## Return to Answer

2 added link, fixed typo

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

We have the following proof sketch.

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

The only fancy thing that happened is in the second last line, where I used the formula for the Gaussian integral .(see multivariate section)

1

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.