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

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.

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.