2024 CGMO

Proof. From the given definition, we have $a_2=\frac{a_1}{2}+\frac{1}{2}=\frac{a_1+a_2}{2}+\frac{1}{3}=\frac{7}{12}$. We now prove by induction that $a_n\le \frac{7}{12}$, with equality if and only if $n=3$. The cases $n=1,2,3$ have been verified. Assume the statement holds for all $1,2,\dots ,n-1$ ($n\ge 4$), then $$\begin{array}{rl}{a}_n& =\frac{a_1+\cdots +a_{\lceil \frac{n}{2}\rceil }}{\lceil \frac{n}{2}\rceil }+\frac{1}{n}\\ & \le \frac{0+\frac{1}{2}+\left(\lceil \frac{n}{2}\rceil -2\right)\cdot \frac{7}{12}}{\lceil \frac{n}{2}\rceil }+\frac{1}{n}\\ & =\frac{7}{12}-\frac{2}{3\lceil \frac{n}{2}\rceil }+\frac{1}{n}\\ & =\frac{7}{12}-\frac{2n-3\lceil \frac{n}{2}\rceil }{3n\cdot \lceil \frac{n}{2}\rceil }\\ & \le \frac{7}{12}-\frac{2n-3\left(\frac{n+1}{2}\right)}{3n\cdot \lceil \frac{n}{2}\rceil }\\ & =\frac{7}{12}-\frac{\frac{n}{2}-\frac{3}{2}}{3n\cdot \lceil \frac{n}{2}\rceil }\\ & <\frac{7}{12}.\end{array}$$ Therefore, the maximum term of the sequence $\{a_n\}$ is $a_3=\frac{7}{12}$. $\square$