Geometric program with reciprocal objective function Let $1 \leq m < M$ and let $\alpha_1, \dots, \alpha_n > 0$ be fixed real numbers.  I want to solve the following $n$-dimensional optimization program
$$\begin{array}{ll} \underset{\alpha_1, \dots, \alpha_n}{\text{minimize}} & \displaystyle\sum_{i=1}^n \alpha_i \frac1{x_i}\\ \text{subject to} & \displaystyle\sum_{i=1}^n x_i \leq M\\ & m \leq x_i\end{array}$$
I have done some quadratic programming but I have no idea to solve this type of problem.
 A: Since increasing any of the $x_i$'s will lower the object function, we can replace the first constraint by an equality constraint.
Also, we can define new variables $y_i = x_i-m $, so that the second constraint is $y_i \ge 0.$
We have $$\min \sum_{i=1}^n  \frac{\alpha_i }{y_i+m}$$
$$\textrm{s.t. }\sum_{i=1}^n  y_i  = M-n m = K$$
$$ y_i \ge 0$$
Looks like we must have $M \ge nm$ in your problem statement.
Anyway, using Lagrange multipliers,
Let $$L = \sum_{i=1}^n  \frac{\alpha_i }{y_i+m} + \lambda \sum_{i=1}^n ( y_i -K). $$
Find $\frac{ \partial L }{\partial y_i}$ and $\frac{ \partial L }{\partial \lambda}$ and set these equal to zero.
$$0=\frac{ \partial L }{\partial y_i}= \lambda - \frac{\alpha_i }{(y_i+m)^2},$$
$$\sum_{i=1}^n  y_i =K.$$
Solving for $y_i$, 
$$y_i =\sqrt{\frac{\alpha_i}{\lambda}} - m.$$
Summing this, we get an expression for $\lambda:$
$$K=  \sum_{i=1}^n \sqrt{\frac{\alpha_i}{\lambda}}  - mn$$ 
$$M=  \sum_{i=1}^n \sqrt{\frac{\alpha_i}{\lambda}} $$ 
$$ \lambda = \left( \frac{\sum_{i=1}^n \sqrt{\alpha_i}}{M} \right)^2.$$
Solving for $y_i$ and substituting to find $x_i$:
$$x_i = \frac{M \sqrt{\alpha_i} }{\sum_{i=1}^n \sqrt{\alpha_i}}.$$
Actually, we are not done!  We have to check that each $x_i \ge m$ (or each $y_i\ge 0$).  Could happen not.  In that case, set any negative $y_i$ equal to zero: ($\forall i, \textrm{s.t.} y_i<0,$ $y_i \rightarrow 0$) and solve the problem again for the remaining variables.
