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 $a + b + c \ge 56$.