I am sure that $(3, 2, 1)$ is the only natural triple satisfying the following Diophantine equation $$2(a^3+b^3+c^3)=abc(abc+6)\quad\text{for}\quad a\ge b\ge c.$$ But I can not prove or refute that. Help me, please.
What I've tried so far. Rearrange the equation to get $$ 2(a^3+b^3+c^3-3abc)=a^2 b^2 c^2 $$ and $$ (a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)=a^2 b^2 c^2 $$ Let $a-b=n$ and $a-c=m$. Hence $m \ge n$ and $$ (3a-m-n)(m^2+n^2+(m-n)^2)=[a(a-m)(a-n)]^2 $$ or $$ 2(3a-(m+n))((m+n)^2-3mn)=[a(a^2-(m+n)a+mn)]^2 $$ Now let $m+n=p$ and $mn=q$, then we have $$ 2(3a-p)(p^2-3q)=[a(a^2-pa+q)]^2 $$ Now we have a quadratic in $q$ and I tried to use $\Delta_{q}=k^2$ (where $k$ is an integer) for bounding some of variables, but I was unsuccessful.
HINT
Using the identity $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$ one can get the original equation in the form of $$2(a+b+c)((a+b+c)^2-3(ab+bc+ac)) = (abc)^2,$$ or $$r=\frac{2s^3-p^2}{6s},$$ where $$s=a+b+c,\quad r=ab+bc+ac,\quad p=abc,$$ $$a>0,\quad b>0,\quad c>0,\quad s^3-27p\ge 0.$$ That allows to write the cubic equation for $x\in\{a, b, c\}$ in the form of $$x^3 - sx^2 + \frac{2s^3-p^2}{6s}x - p = 0,$$ or $$6sx^3 - 6s^2x^2 + (2s^3-p^2)x - 6sp =0.$$ For natural solutions $$p=qx,$$ $$(6s-q^2)x^2 - 6s^2x + 2s(s^2-3q) = 0.\qquad(1)$$ The "linear" case in $x$ is $$\begin{cases} q^2 = 6s\\ x_1=\dfrac{s^2-3q}{3s}\\ x_{2,3}^2 - (s-x_1)x_{2,3} + q = 0\\ x>0,\quad s>0,\quad q>0. \end{cases}\qquad(2)$$ System $(2)$ gives the only solution: $$x_1=\frac{q^2}{18} - \frac 6q,\quad q=s=6,\quad x\in\{1,2,3\},$$ $$\boxed{(a,b,c) = (3,2,1)}.$$ The "quadratic" case in $x$ gives: $$\begin{cases} 6s\not= q^2\\ x_1 = \dfrac{3s^2\pm d}{6s-q^2}\\ d^2 = 9s^4 - 2s(s^2-3q)(6s-q^2)\\ x_{2,3}^2 - (s-x_1)x_{2,3} + q = 0\\ x>0,\quad q>0,\quad s^3\ge 27qx_1. \end{cases}\qquad(3)$$
UPD
Using AM-GM inequality: $$x_1 = s-(x_2+x_3)\ge s-2\sqrt{x_2x_3},$$ $$x_1\ge s-2\sqrt q,$$ so $$\dfrac{2s(s^2-3q)}{3s^2\mp d}\in\mathcal N,\quad \dfrac{2s(s^2-3q)}{3s^2\mp d}\ge s-2\sqrt q\ge0,\quad s^2\ge4q.\qquad(4).$$ Attempts to obtain other solutions either of $(1-4)$ in the case of $6s\not= q^2\\$ do not lead to a positive result. One can do substution
$$s=w\sqrt q,$$ obtaining $$6w\not=q\sqrt q,\quad\dfrac{2w(w^2-3)}{3w^2\mp\sqrt{9w^4 - 2w(w^2-3)(6w-q\sqrt q)}}\ge w-2\ge0$$ and closed inequality in form $q(w)\ge0$, but even in the case $$6w<q\sqrt q,\quad \text{"}\mp\text{"}=\text{"}+\text{"}$$ one get infinity set of inequality solutions, so this don't change the situation cardinally.
This gives grounds for considering that solution $(3,2,1)$ is an easy case of the OP task and reduces the chances of success in the general case.