A hard inequality $(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 ) + 11abc \leq 12$

Question

Given $$c=\min⁡(a,b,c)~, \quad a+b+c=3 \\ P=(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 )~,$$ I have to prove that $$P+11abc \le 12~.$$

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I started with

$$b^2-bc+c^2 \le b^2 \quad \text{and} \quad c^2-ca+a^2 \le a^2~,$$ which led to $$P \le a^2 b^2 (a^2-ab+b^2 ) \le \frac{4}{9} \left[\frac {(a+b+c)^2}{3} \right]^3 \le \frac{4}{9} \left[ \frac {(a+b+c)^2}{3} \right]^3=12~.$$

Then I'm stuck.

Answer 1

The variables should be non-negatives, otherwise the inequality is wrong: $(a,b,c)=(4,0,-1).$

Let $a=\min\{a,b,c\}\geq0$, $b=a+u$, $c=a+v$.

Thus, we need to prove that $$\prod_{cyc}(a^2-ab+b^2)+\frac{11abc(a+b+c)^3}{27}\leq\frac{4(a+b+c)^6}{243}$$ or $$567(u^2-uv+v^2)a^4+36(19u^3+3u^2v+3uv^2+19v^3)a^3+$$ $$+18(11u^4+35u^3v-6u^2v^2+35uv^3+11v^4)a^2+$$ $$+18(4a^5+u^4v+10u^3v^2+10u^2v^3+uv^4+4v^5)a+$$ $$+(u^2+11uv+v^2)(2u-v)^2(u-2v)^2\geq0,$$ which is obvious.