Converse to the Duality Theorem of Geometric Programming The standard method of solving an unconstrained geometric program such as $$\min\left\{g(\mathbf{t}) = \frac{40}{t_1 t_2 t_3} + 20 t_1 t_3 + 10 t_1 t_2 + 40 t_2 t_3 : t_1, t_2, t_3 > 0\right\}$$ is to pass to the dual geometric program, which essentially asks: what is the best set of weights with which to apply the AM-GM inequality to put a lower bound on this objective function? In this example, the weights $\delta_1, \delta_2, \delta_3, \delta_4$ to put on each term would need to satisfy
\begin{align*}-\delta_1 + \delta_2 + \delta_3 &= 0 \\ 
-\delta_1 + \delta_3 + \delta_4 &= 0 \\
-\delta_1 + \delta_2 + \delta_4 &= 0
\end{align*}
to get a lower bound independent of $t_1, t_2, t_3$ from AM-GM, and
\begin{align*}\delta_1 + \delta_2 + \delta_3 + \delta_4 &= 1 \\
\delta_1, \delta_2, \delta_3, \delta_4 &>0 \end{align*}
for the AM-GM inequality to apply. We want to maximize the lower bound $$v(\boldsymbol{\delta}) = \left(\frac{40}{\delta_1}\right)^{\delta_1}\left(\frac{20}{\delta_2}\right)^{\delta_2}\left(\frac{10}{\delta_1}\right)^{\delta_3}\left(\frac{40}{\delta_4}\right)^{\delta_4}.$$

The duality result that is usually proved in textbooks (for example, it is Theorem 3.1 in this PDF) is that to an optimal primal solution $\mathbf{t}^*$ corresponds an optimal dual solution $\boldsymbol{\delta}^*$ with $g(\mathbf{t}^*) = v(\boldsymbol{\delta}^*)$ that can be found by the rule $$\delta_j^* = \frac{j^{\text{th}}\text{ term of }g(\mathbf{t}^*)}{g(\mathbf{t}^*)}.$$ When we actually apply duality to solve the geometric program, however, we need the converse of this result to be sure that it will always work: that, given an optimal dual solution, an optimal primal solution exists which is paired to it by this formula.
This could theoretically be false in two ways:


*

*Maybe, despite the existence of an optimal dual solution, there is no optimal primal solution, just solutions that get arbitrarily close to some positive lower bound. (In this case, there might even be a duality gap.)

*Maybe there is a primal solution, with an associated optimal dual solution given by the theorem above, but we've unluckily stumbled on a different optimal dual solution from the one paired with it by the duality result above.



Update: However, I can rule out (2). The objective function $v(\boldsymbol{\delta})$ is strictly concave, because in general $-\log v(\boldsymbol{\delta}) = \sum_{i=1}^n \delta_i \log \frac{\delta_i}{c_i}$, and its Hessian matrix is diagonal with eigenvalues $\frac1{\delta_1}, \dots, \frac1{\delta_n} > 0$, and therefore $-\log v(\boldsymbol{\delta})$ is strictly convex. So the optimal dual solution, if it exists, is unique.
That still leaves out (1). So my question is, what do we know about the dual of a geometric program with no optimal primal solution? 
The answer we want to get is "in such a case, the dual is always infeasible". The textbook I'm using in the class I'm teaching doesn't mention or prove such a result, and neither does the (unrelated) textbook I've linked to. Is it true? If so, is there a proof of it somewhere?
 A: Making the substitution $x_i=\log t_i$, we can write:
$$\begin{align}\min_{t>0} g(t) &= \min_{t>0} \exp\left(\log\left(\frac{40}{t_1 t_2 t_3}\right)\right) + \exp(\log(20 t_1 t_3)) + \exp(\log(10 t_1 t_2)) + \exp(\log(40 t_2 t_3)) \\
       &= \min_{x}  \exp(\log(40) - x_1 - x_2 - x_3)) + \exp(\log(20)\\
&\qquad \qquad \qquad + x_1 + x_3)) + \exp(\log(10) + x_1 + x_2)) + \exp(\log(40) + x_2 + x_3))\end{align}$$
This is a GP in convex form. For convex optimization, we know that under Slater's condition there is no optimality gap and that both the primal and dual optimal solution are attained. For the primal, Slater's condition boils down to "there is an $x$ such that all expressions within the $\log$ operators are positive".
The Lagrange dual is (source, slide 10, dropping the subscript 0 for convenience):
$$\begin{align}\max_{v\geq 0} \quad & \log(40) v_1 + \log(20) v_2 + \log(10) v_3 + \log(40) v_4 - \sum_{k=1}^4 v_i \log v_i \\
\text{s.t.} \quad & v_1 + v_2 + v_3 + v_4 = 1 \\
& - v_1 + v_2 + v_3 = 0 \\
& - v_1 + v_3 + v_4 = 0 \\
& - v_1 + v_2 + v_4 = 0 \\
\end{align}$$
This is exactly the problem that arises from the AM-GM inequality (after performing the logarithmic transformation you mention in your update, and after taking the closure).
All properties you are looking for follow from results on Lagrange duality. If the dual is not feasible, then the primal (which is the dual of the dual) is either infeasible or unbounded (this follows from weak duality). However, if a dual optimal solution exists, then the dual satisfies Slater's condition, which means that the primal has an optimal solution, and that the optimal solution is attained.
