Upper bound on $\ln(\frac{1}{1-x})$ for $0\leq x\leq 1/2$ Prove that $$\ln\left(\frac{1}{1-x}\right)\leq x+2x^2$$ for $0\leq x\leq 1/2$.
I thought about the Taylor series $\ln(1+x)=x-x^2/2+x^3/3-\ldots$. For small $x$, the values $1+x$ and $1/(1-x)$ are very close to each other, so the inequality should hold since in the Taylor expansion we have $-x^2/2$ while in the desired inequality we have $2x^2$. However, we need to prove the inequality up to $x\leq 1/2$, so something more is needed.

Using series expansion, you may get better bounds..  E.g. using $\log (1+t) = t-\frac12t^2+\frac13t^3-\cdots \ge t-\frac12t^2$, 
$$\log \frac1{1-x} = -\log (1-x) \le x+\frac12x^2 \le x+2x^2$$
