Romanian Mathematical Olympiad

Denote $x_1 = a_1$, $x_k = a_k - (a_{k-1} + \dots + a_1)$, $k = \overline{2, n}$, and observe that $x_{k+1} - x_k = a_{k+1} - 2a_k$, $k = \overline{1, n-1}$. It results
$$2 \left( \frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_4} + \dots + \frac{a_{n-1}}{a_n} \right) = \sum_{i=1}^{n-1} \left( 1 - \frac{x_{i+1} - x_i}{a_{i+1}} \right).$$

$$n - \frac{x_1}{a_1} - \sum_{i=1}^{n-1} \frac{x_{i+1} - x_i}{a_{i+1}} = n - \frac{x_n}{a_n} - \sum_{i=1}^{n-1} x_i \left( \frac{1}{a_i} - \frac{1}{a_{i+1}} \right) \le n,$$
since $x_i \ge 0, \forall i = \overline{1, n}$ and $a_i \le a_{i+1}, \forall i = \overline{1, n-1}$.

Equality holds if and only if $x_2 = x_3 = \dots = x_n = 0$, that is, $a_2 = a_1$, $a_3 = 2a_1, \dots, a_n = 2^{n-2}a_1$.