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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...

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    @CameronBuie. Yes! Thanks a lot. I fixed that and simplified the problem a bit.2012-11-07

2 Answers 2

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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.)

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    Note that the original problem has a different answer than the simplified version. See my updated answer.2012-11-07
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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.

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    @Diego I actually answered first, and we added the negative log almost simultaneously but it's alright with me ;)2012-11-07