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

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 $x>1$.
