Short and slick proof of the strong duality theorem for linear programming
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 $y^Tb &lt; 0$
  


Now let $\delta$ be the optimal objective value of the primal.
Let $\epsilon &gt; 0$ 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
  $y^Tb &lt; \alpha (\delta + \epsilon)$.


Note that if $\epsilon  = 0$ we are in the other alternative of Farkas.  Therefore $\alpha &gt; 0$.

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