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.

A simple way to prove this inequality is the following :
let $f(x)= 2x^2+ x-\ln (\frac{1}{1-x})$. Then $f^{'}(x)=4x+1-\frac{1}{1-x} \geq 0$ for all $ x \in [0, \frac{1}{2}]$. Thus $f$ is increasing on $[0, \frac{1}{2}]$, and so $f(x) \geq f(0)$ for all $ x \in [0, \frac{1}{2}]$. Hence $2x^2+ x-\ln (\frac{1}{1-x}) \geq 0$  for all $ x \in [0, \frac{1}{2}]$. 
