Prove: The inequality can be rewritten in the following form
$$\begin{array}{l} \sum_{c y c} a^{6}+2 \sum_{c y c} a^{3} b^{3} \geq 2 \sum_{c y c} a^{5} b+a b c \sum_{c y c} a^{3} \\ \Leftrightarrow 2\left(\sum_{c y c} a^{6}+\sum_{c y c} a^{4} b^{2}-2 \sum_{c y c} a^{5} b\right)+\left(\sum_{c y c} a^{4} b^{2}+\sum_{c y c} a^{4} c^{2}-2 a b c \sum_{c y c} a^{3}\right)+ \\ \left(2 \sum_{c y c} a^{3} b^{3}-\sum_{c y c} a^{4} b^{2}-\sum_{c y c} a^{2} b^{4}\right) \geq\left(\sum_{c y c} a^{4} b^{2}-\sum_{c y c} a^{4} c^{2}\right) \\ \Leftrightarrow \sum_{c y c}\left(2 a^{4}+c^{4}-a^{2} b^{2}\right)(a-b)^{2} \geq\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(a^{2}-c^{2}\right) \quad \end{array}$$

Let $M=(a-b)(b-c)(a-c)$. Of course, we can assume $a \geq b \geq c, a>c \Rightarrow M \geq 0$. We use the following results to prove the inequality.
(1) $\sum_{\text {cyc }}(3 a+2 c-b)(a-b)^{2} \geq 4 M$
(2) $\sum_{c y c}\left(11 a^{2}+6 c^{2}-b^{2}-4 a b\right)(a-b)^{2} \geq 8(a+b+c) M$
(3) $\sum_{c y c}\left(4 a^{3}+2 c^{3}-a^{2} b-b^{2} a\right)(a-b)^{2} \geq\left(a^{2}+b^{2}+c^{2}+3 a b+3 b c+3 c a\right) M$

These inequalities have an interesting relationship: $(1) \Rightarrow(2) \Rightarrow(3)$. To prove (2) and (3), we first prove (1).
Proof of (1). Of course, it is sufficient to prove the inequality when $c=\min \{a, b, c\}=0$. Because if we reduce $a, b, c$ by the smallest positive real number $c$, the expression on the right side of (1) does not change, while the expression on the left side decreases. If $c=0$, the inequality (1) becomes
$$\begin{array}{l} (3 a-b)(a-b)^{2}+(3 b+2 a) b^{2}+(-a+2 b) a^{2} \geq 3 a b(a-b) \\ \Leftrightarrow 2 a^{3}-8 a^{2} b+10 a b^{2}+2 b^{3} \geq 0 \Leftrightarrow h(x)=2 x^{3}-8 x^{2}+10 x+2 \geq 0 \end{array}$$

where $x=\frac{a}{b}$. Note that $h^{\prime}(x)=2(x-1)(3 x-5)$, so it is easy to get
$$h(x) \geq f\left(\frac{5}{3}\right)>5>0$$

Thus (1) is proved. Now let's prove (2) from (1).
Proof of (2). Let $a=a_{1}+t=f_{1}(t), b=b_{1}+t=f_{2}(t), c=c_{1}+t=f_{3}(t)$, then (2) is equivalent to
$$f(t)=\sum_{c y c}\left[11 f_{1}(t)^{2}-f_{2}(t)^{2}-4 f_{1}(t) f_{2}(t)+6 f_{3}(t)^{2}\right](a-b)^{2}-8\left(f_{1}(t)+f_{2}(t)+f_{3}(t)\right) M \geq 0$$

According to (1), we have
$$f^{\prime}(t)=\sum_{c y c}\left[18 f_{1}(t)-6 f_{2}(t)+12 f_{3}(t)\right](a-b)^{2}-24 M=6 \sum_{c y c}(3 a-b+2 c)(a-b)^{2}-24 M \geq 0$$

So $f(t) \geq f(-c)$ (since $t \geq-c$). This property indicates that it is sufficient to prove (2) when $c=0$.
$$\begin{array}{l} \left(11 a^{2}-b^{2}-4 a b\right)(a-b)^{2}+\left(11 b^{2}+6 a^{2}\right) b^{2}+\left(-a^{2}+6 b^{2}\right) a^{2} \geq 8(a+b) a b(a-b) \\ \Leftrightarrow 11 a^{4}-35 a^{3} b+30 a^{2} b^{2}+6 a b^{3}+10 b^{4} \geq 0 \end{array}$$

This inequality is certainly true, because the AM-GM inequality indicates $11 a^{4}+30 a^{2} b^{2} \geq 2 \sqrt{11 \cdot 30} a^{3} b>35 a^{3} b$. (2) is proved.
Proof of (3). Similarly, according to (2) and for the same reason, we assume that (3) only needs to consider $\min \{a, b, c\}=c=0$. At this point, inequality (3) becomes
$$\begin{array}{l} \left(4 a^{3}-a^{2} b\right)(a-b)^{2}+\left(4 b^{3}+2 a^{3}\right) b^{2}+2 b^{3} a^{2} \geq\left(a^{2}+b^{2}+3 a b\right) a b(a-b) \\ \Leftrightarrow 4 a^{5}-10 a^{4} b+6 a^{3} b^{2}+3 a^{2} b^{3}+a b^{4}+4 b^{5} \geq 0 \end{array}$$

If $2 a \geq 3 b$, since $4 a^{5}-10 a^{4} b+6 a^{3} b^{2}=2 a^{3}(a-b)(2 a-3 b) \geq 0$, the inequality holds. Otherwise, if $2 a \leq 3 b$, then
$$4 a^{5}-10 a^{4} b+6 a^{3} b^{2}+3 a^{2} b^{3} \geq 4 a^{5}-10 a^{4} b+8 a^{3} b^{2} \geq(2 \sqrt{32}-10) a^{4} b \geq 0 \text{. (3) is proved. }$$
Proof of (*). Similarly, according to (3) and for the same reason, we assume $c=0$. Then the inequality becomes a simpler form: $\left(a^{3}+b^{3}\right)^{2} \geq 2 a^{5} b$, or $g(a)=a^{6}-2 a^{5} b+2 a^{3} b^{3}+b^{6}$. It is easy to prove $g^{\prime}(a) \geq 0$, so $g(a) \geq g(b) \geq 0$. The inequality is proved, with equality holding if and only if $a=b=c$.