it’s related to this If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$ :
My try with $a\geq b \geq c$ :
We start with a substitution : $x=\frac{a}{b}$$\quad$$y=\frac{b}{c}$$\quad$$z=\frac{c}{a}$ We get (from the original inequality) : $$\frac{x^3}{x+1}+\frac{y}{y+1}+\frac{y^2}{z+1}\ge 1.5(3)^{\frac{-2}{3}}(x^3+1+z^3)^{\frac{2}{3}}$$ With the condition : $$\frac{1}{x^3+1+\frac{1}{y^3}}+\frac{1}{y^3+1+\frac{1}{z^3}}+\frac{1}{z^3+1+\frac{1}{x^3}}=1$$ Wich is equivalent to : $$xyz=1$$ So with the condition we get : $$\frac{x^3}{x+1}+\frac{y}{y+1}+\frac{y^2}{\frac{1}{xy}+1}-1.5(3)^{\frac{-2}{3}}(x^3+1+\frac{1}{(xy)^3})^{\frac{2}{3}}\ge 0$$We intitulate this inequality$ (E)$
With the new condition $x\ge y \ge 1$ : So we have : $$(E)\ge\frac{x^3}{x+1}+\frac{1}{1+1}+\frac{x}{x+1} -1.5(3)^{\frac{-2}{3}}(x^3+1+\frac{1}{(1)^3})^{\frac{2}{3}}=f(x)\geq 0$$ Because we have : $$\frac{x}{x+1}\leq \frac{y^2}{\frac{1}{xy}+1}$$ The study of the function $f(x)$ is omitted but a graphic could convince you .
Thanks for your advices .
plugging $$a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$$ in the equation $$a^3+b^3+c^3-3=0$$ we get $$\left( {x}^{2}z+x{y}^{2}+{z}^{2}y \right) \left( {z}^{2}{x}^{4}-{y}^ {2}z{x}^{3}+{y}^{4}{x}^{2}-y{z}^{3}{x}^{2}-{y}^{3}{z}^{2}x+{y}^{2}{z}^ {4} \right) =0$$ we have the equation $$(x^2z)^2+(xy^2)^2+(z^2y)^2=x^3y^2z+y^3z^2x+z^3x^2y$$ remember that $$X^2+Y^2+Z^2\geq XY+XZ+ZX$$