The problem: Prove that
$(1+1/a)^{a+1}<(1+1/b)^{b+1}$ if $a>b>0$.
I have tried to prove this by obtaining the derivate of $(1+1/x)^{x+1}$:
$\frac{d}{dx}(1+1/x)^{x+1}=((1 + 1/x)^x (1 + x) (-1 + x \ln(1 + 1/x)))/x^2$
and showing $x \ln(1 + 1/x)$ is less than $1$ on $x>0$. But I failed.
It is curious that $(1+1/x)^{x+1}$ decrease on $x>0$ but $(1+1/x)^x$ is increase on $x>0$.
How can we prove this?
You have to prove that the function $$ \left(1+\frac{1}{x}\right)^{x+1} = \exp f(x), \quad \text{with}\quad f(x) := (x+1)\log(1+1/x),\quad x>0, $$ is decreasing in $(0, +\infty)$, which amounts to prove that $f$ is decreasing in the same interval.
Computing the derivative you get $$ f'(x) = \log(1+1/x) - 1/x < 0 \qquad \forall x>0, $$ since $\log(1+t) \leq t$ for every $t> -1$ (with strict inequality for $t\neq 0$).
