75th Romanian Mathematical Olympiad

a) We notice that $(b_n{)}_{n\ge 1}$ is non-decreasing. Also, the sequence $(a_n{)}_{n\ge 1}$ is non-increasing, since $0\le a_{mn}\le a_n-a_{n+1}$. Moreover, $$\sum _{k=1}^n{a}_{mk}\le a_1-a_{n+1}\le a_1.$$

Using the monotonicity of $(a_n)$ and $(b_n)$, we have: $$b_n\le b_{mn}=\sum _{k=1}^{mn}{a}_k=\sum _{i=1}^{m-1}{a}_i+\sum _{k=1}^n{a}_{mk}+\sum _{k=1}^{n-1}\sum _{j=1}^{m-1}{a}_{mk+j}.$$ By monotonicity, $$\sum _{k=1}^{n-1}\sum _{j=1}^{m-1}{a}_{mk+j}\le (m-1)\sum _{k=1}^{n-1}{a}_{mk}\le (m-1)a_1,$$ hence, $$b_n\le \sum _{i=1}^{m-1}{a}_i+\sum _{k=1}^n{a}_{mk}+(m-1)a_1\le \sum _{i=1}^{m-1}{a}_i+m{a}_1,$$

b) We first prove that the sequence $d_n={\sum }_{k=1}^{n}k{a}_k$ is bounded above. Clearly, $(d_n{)}_{n\ge 1}$ is non-decreasing. Moreover, $$\sum _{k=1}^{n}k{a}_{mk}\le \sum _{k=1}^{n}k(a_k-a_{k+1})=a_1+\sum _{k=2}^n(k-(k-1))a_k-n{a}_{n+1}\le b_n,$$ so the sequence ${\left({\sum }_{k=1}^{n}k{a}_{mk}\right)}_{n\ge 1}$ is bounded above. On the other hand, $$\begin{array}{rl}{d}_n\le d_{mn}& =m\sum _{k=1}^{n}k{a}_{mk}+\sum _{k=1}^{m-1}k{a}_k+\sum _{j=1}^{m-1}\sum _{k=1}^{n-1}(mk+j)a_{mk+j}\\ & \le m{b}_n+d_{m-1}+(m-1)\sum _{k=1}^{n-1}m(k+1)a_{mk}\\ & \le m{b}_n+d_{m-1}+m(m-1)(b_{n-1}+a_1),\end{array}$$ hence $(d_n{)}_{n\ge 1}$ is bounded above.

Again, $(c_n{)}_{n\ge 1}$ is non-decreasing. Similarly, $$\begin{array}{rl}\sum _{k=1}^n{k}^2{a}_{mk}& \le \sum _{k=1}^n{k}^2(a_k-a_{k+1})=a_1+\sum _{k=2}^n(k^2-(k-1{)}^2)a_k-n^2{a}_{n+1}\\ & \le a_1+\sum _{k=2}^{n}2k{a}_k\le 2d_n,\end{array}$$ so the sequence ${\left({\sum }_{k=1}^n{k}^2{a}_{mk}\right)}_{n\ge 1}$ is bounded above. Finally, $$\begin{array}{rl}{c}_n\le c_{mn}& =m^2\sum _{k=1}^n{k}^2{a}_{mk}+c_{m-1}+\sum _{j=1}^{m-1}\sum _{k=1}^{n-1}(mk+j{)}^2{a}_{mk+j}\\ & \le 2m^2{d}_n+c_{m-1}+m^2(m-1)\left(\sum _{k=1}^{n-1}{k}^2{a}_{mk}+2\sum _{k=1}^{n-1}k{a}_{mk}+\sum _{k=1}^{n-1}{a}_{mk}\right),\end{array}$$ which is a finite sum of bounded above sequences, hence $(c_n{)}_{n\ge 1}$ is bounded above.