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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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Do you mean $$(-e^{-x_1})(-e^{-x_2})e^{-x_3},$$ perhaps? –  Cameron Buie Nov 7 '12 at 19:41
    
@CameronBuie. Yes! Thanks a lot. I fixed that and simplified the problem a bit. –  Diego Nov 7 '12 at 20:11
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2 Answers

up vote 1 down vote accepted

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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Arrrgh... My math so rusty! –  Diego Nov 7 '12 at 20:21
    
Note that the original problem has a different answer than the simplified version. See my updated answer. –  Cameron Buie Nov 7 '12 at 20:28
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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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Thanks Jean. I have to give it to Cameron, though, because he answered it first. –  Diego Nov 7 '12 at 20:20
    
@Diego I actually answered first, and we added the negative log almost simultaneously but it's alright with me ;) –  Jean-Sébastien Nov 7 '12 at 20:24
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