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

with loss of generality, assume $x\ge y $
consider $f(x, y, z) = (x^2 + x + 1)(y^2 + y + 1) (z^2 + z + 1) - 9 (z(x+y) + xy)$


*

*we first show that 


$$f(x, y, z) \ge f((x+y)/2, (x+y)/2, z)$$
since
$$f(x,y,z) - f((x+y)/2, (x+y)/2, z) = (x-y)((z^2 + z + 1)(z -2 - 2xy) + 9)$$
we simply show that 
$$(z^2 + z + 1)(z -2 - 2xy) + 9 \ge 0$$
while $z^2 + z + 1 > 0$, 
$$(z^2 + z  + 1)(z - 2 - 2xy) + 9 \ge (z^2 + z + 1)(z - 2 - \frac{(3 - z)^2}{2}) + 9\\
= \frac{1}{2}(18 - (z^2 + z + 1)(z^2 - 8z +13))$$
and it is quite easy to show that when $0\le z\le 3$,
$$(z^2 + z + 1)(z^2 - 8z +13)\le 18$$ with maximum at $z = 1$.


*If we achieve minimum of $f(x, y, z)$ at any point other than $(1,1,1)$, use the above argument, we can show a contradiction.

