<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] Minimize $P = (a-b)(b-c)(c-a)$
given $(a - b)^2 + (b - c)^2 + (c - a)^2 = 12$
WLOG, say $a \ge b \ge c, \, a - b = x, a - c = y$ where $x \ge 0, y \ge x$
which gives,
$P (x,y) = xy(x-y)$ ...(i)
$\, G(x,y) = x^2 + y^2 + (x-y)^2 - 12 = 0$ ...(ii)
Using Lagrange,
$Q(x,y) = P (x,y) + \lambda G(x,y) = xy(x-y) + \lambda (2x^2 + 2y^2 -2xy - 12)$
You get $\, 2xy-y^2 + \lambda(4x - 2y) = 0$ ..(iii)
$\, x^2-2xy + \lambda(4y - 2x) = 0$ ...(iv)
From (iii) and (iv),
You get $y = 2x, x = 2y$
Taking $y = 2x$ (as $y \ge x$) and substituting in (ii), you get $x = \sqrt 2$
From (i) that gives min $\, P = -4\sqrt2$
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