<s>[INST] Find the minimum of $P = (a - b)(b - c)(c - a)$ 
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$

My solution:

*

*We have:

$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$

*

*Using AM-GM Inequality, we have:

$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$

*

*Therefore, $\min P = -8$

Is this solution correct? If not, then why?

 [/INST] We need to show that the value $-8$ occurs, otherwise, we can not say that it's a minimal value.
By the way, we can prove that the minimum is $-4\sqrt2.$
Indeed, we need to prove that
$$(a-b)(b-c)(c-a)\geq-4\sqrt2\left(\sqrt{\frac{\sum\limits_{cyc}(a^2-ab)}{6}}\right)^3,$$  which easy to get by AM-GM after substitution $a-b=x$, $b-c=y$.
</s>