If $\left|z^3 + {1 \over z^3}\right| \le 2$ then $\left|z + {1 \over z}\right| \le 2$ 
$\displaystyle \left|z^3 + {1 \over z^3}\right| \le 2$ prove that $\displaystyle \left|z + {1 \over z}\right| \le 2$


$$\left|z^3 + {1 \over z^3}\right| = \left(z^3 + {1 \over z^3}\right)\left(\overline z^3 + {1 \over \overline{z}^3}\right) = \left(z + {1\over z}\right)\left(z^2 - 1 + {1\over z^2} \right)\left(\overline z + {1\over \overline z}\right)\left(\overline z^2 - 1 + {1\over\overline z^2} \right)$$
$$=\left|z + {1\over z}\right|^2\left|z^2 - 1 + {1\over z^2} \right|^2 \le 2$$
$$\therefore \left|z + {1\over z}\right| \le \sqrt{2}$$ where $\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1$. 

But I am not able to prove $\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1$, need some help on this. 
 A: Let $a = \lvert z + \frac 1z \rvert$. Then
$$
 a^3 = \left\lvert \left( z + \frac 1z \right)^3 \right\rvert =
 \left\lvert z^3 + 3z + \frac 3z + \frac{1}{z^3}  \right\rvert 
 \le \left\lvert z^3 + \frac{1}{z^3}  \right\rvert
 + 3 \left\lvert z + \frac{1}{z}\right\rvert \le 2 + 3a
$$
so that
$$
0 \ge a^3 - 3a - 2 = (a-2)(a+1)^2
$$
and therefore $a \le 2$.
A: It's all about identities.
Note that $\left( z + \frac 1z\right)^3 = z^3 + \frac 1{z^3} + 3\left(z + \frac 1z\right)$.
Apply the triangle inequality:
$$
\left|\left( z + \frac 1z\right)^3 \right| \leq \left|z^3 + \frac 1{z^3}\right| + 3\left|\left(z + \frac 1z\right)\right|
$$
Using what you know:
$$
\left|\left( z + \frac 1z\right) \right|^3 -  3\left|\left(z + \frac 1z\right)\right| \leq  \left|z^3 + \frac 1{z^3}\right| \leq 2
$$
Suppose that $x = \left|\left( z + \frac 1z\right) \right|$, then $x^3 - 3x \leq 2$ is true, along with $x \geq 0$. 
To solve this, note that $x^3-3x = x(x^2-3)$, which are both increasing functions for $x\geq 0$. So, $x^3-3x$ is also increasing. Hence, we only need to find when $x^3 - 3x=2$, which happens at $x=2$. Hence, we can conclude, by the increasing property, that $0 \leq x \leq 2$ is true, but this is the conclusion of  the problem.
