Let $a,b,c\ge 0: ab+bc+ca=abc+2.$ Find the minimum$$P=(a^3+1)(b^3+1)(c^3+1)$$By $a=b=c=1,$ we will prove $P\ge 8$ or $$(a^3+1)(b^3+1)(c^3+1)\ge 8.$$ After full expanding, it remains to prove $$(abc)^2+a^3+b^3+c^3+(ab)^3+(bc)^3+(ca)^3\ge 7$$We can replace the condition but it leads to something complicated.
I hope to see more simple ideas. Thanks
Proof.
Generally, we'll prove the following inequality$$abc+\sqrt[3]{(a^3+1)(b^3+1)(c^3+1)}\ge ab+bc+ca, \tag{*}$$holds for all $a,b,c\ge 0.$
WLOG, by Dirichlet rule we can assume that $(b-1)(c-1)\ge 0.$
It implies$$a(b-1)(c-1)\ge 0\iff abc+a+bc\ge ab+bc+ca.$$Now, using Minkowski on $(*)$ \begin{align*} abc+\sqrt[3]{(a^3+1)(b^3+1)(c^3+1)}&=abc+\sqrt[3]{(a^3+1)(1+b^3)(1+c^3)}\\& \ge abc+\sqrt[3]{a^3}+\sqrt[3]{(bc)^3}\\&= abc+a+bc\ge ab+bc+ca. \end{align*} Thus, the proof is done. Equality holds at $a=b=c=1.$
About Minkowski inequality, see here.