Show that $\exp\left(\frac{1}{x}\log\frac{e^{x}-1}{x}\right)$ is increasing. 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?
 A: 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.
A: Write
$$ f(x) = \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} = \left( \int_{0}^{1} e^{xs} \, \mathrm{d}s \right)^{\frac{1}{x}}. $$
Now let $0 < x < y$ be arbitrary and write $p = \frac{y}{x} > 1$. Then by the Jensen's inequality applied to the strictly convex function $\varphi(t) = t^p$ over $[0, \infty)$, we get
$$ f(x)^{y} = \varphi\left( \int_{0}^{1} e^{xs} \, \mathrm{d}s \right) < \int_{0}^{1} \varphi(e^{xs}) \, \mathrm{d}s = f(y)^{y}, $$
and therefore $f(x) < f(y)$ as desired.

Remarks.

*

*This is a particular instance of a more general observation that the $L^p$-norm $$\| X\|_{L^p} := (\mathbb{E}[|X|^p])^{1/p}$$ of a random variable $X$ is non-decreasing in $p$.


*We may instead use the Hölder's inequality in the proof.
A: Hint : If we have the power series at $x=0$ $$f(x)=e^{x}(x-1)+1-(e^{x}-1)\log(e^{x}-1)+(e^{x}-1)\log x=\sum_{i=0}^{\infty}a_ix^i$$
Prove that $a_i\geq 0$
As underlined by the users the part above is false I propose another proof :
The problem is equivalent to prove that :
$$f(x)=x\ln(x(e^{\frac{1}{x}}-1))$$
Is decreasing .
The second derivative is :
$$f''(x)=\frac{1}{x} - \frac{\operatorname{csch}^2\Big(\frac{1}{2x}\Big)}{4 x^3}$$
Where $\operatorname{csch}$ is the
Hyperbolic Cosecant .
Putting $y=\frac{1}{2x}$ we get :
$$2y - 2y^3\operatorname{csch}^2\Big(y\Big)$$
We want to show that :
$$2y - 2y^3\operatorname{csch}^2\Big(y\Big)\geq 0$$
Or :
$$2 \geq 2y^2\operatorname{csch}^2\Big(y\Big) $$
It's not hard to prove .
Furthermore it's not hard to see that $\lim_{x\to \infty}f''(x)=0$
We conclude that $\lim_{x\to \infty}f'(x)=l$
But again it's not hard to show that $\lim_{x\to \infty}f(x)=l'$ $l'\neq \infty$ so  $\lim_{x\to \infty}f'(x)=l=0$ and $f'(x)$ is increasing on $(0,\infty)$ so the derivative of $f(x)$ is negative and the conclusions follows .
