Prove or disprove the inequality $ -x\ln (x)\leq \ln(1-x)(\ln (x) + 1 - x)$ how can one disprove or prove the equation below? I have already reduced it to its simplest form (according to me) and I am kind of stuck at this point:
$$ -x\ln (x)\leq \ln(1-x)(\ln (x) + 1 - x) \qquad\text{for}\qquad 0\lt x\lt 1$$
Any help will be quite useful to me. Thanks.
 A: The inequality
 $$ -x\ln x\leq \ln(1-x)(\ln x + 1 - x) \qquad\text{for}\qquad 0\lt x\lt 1$$
is not true.
For example, if we put in $x = \frac{1}{2}$ we get
$$
- \frac{1}{2} \ln \frac{1}{2} = \frac12  \ln 2 = 0.34\ldots
$$
and
$$
\ln\left(1 - \frac12\right) 
\left( \ln \frac12 + 1 - \frac12\right)
= \left(- \ln 2 \right) \left( \frac12 - \ln 2 \right) = (\ln 2)^2 - \frac12 \ln 2 = 0.13 \ldots
$$
and $0.34 \ldots > 0.13\ldots$.

You probably made a mistake when you reduced it to this form. If you reverse the inequality, it would be true. 
See the graph below:

We can see that the graph looks to be $> 0$ for $0 < x < 1$.
So we conjecture that the inequality
$$
- x \ln x \ge \ln(1-x) (\ln x + 1 - x)
$$
will be true.
A: Hint: We have $$-\ln(1-x)(\ln(x)+1-x)-x\ln(x)\geq 0$$ for $$0<x<1$$
A: Define
\begin{align*}
f(x) = -x\log(x) - \log(1-x)(\log(x) + 1 - x)
\end{align*}
It suffices to prove $f(x) \ge 0$ for $x \in [0, 1]$. Note that
\begin{align*}
f(0) &= f(1) = 0 \\
f'(x) &= \frac{(x-1)\log(1-x)}{x} + \frac{x\log(x)}{1-x} \begin{cases}
> 0 & x < \frac{1}{2} \\
= 0 & x = \frac{1}{2} \\
< 0 & x > \frac{1}{2}
\end{cases}
\end{align*}
Therefore, $f(x) \ge \min(f(0), f(1)) = 0$ for $x \in [0, 1]$, as desired.
A: Consider the function
$$f(x)=x\ln (x)+ \ln(1-x)\,(\ln (x) + 1 - x) $$ and notice the symmetry which makes that
$$f(x)+f(1-x)=0$$ Compute $f(x)$ anywhere to show that is negative. Using $x=\frac 12$, you have $(\log (2)-1) \log (2)$ which is the minimum value.
