If $abc=1$ and $\prod_{\text{cyc}}(a^2+b^2)=8$, then $(a+b+c)^3\prod_{\text{cyc}}(a^2+b^2)\ge\left (\sum_{\operatorname{cyc}} (a^2b+b^2a) \right) ^3$. 
Let $a,b,c>0 $ where $abc=1$ and $(a^2+b^2)(b^2+c^2)(c^2+a^2)=8$.  Prove that $$ (a+b+c)^3(a^2+b^2)(b^2+c^2)(c^2+a^2) \ge \big((a^2b+b^2a)(b^2c+c^2b)(c^2a+a^2c)\big) ^3.$$

Thanks

By AM-GM, $\left(a^2+b^2\right)\geq 2ab$ and similarly for two other pairs.  Hence, $$8=\prod_{\text{cyc}}\,\left(a^2+b^2\right) \geq \prod_{\text{cyc}}\,(2ab)=8\,(abc)^2=8\,.$$
Therefore, the equality holds, implying that $a=b=c=1$.
