$a^2b^2+a^2c^2+b^2c^2\leq 3$ 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.

Another idea is to notice that:
$$2(a^2-a+1)^2=a^4+1+(a-1)^4\geq a^4+1$$
So, using this and Cauchy-Schwarz:
$$16 = 16(a^2-a+1)^2(b^2-b+1)^2(c^2-c+1)^2 \geq 2(a^4+1)(b^4+1)(c^4+1)=$$
$$=(a^4+a^4+1+1)(b^4+c^4+b^4c^4+1)\geq (a^2b^2+a^2c^2+b^2c^2+1)^2$$
It follows that:
$$a^2b^2+b^2c^2+c^2a^2 \leq 3$$
with equality when $a=b=c=1$.
