$\log_{a+1}(b+1)-\log_ab=(\frac{\log_b(b+1)}{\log_a(a+1)}-1)\log_ab>0$. Let $f(x)=\log_x(x+1)$. $f'(x)<0$, therefore $f(x)$ decrease, when $0<x<1,(f(x)<0)$ and when $x>1,(f(x)>1)$. If $b=1$ for any $a>0 \ \ 0=\log_a1<\log_{a+1}2$. If $a<1,b<1$, then $\frac{f(b)}{f(a)}>1\to f(b)<f(a)\to b>a$. If $a>1,b<1$ or $a<1,b>1$, then hold. If $b>1,a>1$. then $\frac{f(b)}{f(a)}>1\to f(a)<f(b)\to a>b$. Answer $0<a<b<1, 1<b<a, (a<1,b\ge 1),(a>1,b\le 1).$

Approach is quite same as above, but it's a good solution. Consider the function $f(x)=\frac{log(x+1)}{log(x)}$. We know that $log(x+1)\ge log(x)$ and $\frac{1}{x}\ge \frac{1}{x+1} >0$. So we have $f'(x)=\frac{\frac{log(x)}{x+1}- \frac{log(x+1)}{x}}{log^2(x)} \le 0$. So $f$ is decreasing. Also $f(x)\le 0$ whenever $x\le 1$, and $f(x)\ge 0$ when $x\ge1$. The function is discontinuous at $x=1$. If $b\ge 1$, then $f(b)\ge f(a)$ Now if $b\le 1$, then we have $f(b)\ge f(a)$. So the solutions are $(b=1,a\in \mathbb{R}-1$ and $(a>b>1)$ and $(b>1>a)$ and just alternatives. Done.