Romanian Mathematical Olympiad

a)
We easily notice that
$$\begin{aligned} 2|z^{n+1} + \frac{1}{z^{n+1}}| &< |z + \frac{1}{z}| \cdot |z^{n+1} + \frac{1}{z^{n+1}}| \\ & = |z^n + \frac{1}{z^n} + z^{n+2} + \frac{1}{z^{n+2}}| \leq |z^n + \frac{1}{z^n}| + |z^{n+2} + \frac{1}{z^{n+2}}|. \end{aligned}$$

Alternative Solution. Consider the sequence $(\alpha_n)_{n \ge 1}$ given by $\alpha_n = z^n + \frac{1}{z^n}$. Extend to the left with the term $\alpha_0 = z^0 + \frac{1}{z^0} = 2$, and denote $\alpha = \alpha_1$. Clearly $a_n = |\alpha_n|$. We have
$$\alpha\alpha_n = \left(z + \frac{1}{z}\right) \left(z^n + \frac{1}{z^n}\right) = \left(z^{n+1} + \frac{1}{z^{n+1}}\right) + \left(z^{n-1} + \frac{1}{z^{n-1}}\right) = \alpha_{n+1} + \alpha_{n-1}$$
for all $n \ge 1$, so the sequence $(\alpha_n)_{n \ge 0}$ satisfies the linear recurrence relation $\alpha_{n+1} = \alpha\alpha_n - \alpha_{n-1}$. Then for $|\alpha| > 2$ we have
$$a_n = |\alpha_n| = \left| \frac{\alpha_{n+1} + \alpha_{n-1}}{\alpha} \right| \le \frac{|\alpha_{n+1}| + |\alpha_{n-1}|}{|\alpha|} < \frac{a_{n+1} + a_{n-1}}{2},$$
i.e. the sequence $(a_n)_{n \ge 0}$ is convex.
But then, if $a_1 = |\alpha| > 2 = a_0$, any convex sequence is (strictly) increasing, since from $a_n > a_{n-1}$ follows $a_{n+1} > 2a_n - a_{n-1} = a_n + (a_n - a_{n-1}) > a_n$, and the thesis is proved by simple induction. Conversely, if there exists $k \in \mathbb{N}^*$ such that $a_k \le 2$, then $a_1 \le 2$. Therefore the proof for b) comes directly from a), and the nature of the sequence is not anymore relevant.

b)
$$a_2 = |z^2 + \frac{1}{z^2}| = \left|\left(z + \frac{1}{z}\right)^2 - 2\right| \ge \left(z + \frac{1}{z}\right)^2 - 2 = a_1^2 - 2 > a_1,$$
therefore the sequence $(a_n)_n$ is strictly increasing, hence $a_k \ge a_1 > 2$ for all $k$, a contradiction.