# Prove that the function ${f(x)=1+(1-x){\ln(1-\frac{1}{x})} > 1}$, ${\forall x \in \mathbb{R}}$ such that $x>1$

I am attempting to prove that the function $${f(x)=1+(1-x){\ln(1-\frac{1}{x})} > 1}$$, $${\forall x \in \mathbb{R}}$$ such that $$x>1$$.

Just by looking at the equation it is quite obvious that as $$x$$ approaches 1 from the right $$f(x)$$ approaches 1.

There may be an easier way to solve this problem, but as far as I can tell this proof requires two things:

1. A proof that $${\lim \limits_{x \to 1^+} f(x)=1}$$.
2. A proof that $${f'(x)>0}$$, $${\forall x>1}$$

I have solved 1 using l'hopitals rule:

\begin{align} \lim \limits_{x \to 1^+} 1+(1-x){\ln(1-\frac{1}{x})} & = 1+\lim \limits_{x \to 1^+} (1-x){\ln(1-\frac{1}{x})}\\ & = 1+\lim \limits_{x \to 1^+} {\frac{\ln(1-\frac{1}{x})}{\frac{1}{1-x}}}\\ & = 1+\lim \limits_{x \to 1^+} {\frac{\frac{1}{x(x-1)}}{\frac{1}{(x-1)^2}}}\\ & = 1+\lim \limits_{x \to 1^+} {\frac{x-1}{x}}\\ & = 1+0\\ & = 1 \end{align}

Despite solving 1, I am having some difficulty proving number 2, but here is my work so far:

\begin{align} f'(x)&=-\ln(1-\frac1x)-\frac1x\\ & =-\ln(\frac{x-1}{x})+\ln(e^{-\frac1x})\\ & =\ln(\frac{e^{-\frac1r}}{\frac{x-1}{x}})\\ & =\ln(\frac{x}{(x-1)e^{\frac1x}}) \end{align}

Thus to show $$f'(x)>0$$ it would be sufficient to show that $$\frac{x}{(x-1)e^{\frac1x}}>1$$

I am at a complete loss on how to prove this though, and any help or at least a push in the right direction would be much appreciated.

If anybody believes that this approach to the proof is in the completely wrong direction it would be much appreciated as well if you could point me in the right direction, although I do believe this would be an appropriate method anyways.

More simple: $$x>1\ \iff0<1-\frac{1}{x}<1\iff \log(1-\frac{1}{x})<0$$ then: $$(1-x)\log(1-\frac{1}{x})>0$$
Your inequality is equivalent to $$(1-x)\ln\left(1-\frac{1}{x}\right)>0$$ this is true, since $$1-x<0$$ and $$0<1-\frac{1}{x}<0$$ so $$\ln\left(1-\frac{1}{x}\right)<0$$ thus $$(1-x)\ln\left(1-\frac{1}{x}\right)>0$$
• Thus, $0<0 {}$? – Bernard Feb 27 at 10:54
Hints: You can simplify things. First of all, you can cancel $$1$$ from both sides of the inequality in the title, i.e. you want to show that $$(1-x)\ln\left(1-\frac{1}{x}\right) > 0\quad \forall x > 1.$$
Now, if $$x > 1$$, then $$1-x < 0$$, so $$(1-x)\ln\left(1-\frac{1}{x}\right)$$ will be positive iff $$\color{blue}{\ln\left(1-\frac{1}{x}\right) < 0}$$. Thus you just have to show that $$\color{blue}{\ln\left(1-\frac{1}{x}\right) < 0\quad \forall x >1}.$$ Can you finish from here?