<s>[INST] Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.

Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivative of the sought fraction is
\begin{align}
&H(x)^2\frac{dh(x)}{dx} \\
=&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\
=& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\
=& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1
\end{align}
All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get
\begin{align}
g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\
&=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\
&= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt
\end{align}
for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$.
$$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$
Let $p(x):=2x^3+3x^2-2x-7$. It can be shown that $p(x)\le p(1)=-4, \forall x\in[0,1]$. We can take $a=\frac13$ since we can show, with a bit of work, $f''(\frac13)>0$.
(to be continued)
 [/INST] Partial Hints :
We broke the fraction into two part :
Show that :
$$f(x)=x^{-1}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)$$
Is increasing on $[0,2/5]$
And :
$$g(x)=x^{-1}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$
Is decreasing on $(0,1)$
Obviously a increasing function divided by a decreasing function both positive is increasing .
We can make the same thing on $x\in[2/5,3/5]$ with :
$$\frac{1}{1+x}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right),\frac{1}{1+x}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$
Next we use symmetry to get :
$$r\left(x\right)=\frac{\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)}{x},\frac{r\left(1-x^{2}\right)}{x},\frac{t\left(1-x^{2}\right)}{x},t\left(x\right)=\frac{-x\ln\left(x^{2}\right)-\left(1-x\right)\ln\left(\left(1-x\right)^{2}\right)}{x}$$
Currently the function $\frac{r\left(1-x^{2}\right)}{x}$ is decreasing on $[0.5,1)$ and $\frac{t\left(1-x^{2}\right)}{x}$ is increasing both positive so the fraction is decreasing remains to substitute $y=1-x^2$.
For a end see RiverLi's answer I stop here .
</s>