By a little testing, we find the solution
. Now, to prove that this is the minimum:
First, note that
. So, this means that the two factors of
must be kept in separate terms, as well as the two factors of
. Then, since the terms must be distinct, we must distribute the two
s, one to a factor of
, and the other to a factor of
. This gives the solution above. Now, if we were to have that the product is a larger multiple of
, we would need either to multiply the extra factors into one of the terms in our sum, or add it at the end. Both would increase the sum; so, we have found our minimum.
