with this inequality $\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$ 
If $0<x<1$, show that
  $$\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$$

use derivative it's not easy, such 
$$
f(x)=(\ln{x}\ln{(1-x)})^2-x(1-x),
$$
$$
f'(x)=2x-1+\dfrac{2\ln{x}\ln^2{(1-x)}}{x}+\dfrac{2\ln^2{x}\ln{(1-x)}}{x-1}
$$
and we $f(x)=f(1-x)$,then we prove inequality hold in $x\in(0,1/2]$.
can you someone have brief solution?

Let
\begin{equation*}
f(x)=\ln x\ln (1-x)-\sqrt{x(1-x)},\ for\ 0<x\leq \frac{1}{2}.
\end{equation*}
\begin{eqnarray*}
f^{\prime }(x) &=&\frac{\ln (1-x)}{x}-\frac{\ln x}{1-x}+\frac{x-\frac{1}{2}}{%
\sqrt{x(1-x)}} \\
&=&\frac{(1-x)\ln (1-x)}{x(1-x)}-\frac{x\ln x}{x(1-x)}+\frac{(x-\frac{1}{2})%
\sqrt{x(1-x)}}{x(1-x)} \\
f^{\prime }(x) &=&\frac{g(x)}{x(1-x)},\ where
\end{eqnarray*}
\begin{eqnarray*}
g(x) &=&(1-x)\ln (1-x)-x\ln (x)+(x-\frac{1}{2})\sqrt{x(1-x)} \\
g^{\prime }(x) &=&\sqrt{x\left( 1-x\right) }-\ln \left( 1-x\right) -\ln x+%
\frac{\left( -\frac{1}{4}\right) \left( 2x-1\right) ^{2}}{\sqrt{x\left(
1-x\right) }}-2 \\
g^{\prime \prime }(x) &=&\frac{1}{1-x}-\frac{1}{x}+3\frac{\left( \frac{1}{2}%
-x\right) }{\sqrt{x\left( 1-x\right) }}+\frac{\left( -\frac{1}{8}\right)
\left( 2x-1\right) ^{3}}{x\left( 1-x\right) \sqrt{x(1-x)}} \\
&=&h(x)\frac{\left( 2x-1\right) }{x\left( x-1\right) \sqrt{x(1-x)}},\ where\ 
\end{eqnarray*}
\begin{eqnarray*}
h(x) &=&x-x^{2}-\sqrt{x-x^{2}}+\frac{1}{8} \\
h^{\prime }(x) &=&k(x)\frac{\left( \frac{1}{2}-x\right) }{\sqrt{x(1-x)}},\ \
where
\end{eqnarray*}
\begin{eqnarray*}
k(x) &=&\left( 2\sqrt{x-x^{2}}-1\right)  \\
k^{\prime }(x) &=&\frac{2(\frac{1}{2}-x)}{\sqrt{x\left( 1-x\right) }}>0,\
for\ 0<x<\frac{1}{2}.
\end{eqnarray*}
Can you take it from here?
