I have stumbled upon the following claim, and I wonder if it has a simple proof:
Let $P$ be a real $n \times n$ symmetric positive definite matrix. Then for every real skew-symmetric matrix $A$, $\langle P,A^2 \rangle \le 0 $.
($\langle \,, \rangle$ is the standard Frobenius inner product of matrices).
Here are some partial observations:
If $P$ and $A^2$ are commuting, then we can orthogonally diagonalize them simultaneously. Since the inner product is orthogonally-invariant, this reduces the problem to the case where $P$ and $A^2$ are both diagonal. In that case, all the (diagonal) elements of $P$ are positive, and those of $A^2$ are non-positive, since they are squares of the eigenvalues of $A$, which must be imaginary, since $A$ is skew-symmetric.
I am not sure what to do in the general case, where $A^2,P$ are not commuting:
In that case, we can orthogonally diagonalize either of them, but I don't see how to continue from that point. For instance, let us diagonalize $P$: Write $P=V \Sigma V^T$; then $$ \langle P,A^2 \rangle = \langle V \Sigma V^T,A^2 \rangle=\langle \Sigma ,V^TA^2V \rangle=\langle \Sigma ,(V^TAV)^2 \rangle. $$ Since $V^TAV$ is also skew-symmetric, we have reduced the problem to the case where $P=\Sigma$ is diagonal, and $A$ is an arbitrary skew-symmetric matrix.
If we start by diagonalizing $A^2$ instead, we reduce the problem to arbitrary symmetric positive-definite $P$ and diagonal $A^2$ (with non-positive values).