For this try and consider that if (A^(-1) + B^(-1)) is non-singular then (A^(-1) + B^(-1))^(-1) exists.
We know that A^(-1) exists since A is non-singular. We also know that B^(-1) exists since B is non-singular.
We know that (A+B)^(-1) exists since (A+B) is nonsingular.
Let det(A) = a, det(B) = b, det(A+B) = c.
det[A^(-1)] = 1/a, det[B^(-1)] = 1/b and det[A^(-1) + B^(-1)] = c
From this paper
We get 1/a + 1/b <= 1/c
Proof by contradiction: assume that a = -b then 1/a + 1/b = 0 but 1/c can never be zero unless c = infinity which should never happen.
So we rule out this case.
The only other case is that 1/a + 1/b != 0 and thus 1/c is non-zero which implies non-singular matrix.
Thus the statement is proven.
I just realized that another solution could be to assume that 1/c was just non-zero.