Let $P$ be an interior point to the hexagon, therefore also interior to the triangles. Denote by $a$, respectively $b$, the lengths of the sides of the two triangles. The sum of the distances from $P$ to the sides of an equilateral triangle is equal to the altitude of the triangle, hence the sum of the distances from $P$ to the lines $AB$, $BC$, $CA$, $MN$, $NP$, $PM$ is $(a + b) \frac{\sqrt{3}}{2}$.

On the other hand, the sum of the distances from $P$ to the parallel lines $AB$ and $MN$ is precisely the distance between those lines, hence at most $1$. It follows that $(a + b) \frac{\sqrt{3}}{2} \le 3$, so $a + b \le 2\sqrt{3}$, whence $a \le \sqrt{3}$ or $b \le \sqrt{3}$, which is what was asked to be proved.