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.

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. 
$$\Delta = (3+2ab+3a^2b^2)^2-4\cdot (3+a^2b^2)\cdot (1+3a^2b^2) = -3(1-ab)^4 \leqslant 0$$ 
