<s>[INST] Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x)$ has a zero at $x=\frac{1}{2}$. But can we conclude that this is unique？
 [/INST] Note that the derivative is
$$f’(x) = \frac{\ln(1 - x)}{x} - \frac{\ln x}{1 - x}$$
for $0 < x < 1.$ We want to show that $f(x)$ is maximized when $x = 1/2.$
Since we already know that $f’(1/2) = 0,$ it remains to show that $f’’(1/2) < 0$ and that $f’(x) \neq 0$ for all other $x$ in the interval.
We have
$$\begin{align}
f’’(x) &= \frac{1}{x}\cdot \frac{-1}{1 - x} - \frac{\ln(1 - x)}{x^2} - \frac{1}{x}\cdot\frac{1}{1 - x} - \frac{\ln x}{(1 - x)^2} \\
&= -\frac{2}{x (1 - x)} - \frac{\ln(1 - x)}{x^2} - \frac{\ln x}{(1 - x)^2},
\end{align}
$$
from which we can compute that $f’’(1/2) = 8 \ln 2 - 8.$ Since $\ln 2 < 1,$ it follows that $f’’(1/2) < 0.$
If we can show that the function $g$ given by $$g(x) = \frac{\ln(1 - x)}{x}$$ for $0 < x < 1$ is injective, then that will be sufficient to prove that $f’(x)$ is only $0$ for $x = 1/2.$ To do that, we will apply the inequality $$\ln x > 1 - \frac{1}{x}$$ for $x$ in this interval. If you are unfamiliar with this inequality, you can see this question.
From
$$\begin{align}
g’(x) &= -\frac{1}{x(1 - x)} - \frac{\ln(1 - x)}{x^2} \\
&< -\frac{1}{x(1 - x)} - \frac{1}{x^2}\left(1 - \frac{1}{1 - x}\right) \\
&= -\frac{1}{x(1 - x)} + \frac{1}{x(1 - x)} \\
&= 0,
\end{align}
$$
it follows that $g$ is strictly decreasing. Thus, the only zero of $f’(x)$ occurs when $x = 1/2$.
By the second derivative test, we have $$\ln x \ln(1 - x) \leq \ln^2(1/2) = \ln^2 2.$$
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