I have the following question:
Let $A$ be an $(m \times n)$-matrix and $b$ a vector in $\mathbb{R}^m$. The system of inequalities $Ax \leq b$ has a solution $x \geq 0$, if and only if $yb \geq 0$ for every row-vector $y \geq 0, y \in \mathbb{R}^m$ which satisfies $yA \geq 0 $.
I know how to prove this statement in one direction: ($\implies$)
Assume there exists some $x \in \mathbb{R}^n$ with $x \geq 0$ such that $Ax \leq b$. Choose some $y \in \mathbb{R}^m$ with $y \geq 0$ and $yA \geq 0$. If we multiply the inequality $Ax \leq b$ by $y$ on both sides on the left, we get $y (Ax) \leq yb$ (because $y \geq 0$). Because vector/matrix multiplication is associative, we also obtain $ (yA) x \leq yb $ from this. Since $yA \geq 0$ and $ x \geq 0$, we find $(yA)x \geq 0$. Since $(yA)x \leq yb$, it must also be true that $yb \geq 0$.
My question is, how do we prove this statement in the other direction?
The reverse direction can be solved easily via proof by contradiction.
Let us suppose there exists no $x \geq 0$ such that $A x \leq b$. It implies for all $x \geq 0$, $A x > b$. In that case, for all $ y \geq 0$ such that $y A \geq 0$, we have
$$(yA) x > y b,$$
which for $x = 0$ gives $ y b < 0$, a contradiction.