Short and slick proof of the strong duality theorem for linear programming

Consider the linear programs

\begin{array}{|ccc|}
\hline
Primal: & A\vec{x} \leq \vec{b} \hspace{.5cm} &
\max \vec{c}^T\vec{x} \\
\hline
\end{array}
\begin{array}{|ccc|}
\hline
Dual: & \vec{c} \leq \vec{y}^TA \hspace{.5cm} &
\min \vec{y}^T\vec{b} \\
\hline
\end{array}

The weak duality theorem states that
if $\vec{x}$ and $\vec{y}$ satisfy the constraints then
$\vec{c}^T\vec{x} \leq \vec{y}^T\vec{b}$.
It has a short and slick proof using linear algebra:
$\vec{c}^T\vec{x} \leq \vec{y}^T A \vec{x} \leq \vec{y}^T\vec{b}$.

The strong duality theorem states that if the $\vec{x}$ is an optimal solution for the primal then there is $\vec{y}$ which is a solution for the dual and
$\vec{c}^T\vec{x} = \vec{y}^T\vec{b}$.

Is there a similarly short and slick proof for the strong duality theorem?

Probably not. Here is a conceptual argument based on

Farkas Lemma:
Exactly one of the following alternatives has a solution:

• $Ax \le b$ and $x \ge 0$

• $y^TA\ge 0$ and $y^Tb

Now let $\delta$ be the optimal objective value of the primal.
Let $\epsilon > 0$ be arbitrary.
Let $A'$ to be $A$ with an additional $-c^T$ as the last row.
Let $b'$ to be $b$ with an additional $-\delta - \epsilon$ as the last value.

The system $A'x'\le b'$ has no solution.
By Farkas, there is a $y' = (y,\alpha)$ such that:


$y^TA\ge \alpha c$ and
$y^Tb 0$.

Scale $y'$ so that $\alpha = 1$.
$y$ is dual feasible.
The weak duality implies $\delta \le y^Tb < \delta + \epsilon$.

StackExchange Computer Science Q#18151, answer score: 3