I have two real sequences $a\_1,a\_2,\dots,a\_n$ and $b\_1, b\_2, \dots, b\_n$, with $a\_i > 0$ and $1 \leq b\_i < n$, and I'm looking for a lower bound of $\sum\_i \frac{a\_i}{b\_i}$ in terms of $\sum\_i a\_i$ and $\sum\_i b\_i$. There is also an extra contraint that if $b\_i$ is large then $a\_i$ is small (something like $a\_i < b\_i/e^{b\_i})$


They're not necessarily monotonic, so I can't use Chebyshev's inequality. Is there something else that I'm missing?


Edit: 


After thinking a bit more about the problem I get that, as stated, there are instances where $\sum \frac{a\_i}{b\_i} = \frac{\sum a\_i}{\sum b\_i - (n-1)}$. 


That is if $a\_i = \epsilon$ for $1 \leq i \leq n-1$ and $a\_n = \sum a\_i - (n-1)\epsilon$, and $b\_i = 1$ for $1 \leq i \leq n-1$ and $b\_n = \sum b\_i - (n-1)$. Then, as long as $\sum a\_i < \frac{\sum b\_i - (n-1)}{e^{\sum b\_i - (n-1)}} + (n-1)\epsilon$, we get the above as $\epsilon$ goes to $0$. Is this tight as long as the inequality is satisfied? If it isn't satisfied, then $\sum a\_i$ is larger than $\frac{\sum b\_i - (n-1)}{e^{\sum b\_i - (n-1)}}$, implying that the lower bound should be larger. 


So to tweak the problem a bit more, what if the inequality isn't satisfied (for small $ \epsilon $)? Say for $\sum a\_i = 1$ and $\sum b\_i = 2n$?



Ratio of Sequences Sum Inequality
Here is an old cheap trick that may be helpful. Assume that $\sum\_i a\_i=A$, $\sum\_i b\_i=B$ and $a\_i\le b\_if(b\_i)$ where $f$ is a decreasing function tending to $0$. Choose $b$ so that $f(b)=\frac{A}{2B}$. Then $\sum\_{i:b\_i>b}a\_i\le f(b)\sum\_{i:b\_i>b}b\_i\le f(b)B\le\frac A2$, so $\sum\_i\frac{a\_i}{b\_i}\ge \frac A{2b}$.


For $f(x)=e^{-x}$, this results in the estimate $\sum\_i\frac{a\_i}{b\_i}\ge \frac A{2\log\tfrac{2B}A}$. It is not optimal, of course, but may suffice for your purposes. If it doesn't, then tell in what range you'd like to improve it.


