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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 $(x^* ,\infty)$.

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    $\begingroup$ This is not a question at the reseach level, rather an exercise on calculus. $\endgroup$
    – user64494
    Commented Jun 27, 2018 at 19:37
  • $\begingroup$ I think we shouldn't judge the questions, rather we have to respect. Comments are for technical discussions I believe. And FYI, this question is closely related to a research problem. ;-) $\endgroup$
    – James
    Commented Jun 27, 2018 at 21:31
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    $\begingroup$ We certainly should judge questions, James. You've been here for 2 months. user64494 has been here for 5 years, and just might have a better idea of what MO is for and how it's meant to work than you have. $\endgroup$
    – Gerry Myerson
    Commented Jun 27, 2018 at 23:33

1 Answer 1

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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.

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