Let
$$\begin{align}f(x) = 2 x H(x) \left(\frac{\pi ^2}{6}-H\left(x^2,2\right)\right)\\ -H\left(x^2\right) \left(\frac{\pi ^2}{6}-H(x,2)\right)\end{align}\tag{1}$$
Here $H$ are hamonic numbers, c.f. https://en.wikipedia.org/wiki/Harmonic_number
Explicitly, for integer positive $k$, $H(k,m) =H_{k,m} = \sum_{k=1}^{n}\frac{1}{k^m}$ is the generalized harmonic number and, by definition, $H(k) = H(k,1)$. For real $x\gt0$ we have the integral representations $H(x) = \int_{0}^{1}\frac{(1-t^x)}{1-t}\;dt$ and $H(x,2) =\frac{\pi^2}{6} - \int_{0}^{1}\log(\frac{1}{t})\frac{t^x}{1-t}\;dt$.
In this question Prove that $H(x^2)/H(x)$ increases where $H(x)$ is the harmonic number I have shown that $f(x)$ is positive for $x \gt 0$, asymptotically for both small $x$ and large $x$.
Now I want to prove positivity for any $x \gt 0$. I tried to employ inequalities like
$$2 x H(x)-H\left(x^2\right) \gt 0\tag{2a}$$
$$\frac{\pi ^2}{6}-H\left(z,2\right) \gt 0 \text{, }z>0\tag{2b}$$
For $0<x<1$ we have
$$\frac{\pi ^2}{6}-H\left(x^2,2\right)>\frac{\pi ^2}{6}-H(x,2)\tag{3}$$
so that, by $(2)$ and $(3)$,
$$\begin{align}& f(0 \lt x \lt 1) \\ & > \left(2 x H(x) -H\left(x^2\right)\right) \left(\frac{\pi ^2}{6}-H(x,2)\right)\\ & >0\end{align}\tag{4}$$
So we are left with the region $x\gt 1$.
Even though it seems to be a simple task I am moving in circles, and I kindly ask for your help.