Prove $(\sqrt{2})^{\sqrt{3}-1} \gt (\sqrt{3})^{\sqrt{2}-1}$. Prove $(\sqrt{2})^{\sqrt{3}-1} \gt (\sqrt{3})^{\sqrt{2}-1}$
It is part of an exercise where I am given $f,g:(1,\infty) \to \mathbb R,\,\,\, f(x)=x-1-x\ln x, \,\,\,g(x)=\frac{\ln x}{x-1}$.   
I must use of the information that $f(x) \lt 0$ for any $x \in (1,\infty)$ and that $g(x)$ is decreasing. I banged my head against the wall last night trying to solve it, I couldn't see any connection between these function and the inequality and tried to rewrite the inequality in a lot of different ways with not much success. What am I missing?
 A: The inequality can be rewritten as
$$
\frac{\log\sqrt{2}}{\sqrt{2}-1}>\frac{\log\sqrt{3}}{\sqrt{3}-1}
$$
and it's natural to consider the function
$$
g(x)=\frac{\log x}{x-1}
$$
over the interval $(1,\infty)$. Since
$$
g'(x)=\frac{1}{(x-1)^2}\left(\frac{x-1}{x}-\log x\right)=\frac{f(x)}{x(x-1)^2}
$$
where $f(x)=x-1-x\log x$ (defined over $(1,\infty)$ like $g$).
We have
$$
\lim_{x\to1}f(x)=0
$$
and $f'(x)=1-\log x-1=-\log x$, which is negative over $(1,\infty)$. Hence $f$ is decreasing and so everywhere negative. This implies $g$ is decreasing over $(1,\infty)$.
A: We have
$$
\sqrt{2}^{\sqrt{3} - 1} > \sqrt{3}^{\sqrt2-1}\\
\ln\left(\sqrt{2}^{\sqrt{3} - 1}\right) > \ln\left( \sqrt{3}^{\sqrt2-1}\right)\\
(\sqrt{3} - 1)\ln\sqrt{2} > (\sqrt2-1)\ln\sqrt{3}\\
\frac{\ln\sqrt2}{\sqrt2-1} > \frac{\ln\sqrt3}{\sqrt3-1}\\
g(\sqrt2) > g(\sqrt3)
$$
which, since $g$ is (strictly) decreasing, is indeed true.
Strictly speacing, the logic here flows upwards: You start with the known fact $g(\sqrt2)>g(\sqrt(3)$, and work your way through the steps upwards to conclude with the top inequality. But when trying to work out exactly what steps to do, it is often a lot more productive and easier to see what you have to do if you start with what you want to show and work your way towards something you know.
