Romanian Mathematical Olympiad

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

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 $k\in {\mathbb{N}}^{\ast }$ 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.