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

Question

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.

Answer 1

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

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

Answer 3

The Macavity's beautiful idea we can release also by C-S:

$$2(1-a+a^2)(1-b+b^2)\geq2\sqrt{\frac{1+a^4}{2}}\cdot\sqrt{\frac{1+b^4}{2}}\geq1+a^2b^2.$$