I am trying to solve a tricky little problem involving sums of exponentials.
Consider the sequences $\{a_m\}$ and $\{b_m\}$. Both of these sequences are finite and $a_m>0 ~\forall m$ and $b_m>0~\forall m$.
It is always possible to find a $C$ such that \begin{equation} e^{-C} = \sum\limits_{m=1}^N e^{-a_m}. \end{equation} Simple algebraic manipulation shows that $C = -\text{log}\left(\sum\limits_{m=1}^Ne^{-a_m}\right)$.
Now I am interested in the sum \begin{equation} e^{-D} = \sum\limits_{m=1}^Ne^{-a_m-b_m}. \end{equation}
In particular, I wish to simplify the ratio between $e^{-C}$ and $e^{-D}$
Some thoughts.
If $A =\sum_{k=1}^n e^{-a_n} $, $B =\sum_{k=1}^n e^{-b_n} $, and $C =\sum_{k=1}^n e^{-a_n-b_n} $, then, depending of the ordering of $(a_k)$ and $(b_k)$, Chebychev's sum inequality might be of use: https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality
Another possibility might be to use the AM-GM inequality: If $\alpha =\sum_{k=1}^n a_n $,
$\frac{A}{n} \ge \left(\prod_{k=1}^n e^{-a_n} \right)^{1/n} = \left(\exp(\sum_{k=1}^n -a_n) \right)^{1/n} = \left(\exp(-\alpha) \right)^{1/n} = \exp(-\alpha/n) $, and similarly for $B$ and $C$.