Simple (?) inequality with complex numbers and absolute value

Question

I am trying to prove the following: $$ \left| \frac{z^{2n}}{2+z^n+ z^{5n} } \right| \leq \frac{|z|^{2n}}{2\left| 1 - |z| \right| } $$ whenever $|z|<1$. Moreover, I am invited to use the following: $$ ||z|-|w|| \leq |z-w| \leq |z|+|w| $$

Answer 1

Whenever $|z|<1$, we have that higher powers of $z$ are smaller, i.e., $|z^m| < |z|$ for $m>1$. Thus \begin{align*} \Big|\frac{z^{2n}}{2+z^n+z^{5n}}\Big| &= \frac{|z|^{2n}}{|2+z^n+z^{5n}|}\\[5pt] &\leqslant \frac{|z|^{2n}}{||2|-|z^n+z^{5n}||}\\[5pt] &\leqslant \frac{|z|^{2n}}{|2-|z+z||}\\[5pt] &= \frac{|z|^{2n}}{2|1-|z||}, \end{align*} as required.