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: I am not sure you can keep calculus completely out, here is a way using calculus.  First we may note that for $x \in [0, \frac12]$,
$$\frac1{1-x} \le 1+4x \iff x(3-4x) \ge 0$$
which is obviously true.
$$\therefore \int_0^x \frac1{1-x} \le \int_0^x (1+4x) \implies \log \frac1{1-x} \le x+2x^2$$
A: There are many ways. You suggested series, so let's do it that way.
We have 
$$\ln\left(\frac{1}{1-x}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots.$$
The part from $\frac{x^2}{2}$ on is $\le \frac{x^2}{2}\left(1+x+x^2+x^3+\cdots\right)$. Note that $1+x+x^2+\cdots=\frac{1}{1-x}$ and $\frac{1}{1-x}\le 2$ on our interval.  That gives an inequality sharper than the proposed one. 
A: Using series expansion, you may get better bounds..  E.g. using $\log (1+t) = t-\frac12t^2+\frac13t^3-\cdots \ge t-\frac12t^2$, 
$$\log \frac1{1-x} = -\log (1-x) \le x+\frac12x^2 \le x+2x^2$$
A: 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}]$. 
