Does this change in this monotonic function affect ranking? I need to make sure I can take out the one in $(1-e^{-x})e^{-y}$ without affecting a sort order based on this function. I other words, I need to prove the following:
$$
(1-e^{-x})e^{-y} \ >= \ -e^{-x}e^{-y}\quad\forall\ \ x,y> 0
$$
If that is true, then I can take the logarithm of the right hand side above: $\log(-e^{-x}e^{-y}) = x + y$ and my life is soooo much easier...
 A: The left-hand side develops as
$$
e^{-y}-e^{-x}e^{-y}.
$$
The $e^{-y}$ part is strictly positive and so your inequality holds.
Note that you can't take the $\ln$ because you would be taking the logarithm of a negative number.
A: Looking at the current version of your post, we have
$$(1-e^{-x})e^{-y}=e^{-y}-e^{-x}e^{-y}>-e^{-x}e^{-y},$$ since $e^t$ is positive for all real $t$. However, we can't take the logarithm of the right-hand side. It's negative.
Update:
The old version was $$(1-e^{-x_1})(1-e^{-x_2})e^{-x_3}=e^{-x_3}-e^{-x_1-x_3}-e^{-x_2-x_3}+e^{-x_1-x_2-x_3},$$ and you wanted to know if that was greater than or equal to $$(-e^{-x_1})(-e^{-x_2})e^{-x_3}=e^{-x_1-x_2-x_3}$$ for all positive $x_1,x_2,x_3$. Note, then, that the following are equivalent (bearing in mind the positivity of $e^t$):
$$(1-e^{-x_1})(1-e^{-x_2})e^{-x_3}\geq e^{-x_1-x_2-x_3}$$
$$e^{-x_3}-e^{-x_1-x_3}-e^{-x_2-x_3}\geq 0$$
$$e^{-x_3}(1-e^{-x_1}-e^{-x_2})\geq 0$$
$$1-e^{-x_1}-e^{-x_2}\geq 0$$
This need not hold. In fact, for any $x_2>0$, there is some $x_1>0$ such that the inequality fails to hold. (Let me know if you're interested in a proof of that fact.)
