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This is a follow up question to my previous post "Inequalities of expressions completely symmetric in their variables". An answer provided a counterexample to me reasoning: under the constraints $a,b,c\in\Bbb{R}^+$ and $a+b+c=3$,

$$ (a^2-ab+b^2)(c^2-ac+a^2)(b^2-bc+c^2) \le 12. $$

I demanded a proof for this inequality, however since it was an entirely different question, I felt the need for a new post.

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    This is not symmetric. Is the second factor supposed to be $c^2-bc+b^2$? – 2017-02-06
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    Corrected it. It should be fine now. – 2017-02-06
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    @Nilabro Saha The variables should be non-negatives. – 2017-02-06

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Let $a\geq b\geq c\geq0$. Hence, $$\prod_{cyc}(a^2-ab+b^2)\leq(a^2-ab+b^2)a^2b^2=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2=$$ $$=12(3-ab)\cdot\frac{ab}{2}\cdot\frac{ab}{2}\leq12\left(\frac{3-ab+\frac{ab}{2}+\frac{ab}{2}}{3}\right)^3=12$$

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    I'm sorry, I don't quite understand how you asserted that $\prod_{cyc}(a^2-ab+b^2) \le (a^2-ab+b^2)a^2b^2$. Mind explaining a bit more? – 2017-02-06
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    @Nilabro Saha Because $a^2-ac+c^2\leq a^2$ and $b^2-bc+c^2\leq b^2$. – 2017-02-06
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    Also, why should $(9-3ab)a^2b^2\le 12$? – 2017-02-06
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    @Nilabro Saha It's Just AM-GM. See my fixing. – 2017-02-06
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    @MichaelRozenberg For this to be complete, you should also note that the maximum $12$ is in fact attained. Nice example, btw. – 2017-02-06
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    Now I see it! Thanks a lot! – 2017-02-06
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    Just asking out of curiosity.. Does this expression have a minimum value? – 2017-02-06
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    @NilabroSaha $a=b=0\,$. – 2017-02-06
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    Oh! Yea, my bad. – 2017-02-06
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    If $a=2$, $b=1$, $c=0$, then then the maximum of 12 is attained. From the answer it follows that there is no other way to get the maximum with $a\ge b\ge c\ge 0$. That's because to get $=$ instead of $\le$ everywhere in the answer, we need $c=0$, $a+b=3$, $3-ab=ab/2$. – 2018-02-13