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.

Note: This answer is very similar with Michael Rozenberg's, maybe a little more direct, but I think it's worth posting.
Using Holder's inequality:
$$8(x^8+1)=(1^4+1^4)(1^4+1^4)(1^4+1^4)[(x^2)^4+1] \geq (x^2+1)^4$$
and from Cauchy-Schwarz:
$$2(x^4+1) = (1^2+1^2)((x^2)^2+1) \geq (x^2+1)^2$$
Multiplying with similar inequalities, we get:
$$(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1) \geq \frac{1}{2^{12}} \left[(x^2+1)(y^2+1)(z^2+1)\right]^6$$
Therefore, we are left to prove:
$$\frac{1}{2^{12}} \left[(x^2+1)(y^2+1)(z^2+1)\right]^6\geq \left[(x^2+1)(y^2+1)(z^2+1)\right]^2$$
or equivalent:
$$(x^2+1)(y^2+1)(z^2+1) \geq 8$$
This follows from the Cauchy-Schwarz inequality again:
$$2(x^2+1)(y^2+1)(z^2+1) = (1^2+x^2+x^2+1^2)(1^2+y^2+z^2+y^2z^2)$$
$$\geq (1+xy+zx+yz)^2 = 16$$
and the inequality is proven. Equality occurs when $x = y= z = 1$.
