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.

First solution: We have
$$8(xy+yz+zx)^3 = [ \sum (xy+yz) ]^3 \leq 9( (xy+yz)^3 +  (yz+zx)^3 + (zx+xy)^3  ), $$
so it remains to show that
$$ (xy+yz)^3 +  (yz+zx)^3 + (zx+xy)^3   \leq 3 (x^2+yz)(y^2+zx)(z^2 + xy). $$
Let $ M (a, b, c) = \sum_{sym} x^a y^b z^c$. Then, by expanding everything, we get that
$$3(x^2+yz)(y^2+zx)(z^2 + xy)  - (xy+yz)^3 -  (yz+zx)^3 - (zx+xy)^3  \\ = M(4, 1, 1) + M(3, 3, 0) + M(2, 2, 2) - M (3, 2, 1). $$
By taking the symmetric summations of $ x^4 yz + x^3y^3 + x^2y^2z^2 \geq 3 x^3 y^2 z$, we get that $ M(4, 1, 1) + M(3, 3, 0) + M(2, 2, 2) \geq 3 M(3, 2, 1)$.
Hence we are done.

Alternative solution: Continuing from OP's work, WTS
$$27/2 M(4,1,1) + 19/2 M(3, 3, 0) + 1 M(2, 2, 2) \geq 24 M(3, 2, 1).$$
Muirhead tells us that $M(4,1,1) \geq M(3, 2, 1)$ and $M(3, 3, 0) ) \geq M(3, 2, 1)$.
We just need to AM-GM the $M(2, 2, 2)$ term, which has a very small coefficient, so it likely will work out.
As before, by taking the symmetric summations of $ x^4 yz + x^3y^3 + x^2y^2z^2 \geq 3 x^3 y^2 z$, we get that $ M(4, 1, 1) + M(3, 3, 0) + M(2, 2, 2) \geq 3 M(3, 2, 1)$.
Hence (by summing the relevant inequalities), the result follows.
