<s>[INST] $a^2b^2+a^2c^2+b^2c^2\leq 3$ 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.
 [/INST] We'll prove that
$$(a^2-a+1)(b^2-b+1)\geq\frac{a^2+b^2}{2}.$$ 
Indeed, let $a+b=2u$ and $ab=v^2$.
Thus, we need to prove that
$$a^2b^2-ab(a+b)+a^2+b^2+ab-a-b+1\geq\frac{a^2+b^2}{2}$$ or
$$v^4-2uv^2+2u^2-2u+1\geq0$$ or
$$(v^2-u)^2+(u-1)^2\geq0,$$ which is obvious.
Thus, $$1=\prod_{cyc}(a^2-a+1)=\sqrt{\prod_{cyc}(a^2-a+1)(b^2-b+1)}\geq\sqrt{\prod_{cyc}\frac{a^2+b^2}{2}}.$$
Id est, it's enough to prove that:
$$\sqrt{\prod_{cyc}\frac{a^2+b^2}{2}}\geq\left(\sqrt[4]{\frac{a^2b^2+a^2c^2+b^2c^2}{3}}\right)^3.$$
Now, let $a^2=x$, $b^2=y$ and $c^2=z$.
Thus, we need to prove that
$$27(x+y)^2(x+z)^2(y+z)^2\geq64(xy+xz+yz)^3.$$
Now, since $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)$$ it's
$$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that
$$(x+y+z)^2\geq3(xy+xz+yz)$$ or
$$\sum_{cyc}(x-y)^2\geq0$$ and we are done!
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