Let $w_k > 0$, $k = 1, \ldots, K$ be known constants, and let $f(\mathbf{p}) = \sum_{k=1}^K w_k / p_k$. How do I solve the following optimization problem? $$\min_{\mathbf{p}} f(\mathbf{p}) \ \ \text{s.t.} \ p_k > 0 \ \forall k, \sum_k p_k \leq 1$$
The Lagrangian here is $$L(\mathbf{p}, \mathbf{\lambda}) = \sum_k w_k/p_k + \lambda_0 (\sum_k p_k - 1) + \sum_k \lambda_k p_k$$ where $\lambda_0 \geq 0$, $\lambda_k \leq 0 \ \forall k > 1$. Differentiating with respect to $p_k$ and setting equal to $0$, I get $\hat{p}_k = \sqrt{\frac{w_k}{\lambda_0 + \lambda_k}}$. How do I proceed from here?
Clearly, none of the constraints $p_k > 0$ will be active at an optimal solution and the constraint $\sum_j p_j \leq 1$ will be active at an optimal solution.
To see this, note that the point $p = (\frac{1}{K},\cdots,\frac{1}{K})$ is feasible for the true problem, and any feasible solution that has a component $p_j$ that is too small will have worse objective than this point. The argument that $\sum_j p_j = 1$ at an active solution can be argued by noting that if this isn't the case, then we can decrease the objective by increasing one of the components of "the solution" infinitesimally (while still satisfying all constraints).
Using your notation, the KKT conditions yield $-\frac{w_k}{p^2_k} + \lambda_0 = 0$ for each $k$. I think you can proceed from here by using the active constraint $\sum_j p_j = 1$.
Note that the KKT conditions are necessary for a local minimum because the feasible region is polyhedral.