Solving a logarithmic inequality

Question

For what values of $x$ it holds that:

$$-\log_{e}(1-x) \geq x$$

and how can we prove this? $\log_{e}$ is the natural logarithm $\ln$.

Answer 1

Fill in the details in what follows.

Consider the function $f(x)=x+\log(1-x)$, defined for $x<1$. You have $$ \lim_{x\to-\infty}f(x)=-\infty, \qquad \lim_{x\to1^-}f(x)=-\infty $$ Also $$ f'(x)=1-\frac{1}{1-x}=-\frac{x}{1-x} $$ which is positive for $-x>0$, that is, $x<0$. So the function has a maximum at $0$ and $$ f(0)=0 $$ Thus $f(x)\le0$ for every $x<1$.