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

$ln \left(\frac{x}{x-1}\right)<\frac{100}{x}$
$-ln \left(\frac{x-1}{x}\right)<\frac{100}{x}$
$-ln \left(1-\frac{1}{x}\right)<\frac{100}{x}$
$ln \left(1-\frac{1}{x}\right)^x>-100$
$\left(1-\frac{1}{x}\right)^x>e^{-100}$
$\left(1-\frac{1}{x}\right)^x>\frac{1}{e^{100}}$
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.
