A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$ 
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$

My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations and I have a rather complicated question question that I could not write as the sum of squares:
$$ 0\leq 6x^2y^2z^2 +19(x^3y^3+y^3z^3+z^3x^3)+27(x^4yz+xy^4z+xyz^4)-24(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3).$$
Thanks for any ideas that might help me clarify the issue.

Another solution based on what you have already obtained and AM-GM:
It is easy to check that;
$$x^2y^2z^2+x^4yz+x^3y^3 \geq3x^3y^2z.$$
So we need to show that;
$$17(x^3y^3+x^3z^3+z^3y^3)+25(x^4yz+xy^4z+xyz^4)\geq21(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Now realize that we simply get;
$$x^3y^3+x^3y^3+x^3y^3+x^4yz+x^4yz+xyz^4\geq6x^3y^2z,$$
by adding all the similar inequalities together, we will have;
$$(x^3y^3+x^3z^3+z^3y^3)+(x^4yz+xy^4z+xyz^4)\geq(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Hence, it is just needed to prove that;
$$8(x^4yz+xy^4z+xyz^4)\geq4(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Equivalently, we should show that;
$$2xyz(x^3+y^3+z^3)\geq xyz(xy(x+y)+xz(x+z)+yz(y+z)).$$
And this is almost obvious since $x^3+y^3\geq xy(x+y).$
