Q:

minimizing the sum of reciprocals is equivalent to maximizing the sum

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 
$$\max_{x_i\geq 0, \forall i\in [1...m]}\sum_{i=1}^mx_i\,,~~~~\text{subject to }x_i \in X_i\subseteq \mathbb{R}, \forall i \in [1...m]?$$

A:

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

