XI APMO

Letting $b_i=a_i/i,(i=1,2,\cdots )$, we prove that $$b_1+\cdots +b_n\ge a_n(n=1,2,\cdots )$$ by induction on $n$. For $n=1$, $b_1=a_1\ge a_1$, and the induction starts. Assume that $$b_1+\cdots +b_k\ge a_k$$ for all $k=1,2,\cdots ,n-1$. It suffices to prove that $b_1+\cdots +b_n\ge a_n$ or equivalently that $$n{b}_1+\cdots +n{b}_{n-1}\ge (n-1)a_n$$ $$\begin{array}{rl}n{b}_1+\cdots +n{b}_{n-1}& =(n-1)b_1+(n-2)b_2+\cdots +b_{n-1}+b_1+2b_2+\cdots +(n-1)b_{n-1}\\ & =b_1+\left(b_1+b_2\right)+\cdots +\left(b_1+b_2+\cdots +b_{n-1}\right)+\left(a_1+a_2+\cdots +a_{n-1}\right)\\ & \ge 2\left(a_1+a_2+\cdots +a_{n-1}\right)=\sum _{i=1}^{n-1}\left(a_i+a_{n-i}\right)\ge (n-1)a_n\end{array}$$