Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$ 
Let $x,y,z\in \mathbb R^+$ prove that: 
$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$

The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz \le RHS$. How to prove above inequality ? It is too hard for me.
 A: The inequality is homogeneous, so we may assume that $xyz=1$. Then if we expand left and right hand side, we see that we have to show that
$$\sqrt{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 3} \ge 1 + \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
Let
$$u = \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
If we square both sides of the expanded inequality, we get the equivalent inequality
$$u^3 \ge 2u + u^2,$$
or,
$$u^2 - u - 2 \ge 0.$$
Therefore it is enough to show that $u \ge 2$. But this is certainly true since $x^3 + \frac{1}{x^3}, y^3 + \frac{1}{y^3}, z^3 + \frac{1}{z^3} \ge 2$.
A: $$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{AM-GM}{\ge} 3xyz$$
DOES help, since you can also prove:
$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
Note that this is the same as 
$$2^6(x^2y+ y^2z+z^2x)^3(y^2x+z^2y+x^2z)^3 \geq 3^6x^2y^2z^2(x^2+yz)^2(y^2+xz)^2(z^2+xy)^2 \,.$$
which after long computations and many tears can be reduced to this all junk being positive. If you look to the few negative terms , you can get easely rid of them by invoking the AM-GM inequality.
Now, all this suggests that the inequality
$$2\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \geq 3 \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
is true, all you have to do is find a neater solution :)
