Solution:
Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
$$\begin{aligned} 2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\right)+2 \sqrt[3]{\frac{c}{a}} \\ & \geq \frac{a}{b}+3 \sqrt[3]{\frac{a}{b}} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} \\ & =\frac{a}{b}+3 \sqrt[3]{\frac{a}{c}}+2 \sqrt[3]{\frac{c}{a}} \\ & \left.=\frac{a}{b}+\sqrt[3]{\frac{a}{c}}+2 \sqrt[3]{\frac{a}{c}}+\sqrt[3]{\frac{c}{a}}\right) \\ & \geq \frac{a}{b}+\sqrt[3]{\frac{a}{c}}+2 \cdot 2 \sqrt{\sqrt[3]{\frac{a}{c}} \sqrt[3]{\frac{c}{a}}} \\ & =\frac{a}{b}+\sqrt[3]{\frac{a}{c}}+4 \\ & >4 \end{aligned}$$
which yields the given inequality.