Solution: The inequality equivalent to
$$\begin{array}{c} \Leftrightarrow 2 \sum_{cyc} a^{2} b+4 \sum_{cyc} a b^{2} \leq 12 \sqrt{3} \\ \Leftrightarrow 3 \sum_{sym} a^{2}(b+c)+\left(\sum_{cyc} a b^{2}-\sum_{cyc} a^{2} b\right) \leq 12 \sqrt{3} \\ \Leftrightarrow 3 \sum_{cyc} a^{2}(b+c)+(a-b)(b-c)(c-a) \leq 12 \sqrt{3} \end{array}$$

We only need to prove the inequality in the case $(a-b)(b-c)(c-a) \geq 0$.
$$\begin{array}{l} 3 \sum_{\text {sym }} a^{2}(b+c)+\sqrt{(a-b)^{2}(b-c)^{2}(c-a)^{2}} \leq 12 \sqrt{3} \\ \Leftrightarrow 3(p q-3 r)+\sqrt{p^{2} q^{2}+18 p q r-27 r^{2}-4 q^{3}-4 p^{3} r} \leq 12 \sqrt{3} \\ \Leftrightarrow p^{2} q^{2}+18 p q r-27 r^{2}-4 q^{3}-4 p^{3} r \leq(12 \sqrt{3}-3 p q+9 r)^{2} \\ \Leftrightarrow f(r)=108 r^{2}+\left(4 p^{3}-72 p q+216 \sqrt{3}\right) r+4 q^{3}+8 p^{2} q^{2}-72 \sqrt{3} p q+432 \geq 0 \end{array}$$

Putting $r_{ct}=\frac{216 q-108-216 \sqrt{3}}{108}$
Let two cases
1) $0 \leq q \leq \frac{216 \sqrt{3}+108}{216} \Rightarrow r_{ct} \leq 0$
$$f(0)=4(q+12+6 \sqrt{3})(q+3-\sqrt{3})^{2} \geq 0$$
2) $\frac{216 \sqrt{3}+108}{216} \leq q \leq 3 \Rightarrow r_{ct} \geq 0$
$$f\left(r_{ct}\right)=4 q^{3}-36 q^{2}+108 q+81-108 \sqrt{3} \geq 0, \forall q \in\left[\sqrt{3}+\frac{1}{2} ; 3\right]$$

So $f(r) \geq r \forall r \geq 0$. The solution is end.