<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] Let $x = a - b, y = b-c, z = c-a$. Then we have $x^2 + y^2 + z^2 = 12$ and $x + y + z = 0$.
Also, we have $xy + yz + zx = \frac{(x+y+z)^2 - (x^2+y^2+z^2)}{2} = -6$.
We need to find the minimum of $xyz$.
Let $p = x + y + z = 0, q = xy + yz + zx = -6$ and $r = xyz$.
It is easy to prove that
$$-4p^3r+p^2q^2+18pqr-4q^3-27r^2 = (x-y)^2(y-z)^2(z-x)^2 = 27(32-r^2)\ge 0$$
which results in $-4\sqrt{2} \le r \le 4\sqrt{2}$. Also, when $x = \sqrt{2}, y = -2\sqrt{2}, z= \sqrt{2}$ (e.g. $a = -\sqrt{2}, b= -2\sqrt{2}, c = 0$),
we have $xyz = -4\sqrt{2}$. Thus, the minimum of $xyz$ is $-4\sqrt{2}$.
Remark: Actually, $-4p^3r+p^2q^2+18pqr-4q^3-27r^2$ is the discriminant of $u^3 - pu^2 + qu - r = 0$.
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