I have two positive sequences $a_i$ and $b_i$ with unknown (finite) index set $I$. I also have the following: \begin{aligned} \sum_{i \in I} a_i & = \log(m) 2^n \\ \sum_{i \in I} 2^{a_i} & = m 2^n \\ \sum_{i \in I} 2^{a_i} b_i & \leq m \log{m} 2^n. \end{aligned} Is it possible to deduce that $$\sum_{i \in I} \frac{a_i}{b_i} \geq 2^n ?$$
The intuition behind the bounds is that if all $a_i = \log{m}$ and $|I| = 2^n$ then from the last inequality, the $b_i$ must average to at most $\log{m}$ with the equal $2^{a_i}$ weights, so it should also average to that under the $a_i$ weights, i.e $$\sum_{i \in I} a_i b_i \leq \log{m}^2 2^n$$ and then a Jensen's inequality gives the bound.
I can show that if $|I| 2^{-n} = d$ then $\log{m} \leq d (\log{m} - \log{d})$ giving that $1 \leq d$. (It also gives an upper bound on $d$, not sure if that is relevant). This is from another Jensen's inequality, so the $1$ lower bound should be tight when they are all equally $\log{m}$. And if we had $d$ larger, then my intuition is that the $a_i$ must be generally smaller, while the $b_i$ average is still $\log{m}$, which can only increase the inequality we want to deduce.
I also have the following observation: $$\sum_{i \in I} a_i b_i \leq \frac{1}{e \ln{2}} \sum_{i \in I}{2^{a_i} b_i}$$ This follows by assuming $\sum_{i \in I} b_i = B and then applying another Jensen's with the worst case B, but clearly this gives a really bad bound, since the worst case B is far worse than what I conjecture.
Is there any hope that $$\sum_{i \in I} \frac{a_i}{b_i} \geq 2^n$$ holds? If so, how can we deduce it?