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?