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