Solution: Let $a=2, b=1, c=0 \Rightarrow k \geq \frac{4}{27}$.
We'll prove $k \geq \frac{4}{27}$ is the better constant.
The inequality is equivalent to
$$\begin{array}{c} \frac{4}{27}(a+b+c)^{4} \geq \sum_{c y c} a^{3} b+\sum_{s y m} b^{2} c^{2}+a b c \sum_{s y m} a \\ \Leftrightarrow \frac{8}{27}(a+b+c)^{4} \geq\left(\sum_{c y c} a^{3} b+\sum_{c y c} a b^{3}\right)+2 \sum_{s y m} b^{2} c^{2}+\left(\sum_{c y c} a^{3} b-\sum_{c y c} a b^{3}\right)+2 a b c(a+b+c) \\ \Leftrightarrow \frac{8}{27}(a+b+c)^{4} \geq \sum_{\text {sym }} a^{3}(b+c)+2 \sum_{s y m} b^{2} c^{2}+(a+b+c)(a-b)(b-c)(a-c)+2 a b c(a+b+c) \end{array}$$

We only need to prove the inequality in the case $(a-b)(b-c)(c-a) \geq 0$.
$$\begin{array}{c} \Leftrightarrow \frac{8}{27}(a+b+c)^{4} \geq p^{2} q-2 q^{2}-p r+2 q^{2}-4 p r+2 p r+p \sqrt{p^{2} q^{2}+18 p q r-27 r^{2}-4 q^{3}-4 p^{3} r} \\ \Leftrightarrow p^{2}\left(p^{2} q^{2}+18 p q r-27 r^{2}-4 q^{3}-4 p^{3} r\right) \leq\left[\frac{8}{27} p^{4}-p^{2} q+3 p r\right]^{2} \\ \Leftrightarrow 36 p^{2} r^{2}+\left(\frac{52}{9} p^{5}-24 p^{3} q\right) r+\frac{64}{729} p^{3}+4 p^{2} q^{3}-\frac{16}{27} p^{6} q \geq 0 \\ \Leftrightarrow 324 p^{2}+(1404-648 q) r+36 q^{3}-432 q+576 \geq 0 \end{array}$$

Letting $f(r)=36\left[9 r^{2}+(39-18 q) r+q^{3}-12 q+16\right]$. Case 1: $0 \leq q \leq \frac{13}{6} \rightarrow 39-18 q \geq 0$
We have $f(0)=36(q+4)(q-2)^{2} \geq 0$.
Case 2: $\frac{13}{6} \leq q \leq 3 \rightarrow \Delta=(39-18 q)^{2}-4 \cdot 9 \cdot\left(q^{3}-12 q+16\right)=-36 q^{3}+324 q^{2}-972 q+945 \leq 0$ $\forall q \in\left[\frac{13}{6} ; 3\right]$ The solution is complete.