Is it possible to prove this? $\ln(\frac{x}{x-1}) < \frac{100}{x} $ for $ x > 1$ $-\ln(1-(\frac{1}{x})) < \frac{100}{x} $ for $ x > 1$ is what I want to prove. I pulled a negative sign out and I got  $\ln(\frac{x}{(x-1)}) < \frac{100}{x} $ for $ x > 1$.
How do I continue with this proof?
Or is it actually possible to prove this
Edit : I want this proof Algebrically, Calculus is allowed
 A: It is not true for $x=1+e^{-100}$. We then have
$$ \log\frac{x}{x-1} = \log(x)+100 > 100 $$
but
$$ \frac{100}{x} < 100 $$
because $x>1$.
A: As you almost wrote, making $x=\frac 1y$, the problem is to show that, for $y<1$, $$f(y)=100 y+\log(1-y)>0$$  (the function being undefined at $y=1$). We have $$f'(y)=100-\frac{1}{1-y}$$ which is positive as long as $y<\frac{99}{100}$ and negative for $y>\frac{99}{100}$. On the other hand $$f''(y)=-\frac{1}{(1-y)^2}$$ is always negative. So $f(y)$ increases up to a maximum value equal to $99-\log (100) >0$ and starts decreasing very fast and it exists a value of $y$ slightly below $1$ which makes $f(y)=1$ and negative for larger values. When $y=1-\epsilon$ , $f(y)= 100-100\epsilon+\log(\epsilon)$
So, as Henning Makholm answered, this is not true for all $x>1$.
A: $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$
$-ln \left(\frac{x-1}{x}\right)<\frac{100}{x}$
$-ln \left(1-\frac{1}{x}\right)<\frac{100}{x}$
$ln \left(1-\frac{1}{x}\right)^x>-100$
$\left(1-\frac{1}{x}\right)^x>e^{-100}$
$\left(1-\frac{1}{x}\right)^x>\frac{1}{e^{100}}$
I think that $\frac{1}{e^{100}}$ is almost 0 and LHS is ofcourse $>0$ for all $x>1$
So this seems pretty reasonable.
Edit: It suddenly seems not strictly valid for all $x>1$ ,  but not sure.
Edit:2 This is not true for only 
$x\in(1,root\;of\;(x-1)-\frac{x}{e^{100/x}}=0)$
Since, the root lies very close to $1$ , it is approximately valid. And I think, this is good enough for algebraic solution.
A: Solution to
\begin{align}
\ln\left(\frac{x}{x-1}\right) = \frac{100}{x}
\end{align}
in terms of Lambert W function is
\begin{align}
x&=\frac{100}{W(-100\exp(-100))+100}
\\
&\approx
1.000000000000000000000000000000000000000000037200759760
\end{align}
Btw, WolframAlpha
gives the answer to the inequality as
Solution over the reals

A: Required to prove $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$
ie to prove $f(x) = {{100} \over x} - ln\left( {{x \over {x - 1}}} \right)>0$
${d \over {dx}}\left({{100} \over x} - ln\left( {{x \over {x - 1}}} \right)\right) = $
=......
=${{ - 101x - 100} \over {\mathop x\nolimits^2 (x + 1)}}$
$f'(x)$  is negative for all $x > 1$
Meaning always decreasing.
Easy to show $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$
So The function is always decreasing towards 0
implies $f(x) > 0$ for  $x > 1$
Implies $ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$
