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> Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
> $$P = (a - b)(b - c)(c - a)$$
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**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, $\min P = -8$

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> Is this solution correct? If not, then why?
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