Probably not.  There is a conceptual argument, which is based on

**Farkas Alternative**: Exactly one of $Ax\le b$, $x\ge 0$ or $y^TA\ge 0$ $y^Tb < 0$ has a solution.

Assuming this, take $\delta$ to be the objective value of the primal in an optimal solution.  Define  $A'$ to be the matrix that is $A$ with $-c^T$ as the last row and $b'$ to be the vector $(b, -\delta - \epsilon)$.  For any $\epsilon > 0$, the system $A'x'\le b'$ has no solution, so according to Farkas, there is a $y' = (y,\alpha)$ such that $y^TA\le \alpha c$ and $y^Tb < \alpha (\delta + \epsilon)$.  Moreover, $\alpha > 0$, since, when $\epsilon  = 0$ we are in the other alternative of Farkas.

To finish up, just scale $y'$ so that $\alpha = 1$.  Now $y$ is dual feasible, so weak duality implies $\delta \le y^Tb < \delta + \epsilon$, and we are done.