Yet another way to see this is to note that $A = \overline{Q}^{t}Q$ for some invertible 
matrix $Q$. Then 
${\rm det}(A+B) = |{\rm det}(Q)|^{2}{\rm det}{( I + (\overline{Q}^{-1}}^{t}BQ^{-1}))$.`
Now $(\overline{Q}^{-1})^{t}BQ^{-1}$ is Hermitian, and positive definite. 
It suffices to prove that if $X$ is positive definite and Hermitian, then 
${\rm det}(I+X) \geq (1 + {\rm det}X)$. We may conjugate $X$ by a unitary matrix $U$
and assume that $X$ is diagonal. Let the eigenvalues of $X$ be $\{ \lambda_{1},\ldots, \lambda_{n} \}$ (allowing repetitions). Then ${\rm det}(I+X) = \prod_{i=1}^{n}(1 + \lambda_{i})
\geq 1 + \prod_{i=1}^{n} \lambda_{i} = 1 + {\rm det}X.$ Such an argument appears in some 
proofs by R. Brauer, though I do not know whether it originates with him.