Show that $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$ 
For positive real numbers $a,b,c$ with $abc = 8$ prove that
  $$
\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}
+ \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}
+ \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}}
\geq \frac{4}{3}.
$$

Can we prove this by Cauchy-Schwarz or Jensen's inequality? If not how?

By AM-GM $\sqrt{a^3+1}\leq\frac{a+1+a^2-a+1}{2}=\frac{a^2+2}{2}$.
Hence, it remains to prove that $\sum\limits_{cyc}\frac{a^2}{(a^2+2)(b^2+2)}\geq\frac{1}{3}$, which is
$\sum\limits_{cyc}(a^2b^2+2a^2)\geq72$, which is AM-GM again. 
