For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$

First, note that $$2(1-a+a^2)(1-b+b^2)=1+a^2b^2+(a-b)^2+(1-a)^2(1-b)^2 \geqslant 1+a^2b^2$$

Hence it is enough to show that $$3(1+a^2b^2)(1-c+c^2) \geqslant 2(1+abc+a^2b^2c^2)$$ Considering the above as a quadratic in $c$, it is clear when $c=0$ this is true, so we just need to confirm the discriminant doesn't get positive, i.e. $$\Delta = (3+2ab+3a^2b^2)^2-4\cdot (3+a^2b^2)\cdot (1+3a^2b^2) = -3(1-ab)^4 \leqslant 0$$

Another way.

Since $$3(1-a+a^2)^3-1-a^3-a^6=(a-1)^4(2a^2-a+2),$$ by Holder we obtain: $$3\prod_{cyc}(1-a+a^2)\geq\sqrt[3]{\prod_{cyc}(1+a^3+a^6)}\geq1+abc+a^2b^2c^2.$$