2. (This sub-question 20 points) Let $a, b, c$ be non-negative real numbers, $a+b+c=3$,

Find the maximum value of $S=\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right)$
Solution: Without loss of generality, let $a \geq b \geq c$,
then $b^{2}-b c+c^{2} \leq b^{2}, c^{2}-c a+a^{2} \leq a^{2}$,
thus $S \leq a^{2} b^{2}\left(a^{2}-a b+b^{2}\right)$,
let $a+b=p \leq 3, a b=q \leq\left(\frac{a+b}{2}\right)^{2} \leq\left(\frac{a+b+c}{2}\right)^{2}=\frac{9}{4}$, and $p^{2} \geq 4 q$
thus $S \leq a^{2} b^{2}\left(a^{2}-a b+b^{2}\right)=q^{2}\left(p^{2}-3 q\right) \leq q^{2}(9-3 q)$,
let $f(q)=q^{2}(9-3 q)$, then $f^{\prime}(q)=9 q(2-q)$,
if $q \in[0,2)$, then $f^{\prime}(q)>0$; if $q \in\left(2, \frac{9}{4}\right]$, then $f^{\prime}(q)<0$,
thus $\max _{q[0.0\}]} f(q)=f(2)=12$,
therefore, the maximum value of $s$ is 12, which is achieved when $a=2, b=1, c=0$ and its cyclic permutations.