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.

Later edit: Incidentally, I think that with the arithemetic-geometric mean inequality
and a slightly more careful analysis, you can see by this approach that for $X$ as above,
you do have ${\rm det}(I+X) \geq (1 +({\rm det}X)^{1/n})^{n}$ (a special
case of the inequality of Minkowski mentioned in the accepted answer, but enough to prove
the general case by an argument similar to that above). For set $d = {\rm det}X$. Let $s_{m}(\lambda_{1},\ldots ,\lambda_{n})$ denote the $m$-th elementary symmetric function evaluated at the eigenvalues. Using the arithmetic-geometric mean inequality yields that $s_{m}(\lambda_{1},\ldots ,\lambda_{n}) \geq \left( \begin{array}{clcr}
n\\m \end{array} \right)d^{m/n}$, so we obtain ${\rm det}(I+X) \geq (1+d^{1/n})^{n}.$