Let $H = A + i B \ge 0$ be a positive (semidefinite) $n\times n$ matrix with real parts $A = (H+\bar{H})/2 = A^\intercal$ and imaginary part $B = (H-\bar{H})/(2i) = B^\intercal$. (Bars denote the element-wise complex conjugate.) What constraints hold between $A$ and $B$?
Alternatively phrased, what are necessary and sufficient conditions on real matrices $A$ and $B$ such that $A + i B \ge 0$?
I expect this to take the form of an inequality roughly like "some part of A greater than some part of B", because given an $A$ and $B$ satisfying $A + i B \ge 0$ we have $A + A' + i B \ge 0$ for any $A'\ge 0$. That is, we can always make $A$ bigger.
It's clear $A$ is positive (hence symmetric) and $B$ is skew symmetric (hence not positive semidefinite unless $B=0$). When $n=2$, $B = \left(\begin{array}{cc} 0 & -b\\ b & 0 \end{array}\right)$ has only one free parameter, $b^2 = \det B$, that must satisfy $b^2 \le \det A$. (Indeed, for $n=2$, $A\ge 0$, and $B = -B^\intercal$, $A+iB \ge 0$ if and only if $\det B \le \det A$.) What is the generalization of this condition to higher $n$? (We can tell the condition changes for larger $n$ by taking $A$ and $B$ to be block-diagonal and seeing that $\det B \le \det A$ does not imply a similar condition for the individual blocks.)