Taylor Expansion and Simple Inequality $\ln(1+x) \geq x - \frac{x^2}{2(1-|x|)}$.

Question

Why does Taylor's Expansion Theorem show that $$\ln(1+x) \geq x - \frac{x^2}{2(1-|x|)}$$ for all $-1 < x < 1$. So far I have that \begin{align} \ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \dots \end{align} and for $x \geq 0$ \begin{align} \ln(1+x) &= x - \frac{x^2}{2} + \epsilon \end{align} where $\epsilon \geq 0,$ therefore for $0 \leq x < 1$ \begin{align} \ln(1+x) &= x - \frac{x^2}{2} + \epsilon\\ &\geq x - \frac{x^2}{2} \\ &\geq x - \frac{x^2}{2(1-|x|)} \end{align} But I am also searching over $-1 < x < 0$ and don't understand why the equation still holds.

Answer 1

\begin{align} \ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \dots\\ &= x-\dfrac{x^2}{2}\left(1 - \frac{2x}{3} + \frac{2x^2}{4} - \frac{2x^3}{5} \dots\right)\\ &\geqslant x-\dfrac{x^2}{2}\left(1+|x|+|x|^2+ \dots\right)\\ &\geqslant x - \frac{x^2}{2(1-|x|)} \end{align} for $|x|<1$.