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)
 A: Not quite what you asked, but a simple proof of a very related inequality, by Boppana: https://arxiv.org/abs/2301.09664
Theorem. The binary entropy function $h$ satisfies
$$
\frac{h(x^2)}{h(x)} \geq \phi\cdot x, \quad \quad \forall p\in[0,1]
$$
where $\phi=\frac{1+\sqrt{5}}{2}$ denotes the golden ratio.
A: Let
$$u = \ln x, \quad v = \ln(1-x), \quad w = \ln(1 + x).$$
We have
\begin{align*}
 &[H(x)]^2h'(x)\\[6pt]
  =\,& [(1-x)^2v - (1+x^2)u]w + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2\\[6pt]
  \ge\,& [(1-x)^2v - (1+x^2)u]\cdot (x - x^2/2)\\[6pt]
&\qquad + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2 \tag{1}\\[6pt]
  =\,& 2u^2x^2 - [x(1-x/2)(1+x^2) + (1-3x)(1-x)v]u\\[6pt]
&\qquad + v(1-x)^2(v + x - x^2/2)\\[6pt]
  \ge\,& 0. \tag{2}
\end{align*}
Explanations:
(1): $\ln(1+x) \ge x - x^2/2$ for all $x \ge 0$;
Using $(1-x)\ln(1-x) > - 1$, we have
\begin{align*}
 (1-x)^2v - (1+x^2)u &= (1-x)^2\ln(1-x) - (1+x^2)\ln x\\
 &\ge -(1-x) - \ln x\\
 &\ge 0.
\end{align*}
(2):
$\ln(1-x) + x - x^2/2\le 0$ for all $x\in [0, 1)$;
If $1/3 \le x < 1$, clearly
$$x(1-x/2)(1+x^2) + (1-3x)(1-x)\ln(1-x) \ge 0$$
and if $0 \le x < 1/3$, using $(1-x)\ln(1-x) \ge -x + x^2/2$ for all $0 \le x < 1$, we have
\begin{align*}
 &x(1-x/2)(1+x^2) + (1-3x)(1-x)\ln(1-x)\\
 \ge\,& x(1-x/2)(1+x^2) + (1-3x)(-x + x^2/2)\\
 =\,& x^2(3+x)(1-x/2)\\
 \ge\,& 0.
\end{align*}
We are done.
A: 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)
A: I thought about using the general properties $H(0)=H(1)=1$, $H(x)=H(1-x)$, $H''(x)<0$ and $H'(0)=+\infty$.
However, if you plug in $f(x)=\sin(\pi\sqrt{x})$ and $g(x)=f(x)+f(1-x)$ on a graphing calculator,
$\dfrac{g(x^2)}{g(x)}$ seems to increase from $x=0$ to around $x=0.997$, and then it decreases ever so slightly.
Therefore, you probably need to use the precise expression of $H(x)$, and I don't expect a proof much more elegant than the one you sketched.
A: 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 .
