Because y\ge 0 and Ax\ge b, one has

b^\top y\le y^\top Ax.

Transposing the scalar gives

y^\top Ax=x^\top A^\top y.

Since A^\top y\le c coordinatewise and x\ge 0 coordinatewise,

x^\top A^\top y\le x^\top c.

Therefore

b^\top y\le c^\top x.

For the primal side, introduce slack variables s\ge 0 and rewrite

Ax\ge b,\qquad x\ge 0,\qquad c^\top x=r

as

Ax-s=b,\qquad x\ge 0,\qquad s\ge 0,\qquad c^\top x=r.

Thus r\in P iff (b,r) lies in the image of the nonnegative orthant under the linear map

(x,s)\longmapsto \binom{Ax-s}{c^\top x}.

Equivalently, P is an affine slice of a finitely generated cone, hence is closed. The dual side is identical after introducing slack variables t\ge 0 for the inequalities A^\top y\le c.

Let

p^*:=\inf\{c^\top x: Ax\ge b,\ x\ge 0\}.

By hypothesis, p^*\in\mathbb{R} and the primal feasible set is nonempty. Fix any scalar \mu<p^*. Then the system

Ax\ge b,\qquad x\ge 0,\qquad c^\top x\le \mu

is infeasible. Rewrite this as

\begin{pmatrix} A\\ -c^\top \end{pmatrix}x \ge \binom{b}{-\mu}.

By Theorem 3.4, there exists (y,\alpha)\in\mathbb{R}^m\times\mathbb{R} such that

\begin{pmatrix} A\\ -c^\top \end{pmatrix}^\top \binom{y}{\alpha} \le 0,\qquad \binom{y}{\alpha}\ge 0,\qquad \binom{b}{-\mu}^\top\binom{y}{\alpha}>0.

Equivalently,

A^\top y-\alpha c\le 0,\qquad y\ge 0,\qquad \alpha\ge 0,\qquad b^\top y-\alpha\mu>0.

We claim that \alpha>0. Indeed, if \alpha=0, then A^\top y\le 0, y\ge 0, and b^\top y>0, which contradicts primal feasibility by Theorem 3.4. Thus \alpha>0. Define

\bar y:=\frac{y}{\alpha}.

Then

A^\top \bar y\le c,\qquad \bar y\ge 0,

so \bar y is dual feasible, and

b^\top \bar y=\frac{b^\top y}{\alpha}>\mu.

Because \mu<p^* was arbitrary, this shows that for every \mu<p^* there exists a dual-feasible point with objective value strictly larger than \mu. Therefore the dual optimal value d^* satisfies

d^*\ge p^*.

On the other hand, weak duality gives d^*\le p^*. Hence

p^*=d^*.

To conclude attainment, let

P:=\{c^\top x:Ax\ge b,\ x\ge 0\},\qquad D:=\{b^\top y:A^\top y\le c,\ y\ge 0\}.

By Lemma 3.2, both P and D are closed subsets of \mathbb R. The set P is nonempty and bounded below, so p^*=\inf P\in P; hence there exists a primal-optimal point x^\star. The previous cutoff argument shows that for every \mu<p^* there exists a dual-feasible point with objective value strictly larger than \mu, so d^*=p^*=\sup D. Weak duality implies D\subseteq(-\infty,p^*], so D is also bounded above. Since D is closed and nonempty, its supremum belongs to D. Therefore there exists a dual-optimal point y^\star satisfying

A^\top y^\star\le c,\qquad y^\star\ge 0,\qquad b^\top y^\star=p^*.

Thus the dual optimum is attained.

We first show that the two statements cannot hold simultaneously. Assume that there exist x\in\mathbb{R}^n and y\in\mathbb{R}^m such that

Ax\ge b,\quad x\ge 0,\quad A^\top y\le 0,\quad y\ge 0,\quad b^\top y>0.

Then

b^\top y\le y^\top Ax=x^\top A^\top y\le 0,

which contradicts b^\top y>0. Thus at most one statement can hold.

We now show that at least one statement holds. Define

K:=-A\mathbb{R}_+^n+\mathbb{R}_+^m =\{-Ax+s:x\in\mathbb{R}_+^n,\ s\in\mathbb{R}_+^m\}.

Because \mathbb{R}_+^n\times\mathbb{R}_+^m is a finitely generated cone and (x,s)\mapsto -Ax+s is linear, K is a closed convex cone. Statement (1) is equivalent to b\in A\mathbb{R}_+^n-\mathbb{R}_+^m, hence to -b\in K. If statement (1) fails, then -b\notin K.

Applying the cone separation theorem from Lecture 2 to the closed convex cone K and the point -b\notin K, we obtain a vector y\in\mathbb{R}^m such that

\forall z\in K,\qquad \langle y,z\rangle\ge 0, \qquad\text{and}\qquad \langle y,-b\rangle<0.

For each standard basis vector e_i\in\mathbb{R}^m, one has e_i\in\mathbb{R}_+^m\subseteq K, and therefore

\langle y,e_i\rangle=y_i\ge 0.

Thus y\ge 0. Moreover, for every x\in\mathbb{R}_+^n, one has -Ax\in K, so

\langle y,-Ax\rangle\ge 0.

Equivalently,

x^\top A^\top y\le 0 \qquad \forall x\in\mathbb{R}_+^n.

Taking x=e_j shows (A^\top y)_j\le 0 for every j, so A^\top y\le 0. Finally,

b^\top y=-\langle y,-b\rangle>0.

Thus statement (2) holds. This proves that exactly one of the two statements is true.

Assume first that item (1) holds. Then x^\star is primal feasible, y^\star is dual feasible, and both are optimal. By weak duality,

b^\top y^\star\le c^\top x^\star.

Because both points are optimal, equality must hold:

c^\top x^\star=b^\top y^\star.

Using dual feasibility,

c^\top x^\star-b^\top y^\star = x^{\star\top}(c-A^\top y^\star)+y^{\star\top}(Ax^\star-b).

Equivalently,

c^\top x^\star-b^\top y^\star = \sum_{j=1}^n x_j^\star\bigl(c_j-(A^\top y^\star)_j\bigr) \;+\; \sum_{i=1}^m y_i^\star\bigl((Ax^\star)_i-b_i\bigr).

Each term in the sum is nonnegative because x^\star\ge 0, c-A^\top y^\star\ge 0, y^\star\ge 0, and Ax^\star-b\ge 0. Since the sum equals zero, every term must be zero. Thus item (2) holds.

Assume now that item (2) holds. Then x^\star is primal feasible, y^\star is dual feasible, and the complementary-slackness products vanish. Therefore

c^\top x^\star-b^\top y^\star = x^{\star\top}(c-A^\top y^\star)+y^{\star\top}(Ax^\star-b)=0.

Hence

b^\top y^\star=c^\top x^\star.

By weak duality, every dual-feasible objective value is at most every primal-feasible objective value. Since equality is achieved by (x^\star,y^\star), both points are optimal. Thus item (1) holds.