TITLE: Characterizing the dual cone of the squares of skew-symmetric matrices
QUESTION [5 upvotes]: Let $X$ be the set of all real $n \times n$ diagonal matrices $D$ satisfying $\langle D,B^2 \rangle \le 0$ for any (real) skew-symmetric matrix $B$. (I am using the Frobenius Euclidean product here).
$X$ is a convex cone.

Can we give an explicit characterization of $X$? 

Comment:
If we denote by $C$ the space of all  squares of skew-symmetric matrices, we can characterize its dual cone as follows:
Since every square of a skew-symmetric matrix is symmetric, and the symmetric and the skew-symmetric matrices are orthogonal, we know that every skew-symmetric matrix belongs to the dual cone of $C$. So, the question whether a given matrix $A$ belongs to the dual cone of $C$ depends solely on the symmetric part of $A$. Since $C$ is invariant under orthogonal conjugation, we can orthogonally diagonalize $\text{sym}(A)$ and deduce that $A$ lies in $C^*$ if and only if the diagonal matrix whose entries are the eigenvalues of $\text{sym}(A)$ is in $C^*$. Thus, the question reduces to determining the case of diagonal matrices.
Edit:
Omnomnomnom proved in this answer that every $D$ in $X$ has at most one negative entry, and the absolute value of the negative entry is less than or equal to the next-smallest entry.
I have a strangely complicated proof for the converse, namely I can prove that every diagonal matrix satisfying the condition above is in $X$. 
I would like to find a "direct" proof based on linear algebra\matrix analysis. (my proof is based on rather convoluted variational considerations).

REPLY [2 votes]: Here is a slightly different proof of the sufficiency of the condition $d_i+d_j\geq 0$ for all $i\neq j,$ which is the same as the condition in Omnomnomnom's answer.
Note that
\begin{align*}
(B^2)_{ii}
&=\sum_{j} b_{i,j}b_{j,i}\\
&=-\sum_{j:i\neq j} b_{i,j}^2
\end{align*}
So 
$$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-d_i b_{i,j}^2.\tag{1}$$
Swapping the roles of $i$ and $j,$
$$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-d_j b_{i,j}^2.\tag{2}$$
Averaging (1) and (2) gives
$$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-\tfrac12(d_i+d_j) b_{i,j}^2\leq 0.$$