This is a rather long problem...

Consider the following:
maximize \sum c_{j}x_{j} with j = 1 as it goes to n
subject to \sum a_{ij}x_{j} \leq b_{i} where j = 1 as it goes to n, i = 1, 2,...,m
x_{j} \geq 0 where j = 1,2,...,n

first form its dual:
minimize \sum b_{i}y_{i} with i = 1 as it goes to m
subject to \sum y_{i}a_{ij} \geq c_{j}with i = 1 as it goes to m, j = 1,2,...,n
y_{i} \geq 0 where i = 1,2,...,m

Then write that minimization to a maximization in standard form:
maximize \sum b_{i}y_{i} with i=1 as it goes to m
subject to \sum -y_{i}a_{ij} \leq -c_{j} with i=1 as it goes to m, j = 1,2,...,n
y_{i} \geq 0 with i=1,2,...,m

The above process can be thought of as a Transformation T on the space of data defined by: (a_{ij},b_{i},c_{j}) \longrightarrow^T (-a_{ji}, -c_{j},b_{i})

Let \zeta^* (a_{ij},b_{i},c_{j}) denote the optimal objective function value of the standard-form linear programming problem haing data (a_{ij},b_{i},c_{j}. By strong duality and the fact that a maximation dominates a minimization, it follows that:
\zeta^* (a_{ij},b_{i},c_{j}) \leq \zeta^* (-a_{ij},-c_{j},b_{i})

If the process is repeated, transformation T becomes:
(a_{ij},b_{i},c_{j}) \longrightarrow^T (-a_{ij},-c_{j},b_{i})
\longrightarrow^T (a_{ij},-b_{i},-c_{j})
\longrightarrow^T (-a_{ij},c_{j},-b_{i})
\longrightarrow^T (a_{ij},b_{i},c_{j})

and hence...
\zeta^* (a_{ij},b_{i},c_{j}) \leq \zeta^* (-a_{ij},-c_{j},b_{i})
\leq \zeta^* (-a_{ij},-c_{j},b_{i})
\leq \zeta^* (a_{ij},-b_{i},-c_{j})
\leq \zeta^* (-a_{ij},c_{j},-b_{i})
\leq \zeta^* (a_{ij},b_{i},c_{j})

But the first and last entry in this chain of inequalities is equal, therefore all these inequalities appear equal. Although possible, this isn't always true.

1. What is the error in this logic?
2. Can you state a (correct) nontrivial theorem that follows from this line of reasoning?
3. Can you give an example where the four inequalities are indeed all equalities?

Yeah... very long... please help...