Prove or disprove the inequality $ -x\ln (x)\leq \ln(1-x)(\ln (x) + 1 - x)$

Question

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.

Answer 1

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 $0.34 \ldots > 0.13\ldots$.

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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.

Answer 2

Hint: We have $$-\ln(1-x)(\ln(x)+1-x)-x\ln(x)\geq 0$$ for $$0<x<1$$

Answer 3

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.

Answer 4

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.