Bounds for $\log(1-x)$ I would like to show the following

$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$

I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequality on the left, the difference is $\frac{x^2}{2}-\left(\frac{x^3}{3}+\frac{x^4}{4} + \cdots\right)$. How do I show this is nonnegative?

$$\frac{d}{dx}\left(\log(1-x)+x+x^2\right) = 1+2x-\frac{1}{1-x}=\frac{x(1-2x)}{1-x} $$
is a non-negative function on $\left[0,\frac{1}{2}\right]$, hence the LHS-inequality follows.
