ÁD BĐT AM-GM:$6(a^{2}+b^{2}+c^{2})+6abc+30-18(a+b+c)=6(\Sigma a^{2})+3(2abc+1)+27-3.2.3(a+b+c)$
$\geq 6(\Sigma a^{2})+9\sqrt[3]{a^{2}b^{2}c^{2}}+27-3\left[ {(a+b+c)^{2}+9} \right]$
$\geq 3(\Sigma a^{2})+\frac{27}{a+b+c}-6(ab+bc+ca)$(*)
Theo BĐT Schur:$\frac{9}{a+b+c}\geq 4(ab+bc+ca)-(a+b+c)^{2}=2(ab+bc+ca)-(\Sigma a^{2})$
$\Rightarrow (*) \geq 0 \Rightarrow$đpcm