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 $x>1$, $0<f(x)<1$.
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 $0<f(x)$ 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.
Thanks for your time.
 A: 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.
A: 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 $g(t) \stackrel{t\to \infty}{\searrow} 0 \stackrel{g(0)=1}{\Rightarrow} \boxed{0 < g(t) < 1}$. 
A: You can simplify the left side inequality by substituting $y=1-\frac 1 x$. The inequality becomes $y\log\,y >y-1$ for $0<y<1$. 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 $y=1$. 
