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
Solution: The minimum of $a+b+c$ is $\sqrt[9]{\frac{531441}{16}}$. Let us prove it.
First, let $p_0 = \sqrt[9]{\frac{531441}{16}}, q_0 = \sqrt[9]{1458}$ and $r_0 = \sqrt[9]{\frac{1}{157464}}$. The cubic equation $x^3 - p_0x^2 + q_0x - r_0 = 0$ has three distinct positive real roots denoted by $x_1, x_2, x_3$. Let $a = x_1, b = x_2, c= x_3$. Then $a + b + c = p_0 = \sqrt[9]{\frac{531441}{16}}$ and (noting that $2p_0^3-243r_0 = 0$ and $2p_0^2-9q_0 = 0$) \begin{align} abc(a-b)^2(b-c)^2(c-a)^2 &= r_0(-4p_0^3r_0+p_0^2q_0^2+18p_0q_0r_0-4q_0^3-27r_0^2)\\ &= \frac{16}{531441}p_0^9\\ &= 1. \end{align}
Thus, it suffices to prove that for $a, b, c > 0$, $$f(a,b,c) = 16(a+b+c)^9 - 531441abc(a-b)^2(b-c)^2(c-a)^2 \ge 0.$$ We use the pqr method. Let $p = a + b + c$, $q = ab+bc+ca$ and $r = abc$. Then \begin{align} f(a,b,c) &= 2125764rq^3-531441p^2rq^2-9565938pr^2q\\ &\quad +2125764p^3r^2+16p^9+14348907r^3. \end{align} We split into two cases:
1) $2 p^3-243 r = 0$: By using $r = \frac{2p^3}{243}$, we have $f(a,b,c) = 6p^3(7p^2+36q)(2p^2-9q)^2\ge 0 $.
2) $2 p^3-243 r \ne 0$: For each fixed $p, r > 0$, let \begin{align} g(q) &= 2125764rq^3-531441p^2rq^2-9565938pr^2q\\ &\quad +2125764p^3r^2+16p^9+14348907r^3. \end{align} $g(q)$ is a cubic polynomial whose discriminant is \begin{align} \mathrm{discr}(g) = -122009559759792 r^2(2 p^3-243 r)^2h(r, p) \end{align} where \begin{align} h(r,p) &= (64p^{12}-4131p^9r+260^2p^6r^2) \\ &\quad + r^2(8612603p^6-330024861p^3r+3486784401r^2)\\ &\ge (2\sqrt{64p^{12} \cdot 260^2p^6r^2} - 4131p^9r)\\ &\quad + r^2(2\sqrt{8612603p^6 \cdot 3486784401r^2} - 330024861p^3r)\\ &> 0. \end{align} Since $\mathrm{discr}(g) < 0$, the cubic equation $g(q) = 0$ has exactly one real root on $(-\infty, +\infty)$. Also, $g(+\infty) = +\infty$ and $g(0) = 16p^9+2125764p^3r^2+14348907r^3 > 0$. Thus, $g(q) > 0$ for any $q > 0$.
We are done.
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.