TITLE: Curious inequality
QUESTION [24 upvotes]: 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

REPLY [9 votes]: Writing $k=\sum_{j=1}^k 1$ and then summing by parts, one has 
$$\sum_{k\geq 1} \frac{(-1)^{k-1}k}{(x^{k}+1)^2}
=\sum_{j\ge 1}(-1)^{j-1}s_j
=\sum_{k\ge0}[s_{2k+1}-s_{2k+2}],
$$
where 
$$s_j:=\sum_{k=j}^\infty(-1)^{k-j} a(k)=\sum_{m\ge0}(-1)^m a(m+j)
=\sum_{k\ge0}[a(2k+j)-a(2k+1+j)] 
$$
and
$a(q):=\dfrac1{(x^q+1)^2}$. 
So, 
$$s_j-s_{j+1}=\sum_{k\ge0}[a(2k+j)-2a(2k+1+j)+a(2k+2+j)]>0,
$$
since $a''>0$ and hence $a$ is strictly convex. 
So, $\sum_{k\geq 1} \dfrac{(-1)^{k-1}k}{(x^{k}+1)^2}>0$, and the result follows by Max Alekseyev's answer.