Find the minimum of $a + b + c$ Let $a, b, c$ be non-negative real numbers such that
$$abc[(a − b)(b − c)(c − a)]^2 = 1$$
Find the minimum of $a + b + c$.
Source: https://cms.math.ca/crux/v43/n5/public_Chow_et_al_43_5.pdf 
My Attempt: 
$a \mapsto \frac{x+y}{2}$
$b \mapsto \frac{y+z}{2}$
$c \mapsto \frac{x+z}{2}$ $\;$ without any success
 A: Remarks: @youthdoo said there is a AM-GM solution. I found it.
Problem: Let $a, b, c\ge 0$ with $abc[(a-b)(b-c)(c-a)]^2 = 1$. Prove that
$$a + b + c \ge \sqrt[9]{\frac{531441}{16}}.$$
Sketch of a proof:
WLOG, assume that $a > b > c$.
Let $0 < x_1 < x_2 < x_3$ be the three real roots of $2x^3 - 54x^2 + 243x - 243 = 0$.
Let
$$x_4 = \frac{x_1x_2}{x_2 - x_1}, \quad x_5 = \frac{9 - x_2}{2} + \frac{x_1x_2}{x_2 - x_1}, \quad
x_6 = \frac{9 - x_1}{2} - \frac{x_1x_2}{x_2 - x_1}.$$
It is easy to prove that $x_4, x_5, x_6 > 0$.
By AM-GM, we have
\begin{align*}
 &x_1x_2x_3x_4^2x_5^2x_6^2\cdot abc[(a-b)(b-c)(c-a)]^2\\
 ={}&x_1a \cdot x_2b \cdot x_3c \cdot x_4(a - b) \cdot x_4(a - b) \cdot x_5(b - c) \cdot x_5(b - c) \cdot x_6(a - c) \cdot x_6 (a - c) \\
 \le{}& \left(\frac{x_1a + x_2b + x_3c + x_4(a-b) \times 2 + x_5(b - c) \times 2 + x_6(a-c)\times 2}{9}\right)^9\\
 ={}& \left(\frac{(x_1+2x_4+2x_6)a + (x_2-2x_4+2x_5)b + (x_3-2x_5-2x_6)c}{9}\right)^9\\
 ={}& (a + b + c)^9
\end{align*}
where we have used $x_1 + 2x_4 + 2x_6 = 9$, $x_2 - 2x_4 + 2x_5 = 9$, and
$x_3 - 2x_5 - 2x_6 = x_1 + x_2 + x_3 - 18 = 9$ (by Vieta's theorem).
It suffices to prove that
$$x_1x_2x_3x_4^2x_5^2x_6^2 = \frac{531441}{16}.$$
Using $x_1x_2x_3 = \frac{243}{2}$ (by Vieta's theorem), it suffices to prove that
$$(x_4x_5x_6)^2 = \frac{2187}{8}$$
which is true. The proof is omitted here.
We are done.
