Consider the primal LP where , , and . Assume the primal is feasible and that . Then we know the dual is Let’s derive this dual using a separating hyperplanes argument.

The first insight is to define the closed convex cone in a higher dimension, A point means that using constraint budget , one can achieve objective value at least . To check that is a closed convex cone:

1. is a cone. If with witness , then for any , we have and .

2. is convex. If with witnesses , respectively, then for any , and

3. is closed. This is a little tricky to show, since the definition of leaves a witness implicit. We can view as a projection of a higher-dimensional polyhedral cone: Since the linear image of a polyhedron is a polyhedron, is a polyhedral cone, hence closed.

The optimal value can be written as and is at the upper boundary of . The second insight is to invoke a separating hyperplane argument at this point.

The strong separation result for closed convex cones says that if is a closed convex cone and , then there exists such that for all and . If is also polyhedral, we have the following stronger result: if and are such that for all , then there exists such that for all , , and . That is, the hyperplane supports at and separates from the ray with .

Now let and . For every , ; otherwise, the definition of would be contradicted. Hence, the polyhedral cone separation result gives a vector such that Now let . Then This means the hyperplane with normal vector supports at .

Notice that is a dual-feasible solution. First, for any with , we have since and . By the supporting inequality, must hold, which means . Next, for every , since and . Hence, or equivalently, Since this holds for all , we must have . Hence, and , and by the supporting equality, .

Combining this with weak duality, which is the easy direction, we conclude strong duality.