75th Romanian Mathematical Olympiad

a. If $m\le 1$, then $5a+6b+7c+6\le 4a+3b+2c+3$, that is $a+3b+5c+3\le 0$ – impossible. So $m\ge 2$.

b. If $n\ge 2$, then $a+2b+3c+5\ge 6a+2b+4c+10$, whence $0\ge 5a+c+5$, false. Since $n>0$, $n$ must be equal to 1. Now $n=1$ yields $a+2b+3c+5=3a+b+2c+5$, hence $b+c=2a$. Suppose $m\ge 3$. Then $5a+6b+7c+6\ge 12a+9b+6c+9$, therefore $c\ge 7a+3b+3$. This gives $b+c\ge 7a+4b+3$, that is $2a\ge 7a+4b+3$, hence $0\ge 5a+4b+3$, false. So $m<3$, and a) implies $m=2$. The above show that the only possibility is $m=2$ and $n=1$. These values are indeed achieved for $a=b=c$.