67th Romanian Mathematical Olympiad

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

$$n-\frac{x_1}{a_1}-\sum _{i=1}^{n-1}\frac{x_{i+1}-x_i}{a_{i+1}}=n-\frac{x_n}{a_n}-\sum _{i=1}^{n-1}{x}_i\left(\frac{1}{a_i}-\frac{1}{a_{i+1}}\right)\le n,$$ 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 $x_2=x_3=\cdots =x_n=0$, that is, $a_2=a_1$, $a_3=2a_1,\dots ,a_n=2^{n-2}{a}_1$.