imc

The path of the fly consists of eight line segments, where each line segment goes from one corner to another corner. The distance of each such line segment is $1$, $\sqrt{2}$, or $\sqrt{3}$. The only way to obtain a line segment of length $\sqrt{3}$ is to go from one corner of the cube to the opposite corner. Since the fly visits each corner exactly once, it cannot traverse such a line segment twice. Also, the cube has exactly four such diagonals, so the path of the fly can contain at most four segments of length $\sqrt{3}$. Hence, the length of the fly's path can be at most $\boxed{4\sqrt{2}+4\sqrt{3}}$. This length can be achieved by taking the path $$A\to G\to B\to H\to C\to E\to D\to F\to A.$$