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.

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.
