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?


WLOG $a\ge b\ge c$ and let $x=a-b,y=b-c,z=c-a$
We observe $x+y+z=0$ with $x,y\ge 0$.and as you have found $x^2+y^2+z^2=12$
Elimination of $z$ results in: ${(x+y)}^2=6+xy ...(1)$
since ${(x+y)}^2\ge 4xy$ which means $0\le xy\le 2$
let $xy=t$
since $0\le t\le 2$
Now $x^2y^2z^2=t^2(6+t)\le 6.2^2+2^3=32$....(using (1) and $z=-(x+y)$)
or $|xyz|\le 4\sqrt{2}$
