Let one of the longer sides be the base and be called L, and the height h.

If the parallelogram is not already a rectangle, slide the top edge in the direction of making it a rectangle. The longer side will get shorter all the way, at which point (when a rectange is formed) the "longer" side's length will match the "shorter" side, as the shorter side was getting larger through the process. If the sliding were to continue past making a rectangle, what had been the shorter side would become the longer side, and be longer than when in the rectangular state.

Now, if the parallelogram, now a rectangle, is not a square (L strictly greater than h), increase h while decreasing L, while maintaining L*h. The diagonals will decrease all the way to the square state, when L = h.

The shortest possible diagonal is that of a square: sqrt(L*h) aka sqrt(A), where A is the area.

If one insists that the parallelogram not be a rectangle, then there is no one shortest possible. Just approach the square configuration as a limit.