Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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
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.


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.)

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.