For the following inequality, I can see numerically that it will hold at around $x > 1.22$.
\begin{align} \frac{(6x)^3(2x+1)^2}{2x(2x-1)} < \sqrt[2x]{(6x+1)^{6x+1}}, \quad x \in \mathbb{R}^+ \end{align}
But is it possible to get the exact $x$? Or is it at least possible to show that this inequality holds for $x \ge 2$ or $x \ge 3$?