Prove that $\forall x\in\mathbb{R}$, such that $x>1$, $0<1+(x-1)\ln (1-\frac1x)<1$

Let $$f(x)=1+(x-1)\ln (1-\frac1x)$$. I am trying to prove that $$\forall x\in\mathbb{R}$$, such that $$x>1$$, $$0.

I have already proved that $$f(x)<1$$ as follows:

To show $$1+(x-1)\ln (1-\frac1x)<1$$ it would be necessary and sufficient to show that $$(x-1)\ln (1-\frac1x)<0$$. We have that $$x>1\iff x-1>0 \iff 1-\frac1x>0$$ and $$1>0 \iff \frac1x>0 \iff 0>-\frac1x \iff 1>1-\frac1x$$, thus $$x-1>0$$ and $$1>1-\frac1x>0 \iff \ln (1-\frac1x)<0$$. $$\therefore$$ Since $$x-1>0$$ and $$\ln (1-\frac1x)<0$$ then $$(x+1)\ln (1-\frac1x)<0$$.$$\blacksquare$$

Proving that $$0 seems to be less trivial. Perhaps using calculus to prove this is a necessity but I am convinced that proving this is possible using similar methods to the first proof (simply using the assumption that $$x>1$$ and the algebra of inequalities).

At the very least a nudge in the right direction would be much appreciated.

Anytime I see log, reciprocal, and inequality I think about the Mean Value Theorem. I was burned by such an inequality on an Honors Calculus exam over 25 years ago.

Note that $$\begin{gather*} 0 < 1 + (x-1) \ln\left(1 - \frac{1}{x}\right) < 1 \\\iff -1 < (x-1)\ln\left(\frac{x-1}{x}\right) < 0 \\\iff -\frac{1}{x-1} < \ln(x-1) - \ln(x) < 0 \\\iff 0 < \ln(x) - \ln(x-1) < \frac{1}{x-1} \end{gather*}$$ Let $$f(t) = \ln t$$ on the interval $$x-1 \leq t \leq x$$. Since $$x>1$$, $$f$$ is continuous on this interval and differentiable on its interior. Therefore by the Mean Value theorem there exists a number $$c$$ such that $$x-1 < c < x$$ and $$\frac{f(x)-f(x-1)}{x-(x-1)} = f'(c)$$ This means that $$\ln(x) - \ln(x-1) = \frac{1}{c}$$ Notice that $$0 < x-1 < c < x$$, so $$0 < \frac{1}{x} < \frac{1}{c} < \frac{1}{x-1}$$ and therefore $$0 < \frac{1}{x} < \ln(x) - \ln(x-1) < \frac{1}{x-1}$$ as needed.

You may also proceed as follows:

• Set $$t = x-1 \Rightarrow f(x(t)) = g(t) = 1+ t\log \frac{t}{1+t}$$ for $$t > 0$$.
• Note that $$\lim_{t\to 0^+}g(t) = 1$$.

So, $$g$$ can be continuously extended into $$t=0$$ with value $$g(0) = 1$$.

• $$g(t)$$ is strictly decreasing as $$\color{blue}{g'(t)}= (t\log t - t\log(1+t))'= \log t + 1 - \log(t+1)- \frac{t}{1+t}$$ $$= \log t - \log (t+1) + \frac{1}{1+t} \stackrel{MVT}{=} - \frac{1}{\tau} + \frac{1}{1+t} \color{blue}{\stackrel{t < \tau <1+t}{<} 0}$$
• $$\lim_{t\to \infty}g(t) = 0$$. Indeed, setting $$t = \frac{1}{u}$$

$$\begin{eqnarray*}1 + t \log \frac{t}{t+1} & \stackrel{t = \frac{1}{u}}{=} & 1+ \frac{\log \frac{1}{1+u}}{u}\\ & = & 1 -\frac{\log (1+u)}{u}\\ & \stackrel{L'Hosp.}{\sim} & 1-\frac{1}{1+u} \\ & \stackrel{u \to 0^+}{\longrightarrow} & 1-1 = 0\\ \end{eqnarray*}$$

So, on $$(0,\infty)$$ we have $$g(t) \stackrel{t\to \infty}{\searrow} 0 \stackrel{g(0)=1}{\Rightarrow} \boxed{0 < g(t) < 1}$$.

You can simplify the left side inequality by substituting $$y=1-\frac 1 x$$. The inequality becomes $$y\log\,y >y-1$$ for $$0. This is easily proved by noting that the derivative of $$y\log\,y -y+1$$ is $$\log\, y <0$$ and the function $$y\log\,y -y+1$$ vanishes at $$y=1$$.