Problem 1326

27. Since $a+b+c=0$, the inequality to be proved is symmetric with respect to $a, b, c$, so we can assume $a \geqslant b \geqslant c$, hence $c \leqslant 0$,
$$\begin{aligned} a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+3-6 a b c= & a^{2} b^{2}+c^{2}\left[(a+b)^{2}-2 a b\right]+3-6 a b c= \\ & a^{2} b^{2}+c^{2}\left(c^{2}-2 a b\right)+3-6 c a b= \\ & a^{2} b^{2}+2\left(c^{2}+3 c\right) a b+c^{4}+3 \end{aligned}$$

Let $f(x)=x^{2}+2\left(c^{2}+3 c\right) x+c^{4}+3$, where $x=a b$, to prove $f(x) \geqslant 0$, it suffices to prove the discriminant $\Delta=4\left[\left(c^{2}+3 c\right)^{2}-\left(c^{4}+3\right)\right] \leqslant 0$, and since $c \leqslant 0$, we have $\left(c^{2}+3 c\right)^{2}-\left(c^{4}+\right.$ 3) $=3(c+1)^{2}(2 c-1) \leqslant 0$. The inequality is proved.