THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Since $a,b,c\in \mathbb{N}$, from $a>b>c$ follows $a\ge b+1\ge c+2$, hence $a-c\ge 2$. If $a-c\ge 4$, then $13c>11a\ge 11(c+4)$, so $c>22$. It follows that $a+b+c>c+c+c>66>56$. So we still need to discuss the cases $a-c=2$ and $a-c=3$. If $a-c=2$, then from $a\ge b+1\ge c+2$ follows $a=b+1=c+2$, hence $12(c+1)>13c>11(c+2)$, where from $12>c$ and $c>11$, which is impossible, and we conclude $a-c\ne 2$. If $a-c=3$, then $a=c+3$ and either $b=c+1$ or $b=c+2$. If $b=c+1$, then $12(c+1)>13c>11(c+3)$, so $12>c$ and $c>\frac{33}{2}$, which is impossible. If $b=c+2$, then $12(c+2)>13c>11(c+3)$, so $24>c$ and $c>\frac{33}{2}$, hence $c\ge 17$. It follows that $b\ge 19$, $a\ge 20$ and $a+b+c\ge 56$.