ so that we above becomes $x\cdot(x^2-3)\leq2$, or equivalently $x^3-3x-2\leq0$.
Using the rational root test if you like, it's easy to see that $x^3-3x-2=(x-2)(x+1)^2$, from which it follows that $x\leq2$ as desired.
A: 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.
