Determinant inequality about positive definite matrices. Assume $A \in M_n(\Bbb{R})$ (not necessarily symmetric), and for $\forall \alpha\not=0$, $\alpha^TA\alpha>0$. Show that $$\det\left(\frac{A+A^T}{2}\right)\leq \det A.$$
 A: The matrix $B=\frac{A+A^T}{2}$ is symmetric and definite positive. So there
is an orthogonal matrix $P$ (i.e. $P^{T}=P^{-1}$) such that $D=P^TBP$ is a diagonal
matrix, say $D=(d_1,\ldots, d_n)$ with all the $d_i$ positive.  
Replacing the initial $A$ with $P^TAP$, we may assume without loss of generality that
$B$ is diagonal, i.e. $B=D=(d_1,\ldots, d_n)$. Then we can write $A=B+C$ where $C$ is an antisymmetric matrix. 
The following two facts are well-known (and proved on Wikipedia) about antisymmetric matrices :
1) If $C$ is an antisymmetric matrix with odd dimension, then ${\sf det}(C)=0$
(indeed, take the determinant on both sides in $C^T=-C$).
2) If $C$ is an antisymmetric matrix with even dimension, then ${\sf det}(C)$ is nonnegative (indeed, the eigenvalues are all purely imaginary and paired in a nice way
by conjugation).
Now let us look at the expansion of ${\sf det}(A)={\sf det}(B+C)$. For any subset
$I\subseteq \lbrace 1,2, \ldots ,n \rbrace$, put
$$
d_{I}=\prod_{i\in I} d_i
$$
and let $C_I$ be the matrix obtained from $C$ when one deletes the lines and columns
whose index is in $I$.
Then, as the $d_i$ appear on the diagonal of $A$ and nowhere else, we have an expansion
$$
{\sf det}(A)=\sum_{I\subseteq  \lbrace 1,2, \ldots ,n \rbrace} d_I {\sf det}(c_I)
$$
When $I$ is full, $I= \lbrace 1,2, \ldots ,n \rbrace$, $C_I$ is an empty matrix
and we have the term $d_1d_2\ldots d_n={\sf det}(B)$. The others terms are nonnegative
by properties (1) and (2). So we have ${\sf det}(A) \geq {\sf det}(B)$ as wished.
A: 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,
$$
\det(A)=|\det(A)|=\prod_{k=1}^n\sigma_k(A)\ge
\prod_{k=1}^n\sigma_k\left(\frac{A+A^T}2\right)
=\det\left(\frac{A+A^T}2\right).
$$
