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？

A proof by someone posted here
\begin{align*} \ln x\ln(1-x)&=\sum_{n=1}^\infty\frac{1-x^n-(1-x)^n}{n^2}\\ &\le \sum_{n=1}^\infty\frac{1-2\left[\frac{x+(1-x)}{2}\right]^n}{n^2}\color{blue}{~~(\text{Jensen's inequality)}}\\ &=\sum_{n=1}^{\infty}\frac{1-\frac{1}{2^{n-1}}}{n^2}\\ &=\zeta(2)-2\rm{Li_2}\left(\frac{1}{2}\right)\\ &=\frac{\pi^2}{6}-2\left(\frac{\pi^2}{12}-\frac{\ln^2 2}{2}\right)\\  &=\ln^2 2. \end{align*}
