For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:

Let $a,b,c$ be non-negative real numbers. Prove that 
  $$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$

Do you have any hints to solve this inequality? Any hints would be fine. 
I tried this:
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
By Cauchy's inequality, $$a^2+1\ge(2a)$$ and I did the same to $b$ and $c$ and applied it to the problem but the results are $$2abc\ge1+a^2b^2c^2$$ and this is wrong.

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:
$$3\prod_{cyc}(1-a+a^2)\geq\sqrt[3]{\prod_{cyc}(1+a^3+a^6)}\geq1+abc+a^2b^2c^2.$$
