For all $a$ and $b$ such that $a > b > 0$, prove that
$$(a+1)^{b}>(b+1)^{a}.$$
It is equal to $\sqrt[a]{a+1} > \sqrt[b]{b+1}$, but the function $f(x) = \sqrt[x]{x+1}$ is decreasing for $x \in \left(0,+\infty\right)$ (actually it decreases on its entire domain), so doesn't it imply that $f(x_0) < f(x_1)$ is true, for all $x_0$ and $x_1$ in the domain where $x_0 > x_1$?
Is that sufficient to prove that the statement is false, or am I totally wrong?
On
As you said, since $f(x)=\sqrt[x]{x+1}$ is decreasing for positive $x$, for $a>b>0$, we can write $$\sqrt[b]{b+1} > \sqrt[a]{a+1}.$$ Raising each side to the $ab$th power will preserve the inequality and give $$(b+1)^a > (a+1)^b.$$ This makes me believe the inequality is supposed to be reversed. Alternatively, it is intended that $b>a>0$ instead of $a>b>0$.
On
The inequality is reversed. If we consider the function $f(x)=\frac{\log(1+x)}{x}$ over $\mathbb{R}^+$, we have: $$ f(x)=\frac{1}{x}\int_{0}^{x}\frac{dt}{t+1} $$ and since $g(t)=\frac{1}{t+1}$ is a decreasing function over $\mathbb{R}^+$, so it is $f(x)$. It follows that $a>b>0$ implies $f(a)<f(b)$ or $(a+1)^b\color{red}{<}(b+1)^a.$
On
Let us prove that for $0<b<a$ holds $$\frac{\ln{(a+1)}}{a}<\frac{\ln{(b+1)}}{b},$$ which is equivalent with $$(a+1)^b<(b+1)^a.$$
Firstly, observe that $$f(t)=\ln{(t+1)}-\frac{t}{t+1}\geq 0$$ for $t\geq 0$.
(Function $f(t)$ is increasing on $[0,\infty)$, since $f'(t)=\frac{t}{(t+1)^2}\geq 0$. Therefore, for every $t\geq 0$ is $f(t)\geq f(0)=0$).
Define $$g(t)=\frac{\ln{(t+1)}}{t}$$ for $t\geq 0$.
Since $g'(t)=-\frac{f(t)}{t^2}\leq 0$ for $t\geq 0$, we can conclude that function $g$ is decreasing on $[0,\infty)$.
Therefore, for $0<b<a$ holds $g(b)>g(a)$, which is equivalent with $$\frac{\ln{(a+1)}}{a}<\frac{\ln{(b+1)}}{b}.$$
The inequality does not always hold. When $a = 2, b= 0.5$, then the inequality fails.