How do I show this inequality that $\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$ for $x>1$? How do I show this inequality that for $x>1$ $$\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$$
So far I have tried to use the inequality $\ln({x+1}) < x$, but it is not good enough.

You can show it using concavity of $\ln x$ as follows:
The given inequality is equivalent to
$$\ln x \geq \frac{x-1}{x}\ln(x+1)$$
Now, since $x>1$ we have
$$\frac{x-1}{x}\ln(x+1) + \frac 1x\ln 1 \stackrel{concavity}{\leq}\ln\left(\frac{x-1}{x}(x+1) +\frac 1x\cdot 1\right) =\ln x$$
Done.
