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](https://mathoverflow.net/questions/217530/a-problem-of-potential-theory-arising-in-biology)



Curious inequality
I am not sure at all, please double check.


Denote $u(t)=(1+t)^{-1}$, it is a decreasing convex function on $[0,1]$, but $u(e^s)$ is concave on $(-\infty,0]$. As Neil Strickland notes, what we have to prove is that $(x-1/x)g(x)$ increases, or, if we denote $y=1/x$, that $(1/y-y)g(1/y)$ decreases. We have $$(1/y-y)g(1/y)=\sum\_{k=0}^{\infty} \frac{y^{2k}-y^{2k+2}}{1+y^{2k+1}},$$
this is a Riemann sum of the function $u(t)$ corresponding to nodes $t\_k=y^{2k}$, $t\_0>t\_1>\dots$, and intermediate points $s\_k=y^{2k+1}=\sqrt{t\_kt\_{k+1}}\in [t\_k,t\_{k+1}]$. The Riemann sums converge to the integral, and they are always more than integral since $\int\_a^b u(t)dt\leq (b-a)(u(a)+u(b))/2\leq (b-a)u(\sqrt{ab})$ as $u$ is convex and $u(e^s)$ is concave on $(-\infty,0]$. Next, we see that when $y$ increases, all nodes become closer to 1. It suggests to move nodes and see how the Riemann sum $R(t\_0,t\_1,\dots):=\sum (t\_i-t\_{i+1})u(\sqrt{t\_it\_{i+1}})$ behaves. Consider three consecutive nodes $a^2<b^2<c^2$ and increase $b$. What happens to $(b^2-a^2)u(ab)+(c^2-b^2)u(bc)$? Its derivative in $b$ equals
$$
-\frac{(c-a)((a-c)^2+3(ac-b^2)+2(abc-b^3)(a+c)+ab^2c(ac-b^2))}
{(ab+1)^2 (bc+1)^2}.
$$
This is strictly negative if $b^2=ac$ (that holds in our case). It follows that $R(1,y^2,y^4,\dots)$ decreases when $y$ increases, as desired, by
$$
\frac{d}{dy} R(1,y^2,y^4,\dots)=\sum\_{k=1}^{\infty} 2ky^{2k-1}\frac\partial{\partial t\_k} R(1,y^2,\dots)\geq 0.
$$


