Complex square matrices. Difficult proof.
Really it is a very difficult question dated 1980-81 (Am. Math. Monthly). Personally I did not find any solution. The simplest method is as follows.
EDIT.
Show the result when $A\bar{A}$ has no negative eigenvalues.
Show that the multiplicity of a negative eigenvalue of $A\bar{A}$ cannot be $1$.
Show that the set of matrices $A$ that satisfy 1. is dense in $M_n(\mathbb{C})$.
Conclude.
Eigenvalues of $A\bar{A}$ have been thoroughly studied in
D.C. Youla, A normal form for a matrix under the unitary congruence group, Canad. J. Math. 13(1961), 694-704.
In the paper, it is proved (in Lemma 5) that for any complex square matrix $A$, those eigenvalues of $A\bar{A}$ that are not real nonnegative must either be
- non-real and occur in conjugate pairs, or
- real, negative and have even multiplicities (so they also occur in "conjugate pairs").
It follows immediately that $\det(I+A\bar{A})$ is real nonnegative.