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

Question

Let $a,b,c\geq 0$ s.t. $$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1.$$ I have to prove that $a^2b^2+b^2c^2+a^2c^2\leq 3$.

I notice that $(a^3+1)(b^3+1)(c^3+1)=(a+1)(b+1)(c+1)$.

The solution from $a^2b^2+a^2c^2+b^2c^2\leq 3$ has a mistake.