Exercise 3. Find all integers $n \geqslant 3$ such that, if $a_{1}, \ldots, a_{n}$ are strictly positive real numbers such that $\max \left(a_{1}, \ldots, a_{n}\right) \leqslant n \cdot \min \left(a_{1}, \ldots, a_{n}\right)$, then there necessarily exist three of these real numbers that are the lengths of the sides of an acute triangle, that is, a triangle whose three angles are strictly acute.
Solution to Exercise 3 Without loss of generality, we assume that $1=a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}$. We then note that for three real numbers $a_{i}, a_{j}$, and $a_{k}$, with $i < j < k$, if $a_{i}^{2} + a_{j}^{2} < a_{k}^{2}$ (which can be seen, for example, using the Law of Cosines), then $a_{k-2}, a_{k-1}$, and $a_{k}$ are also the sides of an acute triangle. Consequently, let us call a sequence of real numbers $1=a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}$ obtuse if $a_{k}^{2} \geqslant a_{k-2}^{2} + a_{k-1}^{2}$ for all $k \in \{3, \ldots, n\}$. The problem then reduces to finding the integers $\mathrm{n} \geqslant 3$ such that if $a_{1}, \ldots, a_{n}$ is an obtuse sequence, then $a_{n} > n$.

Let $F_{n}$ be the $n^{\text{th}}$ Fibonacci number, defined by $F_{1}=F_{2}=1$ and $F_{n+2}=F_{n+1}+F_{n}$ for all $n \geqslant 1$. In an obtuse sequence $a_{1}, \ldots, a_{n}$, we necessarily have $1 \leqslant a_{1} \leqslant a_{2}$, i.e., $F_{1} \leqslant a_{1}^{2}$ and $F_{2} \leqslant a_{2}^{2}$. An immediate induction on $n$ then shows that $a_{n}^{2} \geqslant F_{n}$. Conversely, we observe that the sequence $a_{1}, \ldots, a_{n}$ defined by $a_{k}=\sqrt{F_{k}}$ is obtuse. Therefore, the integers $n \geqslant 3$ we are looking for are those such that $\sqrt{F_{n}}>n$, i.e., $F_{n}>n^{2}$.

The first terms of the Fibonacci sequence are $\mathrm{F}_{1}=\mathrm{F}_{2}=1, \mathrm{~F}_{3}=2, \mathrm{~F}_{4}=3, \mathrm{~F}_{5}=5, \mathrm{~F}_{6}=8$, $\mathrm{F}_{7}=13, \mathrm{~F}_{8}=21, \mathrm{~F}_{9}=34, \mathrm{~F}_{10}=55, \mathrm{~F}_{11}=89, \mathrm{~F}_{12}=144, \mathrm{~F}_{13}=233$ and $\mathrm{F}_{14}=377$. We observe that $F_{n} \leqslant n^{2}$ for all $n \leqslant 12$, and that $F_{n}>n^{2}$ for $n=13$ and $n=14$. Furthermore, for any integer $n \geqslant 13$ such that $F_{n}>n^{2}$ and $F_{n+1}>(n+1)^{2}$, we observe that

$$
F_{n+2}=F_{n}+F_{n+1}>n^{2}+(n+1)^{2} \geqslant 5 n+(n+1)^{2} \geqslant 2 n+3+(n+1)^{2}=(n+2)^{2}
$$

An immediate induction then shows that the integers we are looking for are the integers $\mathrm{n} \geqslant 13$.