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.
 A: Add the following inequalities:
(1) Schur's inequality times $2xyz$:
$$xyz(x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y))\ge 0$$
That is
$$2(x^4yz+y^4zx+z^4xy)+6x^2y^2z^2\ge 2(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
(2) $12.5$ times $x^3+y^3-x^2y-xy^2\ge 0$: that is $12.5\sum_{cyc}xyz(x^3+y^3-x^2y-xy^2)\ge 0$. That is,
$$25(x^4yz+y^4zx+z^4xy)\ge 12.5(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
(3) $9.5$ times $x^3+y^3-x^2y-xy^2\ge 0$: that is $9.5\sum_{cyc}(xy)^3+(yz)^3-(xy)(yz)^2-(xy)^2(yz)\ge 0$. That is,
$$19(x^3y^3+y^3z^3+z^3x^3)\ge 9.5(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
A: Alternative proof:
(pqr method)
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
The desired inequality is written as
$$\frac{8}{27} \le \frac{p^3 r - 6pqr + q^3 + 8r^2}{q^3} = \frac{p^3 r - 6pqr + 8r^2}{q^3} + 1. \tag{1}$$
We split into two cases:
Case 1 : $p^3 r - 6pqr + 8r^2 \ge 0$
Clearly, (1) is true.
Case 2: $p^3 r - 6pqr + 8r^2 < 0$
Using $q^2 \ge 3pr$, it suffices to prove that
$$\frac{8}{27} \le \frac{p^3 r - 6pqr + 8r^2}{q \cdot 3pr} + 1$$
or
$$\frac{9p^3 - 35pq + 72 r}{27pq} \ge 0.$$
Using $r \ge \frac{4pq - p^3}{9}$ (3 degree Schur),
we have
$$9p^3 - 35pq + 72 r
\ge 9p^3 - 35pq + 72\cdot \frac{4pq - p^3}{9}
= p(p^2 - 3q) \ge 0.$$
We are done.
A: 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 $x^3+y^3\geq xy(x+y).$
A: 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.
