if $F(x)=\ln{x}\ln{(1-x)}$ prove $ F'(x)>0$ 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
$$G(x)=(1-x)\ln{(1-x)}-x\ln{x}>0,0<x\le\dfrac{1}{2}$$
 A: $$
F'(x)=\frac{\log(1-x)}{x}-\frac{\log(x)}{1-x}
$$
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 $F'\left(\frac12\right)=0$.
A: If, we consider $\displaystyle G(x) = (1-x)\log (1-x) - x\log x$, when $x \in \left(0,\dfrac{1}{2}\right)$ and compute the derivative:
$$G'(x) = -2 - \log x(1-x) \textrm{ remains positive in the range } x\in \left(0,\frac{1}{2} - \frac{\sqrt{e^2 - 4}}{2e}\right)$$
Since, $G(0) = 0$, we conclude $G(x) > 0$ in the range mentioned above.
Next, $$(1-x)\log (1-x) - x\log x = \int_x^{1-x} (1 + \ln t)\,dt$$
Since, $H(t) = 1+\ln t$ is a concave function, we may apply Hermite-Hadamard Inequality and infer:
$$(1-x)\log (1-x) - x\log x \ge (1-2x)\left(1+\frac{1}{2}\log (x(1-x))\right)$$
The R.H.S. remains positive when $\displaystyle x(1-x) > e^{-2}$, i.e., $\displaystyle x \in \left(\frac{1}{2} - \frac{\sqrt{e^2 - 4}}{2e},\frac{1}{2}\right)$
Combined, this gives us $G(x) > 0$ in $x \in \left(0,\dfrac{1}{2}\right)$.
A: Write your derivative as:
$$F'(x)=\frac{\log\left(\frac{(1-x)^{1-x}}{x^x}\right)}{x(1-x)}$$
Then as $x<\frac{1}{2}$ then $1-x>x$ and hence $\frac{(1-x)^{1-x}}{x^x}>1$ so the numerator is positive.
A: Note that $G(x)=\ln\left((1-x)^{1-x}\right)-\ln(x^x)=\ln\left(\frac{(1-x)^{1-x}}{x^x}\right)$ .
So $G(x)\geq0 \iff \ln\left(\frac{(1-x)^{1-x}}{x^x}\right)\geq0 \iff \frac{(1-x)^{1-x}}{x^x}\geq1$
Now, since $0\leq x\leq\frac{1}{2}$, we have:
$$1-x\geq x\Rightarrow (1-x)^{1-x}\geq x^x\Rightarrow\frac{(1-x)^{1-x}}{x^x}\geq1$$
And we are done. Note that we do not arrive at $F'(x)>0$, but at $F'(x)\geq0$, since
$$F'(\frac{1}{2})=\frac{\frac{1}{2}\ln(\frac{1}{2})-\frac{1}{2}\ln(\frac{1}{2})}{\frac{1}{4}}=0$$
