How prove $f (x)>\sqrt{\frac{x+1}{x}}$ for $f(x)=\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x+1}}-1}$ Let $$f(x)=\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x+1}}-1}.$$
How to prove that for every $x\in (0,\infty)$ we have:
$$f(x)>\sqrt{\frac{x+1}{x}}$$?

We just have to prove that
$$ f(x) = \sqrt{x}\left(e^\frac{1}{x}-1\right) $$
is decreasing on $\mathbb{R}^+$, or that:
$$ g(x) = f\left(\frac{1}{x}\right)=\frac{e^x-1}{\sqrt{x}} $$
is increasing, that is trivial, since:
$$ g(x) = \sqrt{x}\cdot\frac{e^x-1}{x}=\sqrt{x}\left(1+\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\ldots\right) $$
is the product of two positive increasing functions on $\mathbb{R}^+$.
