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

It's enough to prove that
$$\prod_{cyc}\left(x^2+\frac{x(x+y+z)}{3}+\frac{(x+y+z)^2}{9}\right)-\frac{(xy+xz+yz)(x+y+z)^4}{9}\geq$$
$$\geq\frac{1}{2916}\left(\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)\right)^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Since $$\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)=$$
$$=5(27u^3-27uv^2+3w^3)+6(9uv^2-3w^3)-51w^3=135u^3-81uv^2-54w^3$$ and $$54^2=2916,$$
we see that the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove this inequality for an extremal value of $w^3$, which happens
for equality case of two variables.
Since the last inequality is even degree, we can assume $y=z=1$, which gives
$$(x-1)^4(x+2)^2\geq0.$$
Done!
