Use the concavity of $\log (1-x)+x$ to show that $\log (1-x)\leq -x$

Question

I've been reading this post and its accepted answer here. The OP (accepted answer) made a comment that $\log(1-x)\leq -x$ but I've been having issues proving it.

FULL PROOF (EDIT)

With credits to Kavi Rama Murthy, I provide a full proof.

Let $f: ]0,1[\longrightarrow \Bbb{R},\;x \mapsto \log (1-x)+x$. Then,

\begin{align}f''(x) = -\frac{1}{(x-1)^2}\leq 0,\;\forall \;x\in \;]0,1[.\end{align} Hence, $f$ is monotone non-increasing and $f(x)\leq f(0),\forall \;x\in \;]0,1[,$ that is \begin{align}\log(1-x)\leq - x \end{align}

To show that $f(x)\leq f(0),\forall ;x\in ]0,1[,$ i.e. $\log(1-x)+x\leq 0$ — Omojola Micheal, Jan 03 '19 at 09:13

Kavi Rama Murthy · Accepted Answer · 2019-01-03T09:08:53.873

4

The function is concave on $(0,1)$ because $f''(x)=-\frac 1 {(x-1)^{2}} \leq 0$. It is not defined for $x \geq 1$.

edited Jan 03 '19 at 09:08

answered Jan 03 '19 at 08:56

Kavi Rama Murthy

311,013

(+1) But sorry for the stress! – Omojola Micheal Jan 03 '19 at 09:07

Omojola Micheal · Answer 2 · 2021-01-15T08:21:16.943

2

Another way is to consider the Maclaurin series expansion

\begin{align} \log(1-x)=-\sum^{\infty}_{n=1}\dfrac{x^n}{n}=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}\cdots \leq -x,\;\text{for fixed}\;x\in\;]0,1[\end{align}

edited Jan 15 '21 at 08:21

answered Jan 03 '19 at 10:29

Omojola Micheal

4,504

Use the concavity of $\log (1-x)+x$ to show that $\log (1-x)\leq -x$

2 Answers2