A little tricky part 3

Call the expression A, using AM-GM we get these following inequality $\frac{ab}{c^3(2a+b)}+\frac{abc(2a+b)}{9}+\frac{abc^2}{3}\geq ab$ $\frac{bc}{a^3(2b+c)}+\frac{abc(2b+c)}{9}+\frac{a^2bc}{3}\geq bc$ $\frac{ca}{b^3(2c+a)}+\frac{abc(2c+a)}{9}+\frac{ab^2c}{3}\geq ca$ Combining 3 inequalities and we get $A+2(abc)^2\geq ab+bc+ac\geq 3\sqrt[3]{(abc)^2}$ From the condition, we can prove that $abc\geq 1$ , that means there exist 3 number $x,y,z>0$ satisfy $a=\frac{x}{y} ; b=\frac{y}{z} ;c=\frac{z}{x}$ , replace them into the inequality and we have $A+2\geq 3$ $\Leftrightarrow A\geq 1$ The equality holds when $a=b=c=1$ or $x=y=z$