<s>[INST] Curious inequality Set 
$$
g(x)=\sum_{k=0}^{\infty}\frac{1}{x^{2k+1}+1}  \quad \text{for}  \quad  x>1. 
$$
Is it true that 
$$
\frac{x^{2}+1}{x(x^{2}-1)}+\frac{g'(x)}{g(x)}>0 \quad \text{for}\quad x>1? 
$$
The answer seems to be positive. I spent several hours in proving this statement but I did not come up with anything reasonable. Maybe somebody else has (or will have) any bright idea?  
Motivation? Nothing important, I was just playing around this question:
A problem of potential theory arising in biology
 [/INST] The inequality is equivalent to 
$$S := (x^2+1)g(x) + x(x^2-1)g'(x) > 0.$$ 
The left hand side here can be expanded to
$$S = \sum_{k\geq 0} \frac{(x^2+1)(x^{2k+1}+1) - (2k+1)x^{2k+1}(x^2-1)}{(x^{2k+1}+1)^2} $$
$$= \sum_{k\geq 0} \frac{(x^2+1) - (2k+1)(x^2-1)}{x^{2k+1}+1} + \sum_{k\geq 0}\frac{(2k+1)(x^2-1)}{(x^{2k+1}+1)^2}.$$
Now, the first sum here simplifies to
$$\sum_{k\geq 0} \frac{(x^2+1) - (2k+1)(x^2-1)}{x^{2k+1}+1} = \sum_{k\geq 0} \frac{(2k+2)-2k x^2}{x^{2k+1}+1}$$
$$=\sum_{k\geq 1} \left( \frac{2k}{x^{2k-1}+1} - \frac{2k x^2}{x^{2k+1}+1}\right)=(1-x^2)\sum_{k\geq 1} \frac{2k}{(x^{2k-1}+1)(x^{2k+1}+1)}.$$
Hence
$$\frac{S}{x^2-1} = \sum_{k\geq 0}\frac{2k+1}{(x^{2k+1}+1)^2} - \sum_{k\geq 1} \frac{2k}{(x^{2k-1}+1)(x^{2k+1}+1)}$$
$$\geq \sum_{k\geq 0}\frac{2k+1}{(x^{2k+1}+1)^2} - \sum_{k\geq 1} \frac{2k}{(x^{2k}+1)^2} = \sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2}.$$
Here we used AM-GM inequality $x^{2k-1}+x^{2k+1}\geq 2x^{2k}$ and thus 
$$(x^{2k-1}+1)(x^{2k+1}+1)=x^{4k}+x^{2k+1}+x^{2k-1}+1\geq (x^{2k}+1)^2.$$
So it remains to prove that for $x>1$,
$$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2} > 0.\qquad(\star)$$
UPDATE #1. Substituting $x=e^{2t}$, we have
$$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2} = \sum_{k\geq 1} \frac{(-1)^{k-1}ke^{-2tk}}{4 \cosh(tk)^2} = \frac{1}{4}\sum_{k\geq 1} (-1)^{k-1}ke^{-2tk}(1-\tanh(tk)^2)$$
$$ = \frac{e^{2t}}{4(e^{2t}+1)^2} - \frac{1}{4}\sum_{k\geq 1} (-1)^{k-1}ke^{-2tk} \tanh(tk)^2.$$
UPDATE #2. The proof of $(\star)$ is given by Iosif Pinelis.
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