Let $a, b, c > 0$ $;abc=8$ , $$\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \ge \frac43 $$ (ref: original image)
I tried using AM-GM to eliminate the square root on the bottom but I am stuck, what is the general strategy here and maybe I can use its symmetry?
By AM-GM $$1+a^3=(1+a)(1-a+a^2)\leq\left(\frac{1+a+1-a+a^2}{2}\right)^2=\frac{(2+a^2)^2}{4}.$$ Thus, it's enough to prove that $$\sum_{cyc}\frac{a^2}{(2+a^2)(2+b^2)}\geq\frac{1}{3}$$ or $$3\sum_{cyc}a^2(2+c^2)\geq\prod_{cyc}(2+a^2)$$ or $$\sum_{cyc}(a^2b^2+2a^2)\geq72,$$ which is true by AM-GM: $$\sum_{cyc}(a^2b^2+2a^2)\geq3\sqrt[3]{a^4b^4c^4}+6\sqrt[3]{a^2b^2c^2}=3\cdot16+6\cdot4=72.$$ Done!