TITLE: Show that $\exp\left(\frac{1}{x}\log\frac{e^{x}-1}{x}\right)$ is increasing.
QUESTION [6 upvotes]: I want to show that $f(x)= \exp\left(\frac{1}{x}\log\frac{e^{x}-1}{x}\right),x>0$ is increasing. I figure that it is enough to show that $\frac{1}{x}\log\frac{e^{x}-1}{x}$ is increasing. The derivative is
$$
\frac{e^{x}(x-1)+1-(e^{x}-1)\log(e^{x}-1)+(e^{x}-1)\log x}{x^{2}(e^{x}-1)},
$$
so I think I need to show that the numerator is positive. However, I got stuck there.
I am guessing that the inequality $e^x-1\ge x$ would be useful, but have no idea how to use it.
How can I show that $f$ is increasing?

REPLY [3 votes]: First
\begin{eqnarray}
F(x)&:=&e^{x}(x-1)+1-(e^{x}-1)\log(e^{x}-1)+(e^{x}-1)\log x\\
&=&(e^{x}-1)(x-1)+x-(e^{x}-1)\log(\frac{e^{x}-1}{x})\\
&=&(e^x-1)g(x)
\end{eqnarray}
where
$$ g(x)=x-1+\frac{x}{e^x-1}-\log(\frac{e^{x}-1}{x}).$$
Clearly $g(x)=0$. Note
$$ g'(x)=\frac{e^{2x}+1-e^x(2+x^2)}{x(e^x-1)^2}=\frac{e^x}{x(e^x-1)^2}(e^x+e^{-x}-2-x^2). $$
It is very easy to show that
$$ e^x+e^{-x}-2-x^2>0$$
which I omit the detail and hence $g'(x)>0$ for $x>0$. This gives $g(x)>g(0)=0$. So
$F(x)>0$ which implies $f(x)$ is increasing.