It should be an axiom that if a>0 and b>0, then ab>0.

If you can prove that (-a)(-b) = ab, then you are done: Since a<0 and b<0, we have -a>0 and -b>0 (have you proven this?). By the axiom, we then have that (-a)(-b) > 0, and since (-a)(-b) = ab, we are done.

As for why (-a)(-b) = ab, you can prove first that (-a)(-b) is equal to -(a(-b)), i.e. (-a)(-b) is the additive inverse of a(-b):

a(-b) + (-a)(-b) = (a + (-a))(-b) = 0(-b) = 0 [I need only check this from "one side", since addition is commutative].

Then you can continue to prove that a(-b) is equal to -(ab), i.e. it is the inverse of ab:

ab + a(-b) = a(b + (-b)) = a0 = 0.

In conclusion, (-a)(-b) is the inverse of the inverse of ab, and is as such equal to ab. Do you know already that -(-c) = c for all elements c?