Prove: The inequality is equivalent to
$$\sum_{c y c}\left(\frac{1}{c^{2}-c+3}-\frac{1}{3}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{a(a-1)}{a^{2}-a+3} \geq 0$$

or
$$\sum_{c y c} \frac{a-1}{a-1+\frac{3}{a}} \geq 0$$

According to Chebyshev's inequality and the condition $a+b+c=3$, it suffices to prove that if $a \geq b$, then $a-1+\frac{3}{a} \leq b-1+\frac{3}{b}$ or $(a-b)(a b-3) \leq 0$, which is obviously true. Because $a b \leq \frac{(a+b)^{2}}{4} \leq \frac{9}{4}<3$. Equality holds when $a=b=c=1$.