Let $P(z) = 2z - z^3 + 6z^4 + z^6$ with roots $\alpha_1, \alpha_2, \cdots, \alpha_6$. Using the identity $ ||z|-|\omega|| \leq |z+\omega| \leq |z| + |\omega|$ show that if $|z| > 3$, then $|2z - z^3 + 6z^4| < |z^6|$.
In a previous sub-question it was also proven that $P(z)$ must have at least two real roots, but I have no idea where to go from here. I would greatly appreciate any advice as to how to solve this. Thanks in advance.
If $|z| > 3$ then: $|2z - z^3 + 6z^4| = |z|\cdot |2 - z^2 + 6z^3| \leq |z|\cdot \left(2 + |z|^2 + 6|z|^3\right) < |z|\cdot \left(|z|^3 + |z|^3 + 6|z|^3\right) = 8|z|^4 < |z|^2\cdot |z|^4 = |z|^6$.