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Consider the optimization problem $$\min_{x_i\geq 0, \forall i\in [1...m]}\sum_{i=1}^m\frac{1}{x_i}\,,$$ subject to $$x_i \in X_i\subseteq \mathbb{R}, \forall i \in [1...m]\,.$$

Can I find the solution to the above problem by instead solving $$\max_{x_i\geq 0, \forall i\in [1...m]}\sum_{i=1}^mx_i\,,~~~~\text{subject to }x_i \in X_i\subseteq \mathbb{R}, \forall i \in [1...m]?$$

pinochino
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    Yes. Since the feasible set is a of the form $\prod_{i=1}^mX_i$, there is no relation between the variables. The minimum in the first problem, which will exist only if each $X_i$ has a maximum, will be attained at $(\max X_1,..., \max X_m)$. This is the same point at which the maximum in the second problem is attained. –  Apr 12 '18 at 00:18

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Let $A_i = X_i \cap \mathbb R_+$ and notice that since there is no interaction between the variables in the objective function or in the constraints,

$$\min_{ \{x_i\in A_i\}_{i=1}^m}\sum_{i=1}^m\frac{1}{x_i}=\sum_{i=1}^m\min_{ x_i\in A_i}\frac{1}{x_i}.$$

Moreover, the solution to

$$\min_{ x_i\in A_i}\frac{1}{x_i}$$

is simply to choose the maximum value of $x_i$ in $A_i$, that is, it is equivalent to solving

$$\max_{ x_i\in A_i}x_i.$$

Finally, and for the same reasons described above,

$$\sum_{i=1}^m\max_{ x_i\in A_i} x_i=\max_{ \{x_i\in A_i\}_{i=1}^m}\sum_{i=1}^m x_i.$$

mzp
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