Beautiful

By weighted AM-GM: $$\sum_{cyc}\frac{1}{a}(a^a) \geq abc$$Thus we have $$\sum_{cyc} a^a b c \geq (abc)^2 = (\prod_{cyc}(1+\frac{a}{b} +\frac{a}{c}))^2 \geq (3\prod_{cyc} \sqrt[3]{\frac{a^2}{bc}})^2 = 729$$Checking all equality conditions we see $a=b=c=3$ works, therefore the minimum is $\boxed{729}$

MarkBcc168 wrote: Let $a$, $b$, and $c$ be positive real numbers satisfying $ab+bc+ca=abc$. Determine the minimum value of $$a^abc + b^bca + c^cab.$$ https://artofproblemsolving.com/community/c6h2727153p23739587 https://artofproblemsolving.com/community/c6h2727218p23739983

The minimum is $3^6$, which is achivable by $(a,b,c)=(3,3,3)$. Firstly, It's not hard too see that $f(x)=x^{x-1}$ is convex in the interval $(1,\infty)$ and by AM-GM, we get \[abc\geq 27\implies a+b+c\geq 9.\]Lastly, By Jensen we get \begin{align*} P=abc(f(a)+f(b)+f(c)) &\geq abcf(\frac{a+b+c}{3})=abc(\frac{a+b+c}{3})^{\frac{a+b+c}{3}}\geq 27\times 3^3=3^6 \end{align*}as desired.

MarkBcc168 wrote: Let $a$, $b$, and $c$ be positive real numbers satisfying $ab+bc+ca=abc$. Determine the minimum value of $$a^abc + b^bca + c^cab.$$ Bu weighted $AM$$\ge$$GM$ $\frac{a^abc + b^bca + c^cab}{ab+ac+bc}$$\ge$ $\sqrt[ab+bc+ac]{a^{abc}×b^{abc}×c^{abc}}$$=$$abc$ then $a^abc + b^bca + c^cab$$\ge$$(abc)^2$ and we get from $ab+ac+bc=abc$ by $AM$$\ge$$GM$ $abc$$\ge$$27$ then $a^abc + b^bca + c^cab$$\ge$$27^2$$=$$729$