Is it possible to prove that $(x+1)\cdot\ln(x-1) > x\cdot\ln(x)$ for all integer values of $x>4$?

Question

Is it possible to prove that $(x+1)\cdot\ln(x-1) > x\cdot\ln(x)$ for all integer values of $x>4$?

It comes from a broader question about $(x-1)^{x+1} > x^x$ and the logarithm approach seems to be the most promising one.

Answer 1

$$(x+1)\cdot\ln(x-1) - x\cdot\ln(x)=:y$$

$$y' = \frac{x+1}{x-1} + \ln(x-1) - 1 - \ln(x)\\ y'=\frac{2}{x-1}+\ln\frac{(x-1)}{x}\\ y'>0\ \text{when,}\ x\ge5\\ \therefore y \text{ is increasing for}\ x \in[5,\infty) \cdots I$$

Now, $y_{x=5}=6\ln4-5\ln5=\ln(\frac{4096}{3125})\approx0.2719>0 \cdots II$

Hence, $y>0\ \forall x>4, x\in\mathbb Z\ [\text{from I and II}]$

$\therefore (x+1)\cdot\ln(x-1)>x\cdot\ln(x);\ \forall x>4, x\in\mathbb Z\ $