Inequality related AM-GM It looks easy but I cannot find the solution. 

Let $x,y,z$ be positive reals and satisfy $x^2+y^2+z^2=2$, prove that
  $$ \left(x^3+\sqrt{3}xyz\right)\left( y^3-\sqrt{3}xyz\right)\leqslant 1. $$

I invented this inequality about a month ago. Essentially it can be proved by brutal force ( homogenize the inequality and then massive expanding), by SOS ( the usual method of express the polynomial into sum of square), or the Lagrange Multiplier Method (very painful task).
I am looking for a smart solution where the use of computer is minimum. 

If $y^2-\sqrt3xz<0$ then our inequality is obviously true.
But for $y^2-\sqrt3xz\geq0$ by AM-GM we obtain:
$$(x^3+\sqrt3xyz)(y^3-\sqrt3xyz)=xy(x^2+\sqrt3yz)(y^2-\sqrt3xz)\leq$$
$$\leq\frac{1}{27}(x^2+y^2+xy+\sqrt3yz-\sqrt3xz)^3=$$
$$=\frac{1}{216}(2x^2+2y^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$
$$=\frac{1}{216}(6-x^2-y^2-3z^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$
$$=\frac{1}{216}\left(6-(x-y+\sqrt3z)^2\right)^3\leq\frac{6^3}{216}=1.$$
Done!
