Solve: By the AM-GM inequality, we have
$$\begin{aligned} (a+b)^{2}+(a+b+4 c)^{2} & =(a+b)^{2}+[(a+2 c)+(b+2 c)]^{2} \\ & \geqslant(2 \sqrt{a b})^{2}+(2 \sqrt{2 a c}+2 \sqrt{2 b c})^{2} \\ & =4 a b+8 a c+8 b c+16 c \sqrt{a b} \end{aligned}$$

Thus,
$$\begin{aligned} & \frac{(a+b)^{2}+(a+b+4 c)^{2}}{a b c} \cdot(a+b+c) \\ \geqslant & \frac{4 a b+8 a c+8 b c+16 c \sqrt{a b}}{a b c} \cdot(a+b+c) \\ = & \left(\frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{a b}}\right)(a+b+c) \\ = & 8\left(\frac{1}{2 c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{a b}}+\frac{1}{\sqrt{a b}}\right)\left(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c\right) \\ \geqslant & 8\left(5 \sqrt[5]{\frac{1}{2 a^{2} b^{2} c}}\right)\left(5 \sqrt[5]{\frac{a^{2} b^{2} c}{2^{4}}}\right)=100 . \end{aligned}$$

Equality holds when $a=b=2 c>0$. Therefore, the minimum value is 100.