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
$$
\frac{d}{dy} R(1,y^2,y^4,\dots)=\sum_{k=1}^{\infty} 2ky^{2k-1}\frac\partial{\partial t_k} R(1,y^2,\dots)\geq 0.
$$