<s>[INST] Ptove that $3\prod\limits_{cyc}(x^2-x+1)\geq x^2y^2z^2-5xyz+\sum\limits_{cyc}(x^2y+x^2z)$ For all real numbers $x$, $y$ and $z$ prove that:
$$3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2-5xyz+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y$$
I tried $uvw$ and more. For example, this inequality is a quadratic inequality of $x$ and it's enough to prove it for $\Delta\ge0$, but it's nothing.
I found this problem here :artofproblemsolving.com/community/c6h140p779898 
My motivation is the following.
There is the following easy similar inequality.
For all non-negative $x$, $y$ and $z$ prove that:
$$3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq1+xyz+x^2y^2z^2$$ 
 [/INST] Let us prove that
$$3(x^2-x+1)(y^2-y+1)(z^2-z+1) \ge x^2y^2z^2 - 5xyz + x^2(y+z) + y^2(z+x) + z^2(x+y) + 1.$$
Proof: With the substitutions $y = 1 + u, \ z = 1+v$, we have
$$\mathrm{LHS} - \mathrm{RHS} = Ax^2 + Bx + C$$
where
\begin{align}
A &= (2 v^2+v+2) u^2+(v^2-v) u+2 v^2, \\
B &= -3 u^2 v^2-3 u^2 v-3 u v^2-4 u^2+2 u v-4 v^2, \\
C &= (3 v^2+2 v+2) u^2+(2 v^2-v) u+2 v^2.
\end{align}
$A \ge 0$ follows from $2 v^2+v+2 > 0$ and $4(2 v^2+v+2)\cdot 2v^2 - (v^2-v)^2 = 5v^2(3v^2+2v+3) \ge 0$.
$C \ge 0$ follows from $3 v^2+2 v+2 > 0$ and $4(3 v^2+2 v+2)\cdot 2v^2 - (2 v^2-v)^2 = 5v^2(4v^2+4v+3)\ge 0$.
We have
$$4AC - B^2 = 5u^2v^2\big[(3v^2+2v+3)u^2+(2v^2-2v)u+3v^2\big].$$
$4AC - B^2\ge 0$ follows from $3v^2+2v+3 > 0$ and
$4(3v^2+2v+3)\cdot 3v^2 - (2v^2-2v)^2 = 32v^2(v^2+v+1) \ge 0$.
We are done.
</s>