Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:

If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.

Initial attempts at AM-GM are not very helpful for me, I don't think.

$x^8+1≥2x^4,$ and $x^4+1≥2x^2$, so $(x^8+1)(x^4+1)≥4x^2(2x^2-1)$. It is easy to show $(x^8+1)(x^4+1)≥4x^2(2x^2-1)≥(x^2+1)^2$ if $x≥\sqrt{\frac{3}{7}},$ but that's insufficient.

I doubt that this problem turns into a Vieta bash due to the nature of having $x^8$ on the LHS. Initial attempts at Newton's sums were giving what looked like annoying numbers and something more bashy than what I would guess is a more elegant solution.
We begin with the following observation $$(x^8+1)(x^4+1)=((x^4)^2+1^2)((x^2)^2+1^2)\geq(x^4\cdot x^2+1\cdot1)^2=(x^6+1)^2.$$
By the Cauchy-Schwarz Inequality $$\prod_{cyc}(x^8+1)(x^4+1)\geq\prod_{cyc}(x^6+1)^2.$$
Thus, it's enough to prove that
$$\prod_{cyc}(x^4-x^2+1)\geq1$$ and since $$x^4-x^2+1\geq\left(\frac{x^2+1}{2}\right)^2,$$ it's enough to prove that
$$\prod_{cyc}(x^2+1)\geq8$$ and since by the Cauchy-Schwarz Inequality again $$\prod_{cyc}(x^2+1)(1+y^2)\geq\prod_{cyc}(x+y)^2,$$ it's enough to prove that
$$(x+y)(x+z)(y+z)\geq8,$$ which is true because
$$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)=\frac{8}{3}(x+y+z)=$$
$$=\frac{8}{3}\sqrt{(x+y+z)^2}\geq\frac{8}{3}\sqrt{3(xy+xz+yz)}=8$$
Finally, equality is achieved when $$x=y=z=1$$
COMMENT.-I find this inequality quite difficult to prove by ordinary means. However it seems to me we could do something valuable as follows: The maximum of $f(x)=\dfrac{(x^2+1)^2}{(x^8+1)(x^4+1)}$ is equal to $1.695$ and it is taken at $x=0.717$ Putting $f(x)$ and $f(y)$ maximum we must have $z=\dfrac{3-xy}{x+y}=1.7335502$ so $f(z)=0.01936916$. This way we verify that in fact
$$1\ge f(0.717)f(0.717)f(1.73355)=0.0556392$$ what makes quite plausible that $$1\ge f(x)f(y)f(z)$$ which is the proposed inequality.