<s>[INST] this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$
show that:
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$
Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$
so it suffices to prove that
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$
then the problem is solved.   I stuck in here
 [/INST] Assume that $x,y,z$ are roots of the polynomial 
$$ q(w) = w^3-3w^2+s w-p \tag{1}$$
where $s=xy+xz+yz$ and $p=xyz$. 
In order that $x,y,z\in\mathbb{R}$, the discriminant of $q$ must be non-negative, hence:
$$ 54ps + 9s^2 \geq 4s^3+27p^2+108 p.\tag{2}$$
On the other hand, $2x+1,2y+1,2z+1$ are roots of the monic polynomial $8\cdot q\left(\frac{w-1}{2}\right)$, hence $(2x+1)^2,(2y+1)^2,(2z+1)^2$ are roots of the monic polynomial:
$$ w^3 + (8s-51) w^2 + (99-144 p+48 s+16 s^2) w + (-49 - 112 p - 64 p^2 - 56 s - 64 p s - 16 s^2) $$
and $4(x^2+x+1),4(y^2+y+1),4(z^2+z+1)$ are roots of the previous polynomial evaluated at $w-3$. Viete's theorem hence gives:
$$\prod_{cyc}(x^2+x+1) = 13-5 p+p^2+2 s+p s+s^2\tag{3}$$
and our problem boils down to proving that:
$$ 13+p^2+p s+s^2 \geq 7s+5p.\tag{4} $$
However, that is trivial since the quadratic form $p^2+ps+s^2-7s-5p+13$ is positive definite and achieve its minimum, zero, at $(p,s)=(1,3)$.
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