This question came from my misunderstanding of this question Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. where H(x) is an entropy function. I have read H(x) as the harmonic number and found that the claim holds as well.
Define $f(x) = \frac{H(x^2)}{H(x)}$
where $H(x)$ is the harmonic number (https://en.wikipedia.org/wiki/Harmonic_number) defined for positive integers $x=n$ as $H(n)=\sum_{k=1}^{n}\frac{1}{k}$ and for real $x$ as $H(x) = \int_{0}^{1}\frac{1-t^x}{1-t}\;dt$
My first attempts are:
Plotting $f(x)$ indicates that the claim holds.
Now we consider the two limiting cases $x \to 0$ and $x \to \infty $
For $x \to 0$ we have $H(x) = \sum_{k\ge 2}(-1)^k x^{k-1}\zeta(k) \simeq x \; \zeta(2)$ and hence $f(x) \simeq x $ which is obviously increasing.
For $x \to \infty $ we have the well known approximation $H(x) \simeq \gamma + \log(x)$ where Euler's $\gamma$ appears. This leads to $f(x) \simeq \frac{\gamma + \log(x^2)}{\gamma + \log(x)}=\frac{\gamma + 2\log(x)}{\gamma + \log(x)}= 1 + \frac{\log(x)}{\gamma + \log(x)}$. The drivative is $f'(x) \simeq \frac{\gamma }{x (\log (x)+\gamma )^2}$ which is positive showing that $f(x)$ is increasing.