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?


The inequality
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
become equality when $a-b=b-c=c-a,$ or $a=b=c,$ but for this value then
$$(a-b)(b-c)(c-a)=0 \ne -8.$$
This is my solution, we have
$$P^2 = \frac{4(a^2+b^2+c^2-ab-bc-ca)^3-(a+b-2c)^2(b+c-2a)^2(c+a-2b)^2}{27} \quad (1)$$
$$\leqslant \frac{4}{27}(a^2+b^2+c^2-ab-bc-ca)^3.$$
Therefore
$$P^2 \leqslant \frac{4}{27}(a^2+b^2+c^2-ab-bc-ca)^3 = 32,$$
or
$$-4\sqrt 2 \leqslant (a-b)(b-c)(c-a) \leqslant 4\sqrt 2.$$
So $P_{\min} = -4\sqrt 2,$ equality occur when $a=1,\;b=1+2\sqrt 2,\;c=1+\sqrt 2.$
Note. How to find constant $\frac{4}{27}?$
For $(a-b)(b-c)(c-a) \ne 0,$ setting $x=a-b,\,y=b-c,$ then
$$F = \frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2+c^2-ab-bc-ca)^3} = \frac{x^2y^2(x+y)^2}{(x^2+xy+y^2)^3}.$$
From $x^2+xy+y^2 \geqslant \frac{3}{4}(x+y)^2$ and the AM-GM inequality, we have
$$F \leqslant \frac{64}{27} \cdot \frac{x^2y^2}{(x+y)^4} \leqslant \frac{4}{27}.$$
From this proof we get
$$\frac{4}{27}-  \frac{x^2y^2(x+y)^2}{(x^2+xy+y^2)^3}=\frac{(x-y)^2(2y+x)^2(2x+y)^2}{(x^2+xy+y^2)^3}.$$
It's equivalent to  the identity $(1).$
