$a^2b^2+a^2c^2+b^2c^2\leq 3$ Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$ 
Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.
My idea:
I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.
Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$
I have to show that $$y^2-2xz\leq 3$$
I tried to prove it with the sign of trinom, but it doesn't work.

Remark: I just realize that my Fact 1 is the same at Atticus's solution (he gave an elegant solution).
My Fact 1 is derived from the optimization theory (the method of Lagrange Multipliers, penalty functions, etc.).
We want to minimize $f(a,b,c) = 3 - (a^2b^2 + b^2c^2 + c^2a^2)$ subject to $g(a,b,c) = (a^2-a+1)(b^2-b+1)(c^2-c+1)-1 = 0$.
Let $L(a,b,c) = f(a,b,c) + \lambda g(a,b,c)$ be the Lagrangian. At the optimum $(a, b, c) = (1,1,1)$, 
from $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial a} = \frac{\partial L}{\partial a} = 0$, we get $\lambda = 4$.
I use Mathematica Resolve command to see that $L(a, b, c)\ge 0$ is true surprisingly! 
Actually, I first tried penalty functions. Find $\mu > 0$ such that $F(a,b,c) = f(a,b,c) + \mu [g(a,b,c)]^2 \ge 0$. 
But I found it is complicated.   
An alternative solution: 
I found the following fact (the proof is given later):
Fact 1: For any real numbers $a, b, c$, it holds that
$$a^2b^2 + b^2c^2 + c^2a^2 \le 3 + 4\cdot [(a^2-a+1)(b^2-b+1)(c^2-c+1)-1].$$
From Fact 1, the desired result follows.
$\phantom{2}$
Proof of Fact 1: $\mathrm{RHS} - \mathrm{LHS}$ can be expressed as SOS (Sum of Squares). Indeed, 
we have
$$\mathrm{RHS} - \mathrm{LHS} = \frac{1}{6} z^T Q z$$
where $z = [1, a, b, c, ab, ca, bc, abc]^T$ and ($Q$ is positive semidefinite)
\begin{align}
Q = \left(\begin{array}{rrrrrrrr} 18 & -12 & -12 & -12 & 7 & 7 & 7 & -3\\ -12 & 24 & 5 & 5 & -12 & -12 & -3 & 5\\ -12 & 5 & 24 & 5 & -12 & -3 & -12 & 5\\ -12 & 5 & 5 & 24 & -3 & -12 & -12 & 5\\ 7 & -12 & -12 & -3 & 18 & 7 & 7 & -12\\ 7 & -12 & -3 & -12 & 7 & 18 & 7 & -12\\ 7 & -3 & -12 & -12 & 7 & 7 & 18 & -12\\ -3 & 5 & 5 & 5 & -12 & -12 & -12 & 24 \end{array}\right).
\end{align}
The desired result follows.
