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A closed half-space consists of a hyperplane and one of its two sides in a real affine space. Replacing the weak inequality by a strict inequality gives an open half-space. Half-spaces are convex sets, and finite intersections of closed half-spaces are linear polyhedra.

Use the Charnes-Cooper transformation on the positive-denominator region:It gives $Ay\le bt$, $d^{T}y+\beta t=1$ and $c^{T}y+\alpha t=(c^{T}x+\alpha )/(d^{T}x+\beta )$.

We must check that allowing $t=0$ introduces no false feasible solution. If $t=0$, then $Ay\le 0$. The recession cone of the nonempty linear polyhedron $P=\{x:Ax\le b\}$ is $\{y:Ay\le 0\}$. Indeed for $x_0\in P$, $x_0+uy\in P$ for every $u\ge 0$. Boundedness of $P$ therefore forces $y=0$, contradicting $d^{T}y=1$. Thus every transformed feasible point has $t>0$.

Conversely set $x=y/t$. The constraints give $x\in P$ and $d^{T}x+\beta =1/t>0$, with the same objective value. The assumed original optimizer with positive denominator supplies a transformed feasible point. Every transformed point corresponds to an original point and has objective at most that optimizer's value. Hence the transformed linear program attains the original optimum, and its optimizer recovers $x=y/t$. Positivity on every point of $P$ is not needed here; the given positive-denominator optimum suffices.

Setting the third coordinate to zero and the first to four leaves $6x_2\ge 4$ and $4x_2\le 3$. Thus $x_2=2/3$ works. Direct substitution givesHence this point lies in both linear polyhedra.

If a vector $h$ gives the strict separation, a common point would satisfy $h^{T}x<h^{T}x$, which is impossible.

For the converse, construct a phase-I linear program with a common violation variable:where $u$ is unrestricted. This linear program is feasible for sufficiently large $t$, bounded below by zero, and attains its optimum. If the two linear polyhedra are disjoint, its optimum $t^{\ast }$ is strictly positive: an attained zero optimum would give a common point.

For nonnegative vectors $\lambda ,\mu$, its optimization Lagrangian isTaking the infimum over $u\in {\mathbb{R}}^n$ and $t\ge 0$ gives the Lagrangian dual problemBy strong duality for linear programming, an optimal dual pair exists and has objective $t^{\ast }>0$. It supplies the infeasibility certificateTake $h=A^T\lambda =-C^T\mu$. Every $x\in P$ and $y\in Q$ obeysThe assumed nonemptiness of each linear polyhedron also ensures $h\ne 0$: if $h=0$, their feasibility would force both ${\lambda }^{T}b$ and ${\mu }^{T}d$ to be nonnegative. This proves the strict separation of disjoint linear polyhedra using Lagrangian duality.

The recession cone of a nonempty convex set $C$ consists of directions $d$ for which $x+td\in C$ for every $x\in C$ and $t\ge 0$. For a linear polyhedron $P=\{x:Ax\le b\}$ it is exactly $\{d:Ad\le 0\}$. A nonzero recession direction gives an unbounded ray, so a bounded nonempty polyhedron has recession cone $\{0\}$. This excludes spurious zero-scale feasible points in the Charnes-Cooper transformation.

Nonempty disjoint linear polyhedra $P=\{x:Ax\le b\}$ and $Q=\{y:Cy\le d\}$ admit $h$ with a strictly positive separation gap. A phase-I linear program minimizing common constraint violation has positive attained optimum. Its Lagrangian dual problem gives nonnegative $\lambda ,\mu$ with $A^T\lambda +C^T\mu =0$ and ${\lambda }^{T}b+{\mu }^{T}d<0$. Then $h=A^T\lambda$ satisfies $h^{T}x\le {\lambda }^{T}b<-{\mu }^{T}d\le h^{T}y$. The polyhedral hypothesis matters; arbitrary disjoint closed convex sets need not have a positive separation gap.