ta có$:x^2+xy+xz=3yz\Leftrightarrow (x+y)(y+z)=4yz$
đặt$ x+y=a;z+x=b\Rightarrow (a-b)^2=(y-z)^2,ab=4yz$
lại có$:a^3+b^3=(a+b)(a^2-ab+b)^2\leq \sqrt{2(a^2+b^2)}[(a-b)^2+ab]=\sqrt{2[(z+y)^2+4yz]}(y+z)^2\leq \sqrt{4(y+z)^2}(y+z)^2=2(y+z)^3(1)$
tiếp$:3(x+y)(y+z)(z+x)=12yz(y+z)\leq3(y+z)^2(y+z)=3(y+z)^3(2)$
$(1)+(2)=đpcm$