Here is yet another overkill, but hopefully not too bad a way to prove this inequality.

We have the following proof sketch.

$$\begin{eqnarray}
x^T(A+B)x &\ge& x^TAx\quad\forall x\\\\
-x^T(A+B)x &\le& -x^TAx\\\\
\exp(-x^T(A+B)x) &\le& \exp(-x^TAx)\\\\
\int\exp(-x^T(A+B)x)dx &\le& \int\exp(-x^TAx)dx\\\\
\frac{1}{\det(A+B)} &\le& \frac{1}{\det(A)}
\end{eqnarray}
$$

The only fancy thing that happened is in the last line, where I used the formula for the Gaussian integral.