$x,y,z > 0$, prove that $ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $ $x,y,z > 0$, prove that 
$$ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $$
Without using Schur's inequality,

Attempt:
By $C.S$:
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} = x^{3/2}y^{3/2} + y^{3/2}z^{3/2} + z^{3/2}x^{3/2} $$
$$ \le \sqrt{ x^{3} + y^{3} +z^{3}} \sqrt{ y^{3} + z^{3} + x^{3}} = x^{3} + y^{3} + z^{3} $$
then I have to prove 
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2}  \le 3xyz$$
this one is difficult. Another thing that we know by AM-GM: $ x^{3} + y^{3}+ z^{3} \ge 3xyz$.

First, Schur's inequality provides that
$$
x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x).
$$
Then, 
$$
xy(x+y)\ge 2xy\sqrt{xy} 
$$
and hence
$$
x^3+y^3+z^3+3xyz\ge 2\big(xy\sqrt{xy} +yz\sqrt{yz} +zx\sqrt{zx} \big)
$$
