<s>[INST] A hard inequality $(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 ) + 11abc \leq 12$ 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 $$P+11abc \le 12~.$$ 

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.
 [/INST] 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.
</s>