Minimum of $\prod_{1\leq i \leq n} (1-p_i) \left[ \sum_{i=1}^n \frac{p_i}{1-p_i} + \sum_{i=1}^n \sum_{j=i+1}^n \frac{p_i p_j}{(1-p_i)(1-p_j)} \right]$

Question

Let $p_1, ..., p_n \in (0,q)$ such that $\sum_i p_i \geq q$ for $q\in[0,1]$. I think that

$$\prod_{1\leq i \leq n} (1-p_i) \left[ \sum_{i=1}^n \frac{p_i}{1-p_i} + \sum_{i=1}^n \sum_{j=i+1}^n \frac{p_i p_j}{(1-p_i)(1-p_j)} \right]$$ Is minimized when $p_i=\frac{q}{n}$ for all $i$, but I don't know how to show it. Any hints? If not, then what is the minimizer?

I can show $$\prod_{1\leq i \leq n} (1-p_i) \left[ \sum_{i=1}^n \frac{p_i}{1-p_i} \right]$$ is minimized when $p_i=\frac{q}{n}$,

Answer 1

Partial answer, for the case $q=1$, $n>2$, there seems to be no minimizer, but, you can get as close to 0 as you want by choosing $p_i=1-\epsilon$: $$ \epsilon^n \left[ n\frac{1-\epsilon}{\epsilon} + \frac{n(n-1)}2 \frac{(1-\epsilon)^2}{\epsilon^2} \right]\\ =n \epsilon^{n-2} (1-\epsilon)\left(\epsilon+\frac{(n-1)(1-\epsilon)}{2}\right) $$ Since we assumed $n>2$, this scales down with $\epsilon$