ta có:
$\Rightarrow (a-1)(b-1)\geq 0\Rightarrow ab\geq a+b-1\Rightarrow abc\geq ac+bc-c=c(a+b)-c=c(3-c)-c=2c-c^{2}$
$\Rightarrow 3a^{2}+3b^{2}+3c^{2}+4abc\geq \frac{3}{2}(a+b)^{2}+3c^{2}+4(2c-c^{2})=\frac{3}{2}(3-c)^{2}+3c^{2}+4(2c-c^{2})=\frac{c^{2}}{2}-c+\frac{27}{2}=\frac{(c-1)^{2}+26}{2}\geq \frac{26}{3}=13$
Tại sao $(a-1)(b-1)\geq 0$ vậy bạn