ÁD BĐT AM-GM:$6(x^{2}+y^{2}+z^{2})+6xyz+30-18(x+y+z)=6(\Sigma x^{2})+3(2xyz+1)+27-3.2.3.(x+y+z)$
$\geq6(\Sigma x^{2})+9\sqrt[3]{x^{2}y^{2}z^{2}}+27-3\left[ {(x+y+z)^{2}+9} \right]$
$\geq 3(\Sigma x^{2})+\frac{27}{x+y+z}-6(xy+yz+zx)(*)$
Theo BĐT Shur:$\frac{9}{x+y+z}\geq4(xy+yz+zx)-(x+y+z)^{2}=2(xy+yz+zx)-(\Sigma x^{2})$
$\Rightarrow(*) \geq0\Rightarrow đpcm$