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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}

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    $\begingroup$ $log(1-x)$ is not defined for $x\geq 1$ $\endgroup$ – Kabo Murphy Jan 3 at 8:53
  • $\begingroup$ To show that $f(x)\leq f(0),\forall \;x\in ]0,1[,$ i.e. $\log(1-x)+x\leq 0$ $\endgroup$ – Omojola Micheal Jan 3 at 9:13
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The function is concave on $(0,1)$ because $f''(x)=-\frac 1 {(x-1)^{2}} \leq 0$. It is not defined for $x \geq 1$.

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  • $\begingroup$ (+1) But sorry for the stress! $\endgroup$ – Omojola Micheal Jan 3 at 9:07
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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}

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