TITLE: if $F(x)=\ln{x}\ln{(1-x)}$ prove $ F'(x)>0$
QUESTION [6 upvotes]: an anyone please help me with the following proof:
Let $$F(x)=\ln{x}\ln{(1-x)},0<x\le\dfrac{1}{2}$$
show that
$$F'(x)>0$$

because 
$$F'(x)=\dfrac{(1-x)\ln{(1-x)}-x\ln{x}}{x(1-x)}$$
It suffices to show that
$$G(x)=(1-x)\ln{(1-x)}-x\ln{x}>0,0<x\le\dfrac{1}{2}$$

REPLY [2 votes]: $$
F'(x)=\frac{\log(1-x)}{x}-\frac{\log(x)}{1-x}
$$
Since $e^u\ge1+u$ for all real $u$,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{\log(1-x)}{x}
&=\frac{-\frac{x}{1-x}-\log(1-x)}{x^2}\\
&=\frac{1-\frac1{1-x}+\log\left(\frac1{1-x}\right)}{x^2}\\
&=\frac{1+u-e^{u}}{x^2}\\[6pt]
&\le0
\end{align}
$$
where $u=\log\left(\frac1{1-x}\right)$.
Thus, $F'(x)$ is decreasing (decreasing function minus increasing function) and $F'\left(\frac12\right)=0$.