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

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 $x>1$.
