<s>[INST] Determine $\textrm{sign}[x + \ln y]$ I need to determine the sign of 
\begin{align}
f(a,b) = \textrm{sign}\left[\frac{(a-1)(b-1)^2}{ba^2(b(a-1)+1)} + \ln\left(\frac{b(a-1)+1}{ab}\right)\right] = ?
\end{align}
with $a \geq 1$, and $b > 0$. 
We have $f(a,b=1) = 0$. I conjecture also that $f(a,b<1) > 0$ and $f(a,b>1) < 0$. Cannot show it, however, because of the $\ln$. Any hints are appreciated.  
Idea: There is only one root for $f$, namely $f(a,b) = 0$ iff $b=1$ for all $a$. Let $a=1$ such that 
\begin{align}
f(1,b) = \ln\left(\frac{1}{b}\right) 
\begin{cases}
> 0 \quad &\textrm{if } b < 1\\
< 0 \quad &\textrm{if } b > 1.
\end{cases}
\end{align}
 [/INST] for $0<b<1$:
$$
[\ldots]=\left[\underbrace{\ldots}_{> 0}+\ln\Big(\underbrace{1+\frac{1-b}{ab}}_{> 1}\Big)\right]>0.
$$
for $b>1$: use the fact that $\ln(1+x)\le x$, for $x>-1$, i.e.
$$
[\ldots]\le\frac{a-1}{a-1+1/b}\cdot\Big(\frac{b-1}{ab}\Big)^2
-\frac{b-1}{ab}=\epsilon\cdot q^2-q<0
$$
since $0<\epsilon<1$ and $0<q<1$.
</s>