Lp optimal solution question

Question

i have a general question.

if there is a general LP problem $c^Tx$ s.t $A\cdot x \le b$, and $x \ge 0$ and assuming that the components of $c$ are non-zero entries then how can I prove that when $x$ satisfies $a\cdot x < b$ (notice the strict inequality here) and $x > 0$, then $x$ cannot be an optimal solution?

Answer 1

Move from $x$ to $x +\epsilon$, where $\epsilon > 0$. Either $\epsilon$ is too big and breaks your inequality, or it doesn't. If it doesn't, then $c^T(x+\epsilon) > c^Tx$ and you are done.