TITLE: Sum of Difference of anti-diagonal matrix elements
QUESTION [7 upvotes]: Let $A \in \mathbb{R}^{n \times n}$, with elements $a_{ij}$
What conditions on $A$ are required for the following to be true?
There exists some vector $x \in \mathbb{R}^n_+$, $x \neq 0$ such that for all $i=1\dots n$, 
$$\sum_{j=1}^n x_j(a_{ij} - a_{ji}) \geq 0$$
Obviously if $A$ is symmetric this is true. I believe it may be true for all $A$ but I haven't been able to prove it.
(sorry really wasn't sure what to title this post)
I've tried the usual  Farkas' Lemma-style trick but the problem is self-dual

REPLY [3 votes]: The original question asks if we can find a nonzero vector $x \ge 0$ such that $(A-A^T)x \ge 0$, of equivalently, for a skew-symmetric (aka antisymmetric) matrix $M$, the claim is that there always exists an $0 \neq x \ge 0$ such that $Mx \ge 0$. Chasing Gerry Myerson's suggestion above, here is a link to a proof of this claim by invoking Farkas' Lemma.
Lemma 4.5 here.