TITLE: Determinant inequality about positive definite matrices.
QUESTION [5 upvotes]: Assume $A \in M_n(\Bbb{R})$ (not necessarily symmetric), and for $\forall \alpha\not=0$, $\alpha^TA\alpha>0$. Show that $$\det\left(\frac{A+A^T}{2}\right)\leq \det A.$$

REPLY [2 votes]: If $(\lambda, x)$ is a real eigenpair of $A$, we have $\lambda\|x\|^2 = x^TAx>0$. Therefore all real eigenvalues of $A$ are positive. As nonreal eigenvalues of $A$ must occur in conjugate pairs, it follows that $\det(A)>0$. Now, by the given condition, we have $x^T\frac{A+A^T}2x = x^TAx >0$ for all nonzero vector $x$. Therefore the symmetric part of $A$ is positive definite. Hence $\sigma_k(A)$, the $k$-th largest singular value of $A$, is always greater than or equal to $\sigma_k\left(\frac{A+A^T}2\right)$ (see a proof here) and consequently,
$$
\det(A)=|\det(A)|=\prod_{k=1}^n\sigma_k(A)\ge
\prod_{k=1}^n\sigma_k\left(\frac{A+A^T}2\right)
=\det\left(\frac{A+A^T}2\right).
$$