Show a vector $\vec c$ exists such that $\max\{c^Tx:Ax\le b,x\ge 0\}=+\infty$ if and only if it exists $j\in\{1,...,n\}$ such that $\max \{x_j:Ax\le b, x\ge 0\}=+\infty$
I'm only asking for a hint or criticize of my proof attempt
Broadly speaking, I think I have to show that a vector tends to infinity when maximizing the whole program iif the the objective function tends to infinity when maximizing under the constraints (I'm not sure about this part)
I started with assuming that it exists $x_j$ such that $\max \{x_j:Ax\le b, x\ge 0\}=+\infty$
for instance
\begin{cases} \max & x_1\\ &-x_1 &\le -1\\ &x_1&\ge 0 \end{cases}
then $Ax$ must be negative and $b$ positive then $\forall c^T$ such that $c^Tx\ge0$, $\max \{c^Tx:Ax\le b, x\ge 0\}=+\infty$, but is it a rigorous and strict proof?
I then have to prove the revert
Let's assume that it exists $c^Tx$ such that $\max \{c^Tx:Ax\le b, x\ge 0\}=+\infty$