Give me some time to write down, index is hard to be writed

We consider $a_{kt}=: \sum_{i=1}^k m_i+m_t$, where $0 \leq k \leq r$, $k+1 \leq t \leq s$, and we define $\sum_{i=1}^0 m_i=0$. If $a_{kt}=a_{k_1t_1}$, then if $k=k_1$, we have $t=t_1$; if $k<k_1$, we can know $t>t_1$, and $m_t=m_{k+1}+\cdots+m_{k_1}+m_{t_1}$, a contradiction. So these $a_{kt}$ are distinct, the number of $a_{kt}$ is $\sum_{k=0}^r (s-k)=s(r+1)-\frac{1}{2}r(r+1)$, it is clear $a_{sr}$ is the maximal, then we know $a_{sr}=m_s+\sum_{i=1}^r m_i \geq s(r+1)-\frac{1}{2}r(r+1)$. Clearly, $m_r-m_i \geq r-i$, we can get $m_i \leq m_r-(r-i)$($1 \leq i \leq r$). So $$m_s+rm_r-\sum_{i=1}^r (r-i) \geq m_s+\sum_{i=1}^r m_i \geq s(r+1)-\frac{1}{2}r(r+1),$$now we have $$m_s+rm_r \geq s(r+1)-r> (s-1)(r+1).$$