I'm looking for lower bounds on the function $x\mapsto \log(1-x)$ with $0\leq x<1$. Actually, I only need $x\in [0,c]$, $c<1$.
So far, I've tried using Taylor expansion with integral remainder but since all derivatives are negative, I can only get an upper-bound.
Best Answer
From this very handy cheatsheet of inequalities: for $0\leq x<1$, $$\begin{align} \ln(1-x) \geq \frac{-x}{\sqrt{1-x}} \tag{$\dagger$} \end{align}$$
Proof. let $f$ be defined on $(-1,1)$ by $f(x) = \ln(1-x) + \frac{x}{\sqrt{1-x}}$. It is $\mathcal{C}^1$ on its domain, $f(0)=0$, and $$ f'(x) = \frac{2-x-2\sqrt{1-x}}{2(1-x)^{3/2}}\,. $$ It is easy to see that, for $-1< x<1$, $f(x)\geq 0$ (it suffices to analyze the numerator, as the denominator is positive). This gives the inequality.