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- Feb 14, 2012

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Problem:

Assume $x>0$, and $y$ satisfy that $e^y=\dfrac{x}{1-e^{-x}}$, prove that $x>2y$.

Attempt:

I first tried to express $y$ in terms of $x$ and get:

$y=x+\ln x- \ln (e^x-1)$

and I am aware that one of the method to prove the intended result is to rewrite the equation above as

$y-\dfrac{x}{2}=\dfrac{x}{2}+\ln x- \ln (e^x-1)$

and if I can prove $\dfrac{x}{2}+\ln x- \ln (e^x-1)<0$, which also implies $y-\dfrac{x}{2}<0$, then the result is proved.

Now, I see it that no inequality theorems that I know of could be applied to the expression $\dfrac{x}{2}+\ln x- \ln (e^x-1)$ and hence I get stuck, so stuck that I wish to drop this problem behind my mind!

But, I hope to solve it nonetheless and that's why I posted it here and hope someone can chime in to help me out.

Thanks in advance.