If $\left| z^3+\frac{1}{z^3} \right| \leq 2$, prove that $\left| z+\frac{1}{z} \right| \leq 2$.
My approach:
Let $z=x+yi$, where $x, y \in \mathbb{R}$ and $i=\sqrt{-1}$ is the imaginary unit.
Factor the expression$$\left| z^3+\frac{1}{z^3} \right| = \left| \Big(z+\frac{1}{z}\Big)\Big(z^2+\frac{1}{z^2}-1\Big)\right|=\left|z+\frac{1}{z}\right|\left|z^2+\frac{1}{z^2}-1\right|$$
Now use that fact that $$\left|\sum_{i=1}^{k}z_i\right| \leq \sum_{i=1}^{k}\left| z_i \right|$$
Hence, $$\left|z^2+\frac{1}{z^2}-1\right| \leq \left|z^2\right|+\left|\frac{1}{z^2}\right| - 1$$
But $$\left|z^2\right|+\left|\frac{1}{z^2}\right| = x^2+y^2 + \frac{1}{x^2+y^2} \leq 2$$
Which implies $$\left|z^2\right|+\left|\frac{1}{z^2}\right|-1 \leq \implies \left|z^2+\frac{1}{z^2}-1\right| \leq 1$$
Which implies $$\left| z + \frac{1}{z}\right| \leq 2$$
Did I miss something or is there a better way to prove the inequality?
Thank you :)
$$2\ge|a^3+b^3|=|(a+b)^3-3ab(a+b)|$$
$$\ge|a+b|^3-3|ab(a+b)|$$