Consider the following optimization problem in $x_1, \ldots, x_m \in \mathbb{R}$
$$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{i=1}^m \frac{i}{x_i}\\ \text{subject to} & \displaystyle\sum_{i=1}^m (m-i)x_i\leq 1\\ & x_i > 0\end{array}$$
One trivial solution is ofcourse setting $x_i=1/(m(m-i))$, which gives an objective value of
$$m\sum_i i(m-i) \sim m^4$$
Numerical simulations seem to tell me that the answer should be $m^3$ and not $m^4$, but I haven't been able to exhibit a feasible solution.
Can one show that the objective is $\geq m^4$ or give a feasible solution to obtain $m^3$? Since it is not a linear program, I am not sure how to take the dual and analyze it as well.
Introducing optimization variable $t > 0$, in epigraph form, the objective to be minimized becomes $t$ and we append the following inequality constraint
$$\frac{1}{x_1} + \frac{2}{x_2} + \dots + \frac{m}{x_m} \leq t$$
where $x_1, x_2, \dots, x_m > 0$. Using the Schur complement, the inequality above can be rewritten as the following linear matrix inequality (LMI)
$$\begin{bmatrix} \mbox{diag} \left(x_1, \frac12 x_2, \dots, \frac1m x_m \right) & 1_m\\ 1_m^\top & t\end{bmatrix} \succeq \mathrm O_{m+1}$$
where the diagonal matrix is positive definite (and, thus, invertible) because $x_1, x_2, \dots, x_m > 0$. Hence, the original optimization problem can be rewritten as a (convex) semidefinite program.