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?
For $x\in (0,1)$ $$0<-\log(x)<\frac{1-x^2}{2x}.$$ This follows by estimating the integral $$-\log(x)=\int_x^1\frac{dt}{t}$$ with a trapezoid. Also $\log(x)=2\log\sqrt{x}$ so $$0<-\log(x)<\frac{1-x}{\sqrt{x}}$$ on this interval and your inequality follows.
This is in fact a comment with a picture. The point is to obtain a formal symbolic proof of the inequality. Otherwise we could just offer
Use well known inequality,we have $$\sqrt{ba}<\dfrac{b-a}{\ln{b}-\ln{a}},a>0,b>0$$ let $b=x,a=1$,then we have $$\Longrightarrow \ln{x}>\dfrac{x-1}{\sqrt{x}}$$
$$\Longrightarrow -\ln{x}<\dfrac{1-x}{\sqrt{x}}\tag{1}$$ simaler we have $$-\ln{(1-x)}<\dfrac{x}{\sqrt{1-x}}\tag{2}$$ $(1)\times(2)$ we have $$\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$$
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?
