<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] 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,
$$19(x^3y^3+y^3z^3+z^3x^3)\ge 9.5(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3)$$
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