<s>[INST] 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.
 [/INST] 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.
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