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

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}

• $log(1-x)$ is not defined for $x\geq 1$ – Kabo Murphy Jan 3 at 8:53
• To show that $f(x)\leq f(0),\forall \;x\in ]0,1[,$ i.e. $\log(1-x)+x\leq 0$ – Omojola Micheal Jan 3 at 9:13

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