21st Mediterranean Mathematical Competition

For $k,l$ with $0\le k\le r$ and $k+1\le l\le s$, we introduce the auxiliary value $$T(k,l):=m_l+\sum _{i=1}^k{m}_i.$$

We claim that these $\frac{1}{2}(r+1)(2s-r)$ auxiliary values are all pairwise distinct: Let us assume that $T(k,\ell )=T(u,v)$. Without loss of generality $k\le u$, so that this equality turns into $$m_{\ell }=m_v+\sum _{i=k+1}^u{m}_i.$$ But then $m_{\ell }$ can be written as a sum of distinct other numbers in the sequence, unless $\ell =v$ and $k=u$ holds. Hence the auxiliary values indeed are distinct. As altogether there are $\frac{1}{2}(r+1)(2s-r)$ auxiliary values, the largest value $T(r,s)$ must be at least $\frac{1}{2}(r+1)(2s-r)$. This yields $$\frac{1}{2}(r+1)(2s-r)\le T(r,s)=m_s+\sum _{i=1}^r{m}_i\le m_s+\sum _{i=1}^r(m_r-r+i).$$ Here we used $m_i\le m_r-r+i$, which follows as the sequence is increasing. The above inequality can be rewritten into $$rs+s-r\le r\cdot m_r+m_s,$$ which immediately implies the desired inequality from the problem statement.