By $AM-GM$: $x+x+x+y \geq 4\sqrt[4]{x^3y}$ and 2 similar inequalities we get the desired result. Equality when $a=b=c$.

By AM-GM: $3x+y=x+x+x+y \geq 4\sqrt[4]{x^3y}$ $3y+z=y+y+y+z\geq 4\sqrt[4]{y^3z}$ $3z+x=z+z+z+x \geq 4\sqrt[4]{z^3x}$ $ So, (3x+y)(3y+z)(3z+x) \geq 64xyz $ as desired Equality occurs if and only if and only if x=y=z

If $x,y,z>0$. Prove that $$(x+y)(y+z)(3z+x) \ge (4+3\sqrt[3]{3}+3\sqrt[3]{9})xyz$$$$(x+y)(2y+z)(3z+x) \ge (7+3\sqrt[3]{6}+3\sqrt[3]{36})xyz$$