Consider the following function defined on $x \in \mathbb{R}^+ \cup\{0\}$


$$
f(x)= \log\left(1+\frac{r}{x+a}\right) + \log\left(1+\frac{r}{2x+a}\right) - 2r \log \left(1+\frac{x}{x+a+r} \right),
$$
where $a=3/2$ and $r$ is an arbitrary non-negative number. I want to prove the following statement for $f(x)$ but it seems a bit difficult to me. Does anybody have any idea how to do that?


**Statement:** There exists $x^\*$ for which $f(x)$ is positive in the interval $[0, x^\*)$, and $f(x^\*)=0$, and $f(x)$ is negative in the interval $(x^\* ,\infty)$.



Behavior of $f(x)= \log\left(1+\frac{r}{x+a}\right) + \log\left(1+\frac{r}{2x+a}\right) - 2r \log \left(1+\frac{x}{x+a+r} \right)$
If $r=0$, then $f(x)=0$ for all $x\ge0$. If $r>0$, then 
$$f'(x)=-\frac{8 r (x+1) (8 r x+18 r+24 x+27)}{(2 x+3) (4 x+3) (2 r+2 x+3) (2 r+4 x+3)}<0$$
for $x\ge0$, $f(0)=2 \log \left(\frac{2 r}{3}+1\right)>0$, and $f(\infty-)=-r \log (4)<0$. So, your desired result follows.


