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
 A: New proof
I would like to start by looking at the second limitation, $(a^2+b^2)(b^2+c^2)(c^2+a^2)=8$. Multiplying the parentheses and noting that $abc = 1$ gives $a^4b^2 = \frac{a^4b^2c^2}{c^2} = \frac{a^2}{c^2}$, we get
\begin{align}
8 &= (a^2+b^2)(b^2+c^2)(c^2+a^2)\\\\
8 &= a^4 b^2+a^4 c^2+a^2 b^4+a^2 c^4+b^4 c^2+b^2 c^4+2a^2b^2c^2\\\\
9 &=  \frac{b^2 + c^2}{a^2}+\frac{a^2 + c^2}{b^2}+\frac{a^2 + b^2}{c^2} + 3\\\\
9 &=  \frac{b^2 + c^2}{a^2}+\frac{a^2 + c^2}{b^2}+\frac{a^2 + b^2}{c^2} +\frac{a^2}{a^2}+\frac{b^2}{b^2}+\frac{c^2}{c^2}\\\\
9 &= (a^2 + b^2 + c^2) \bigg( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\bigg)
\end{align}
By the Cauchy-Schwarz inequality this means that the vectors $(a, b, c)$ and $(\frac{1}{a}, \frac{1}{b}, \frac{1}{c})$ are linearly dependant, which again means that $a^2 = b^2 = c^2$, and thus $a=b=c = 1$. Now the original inequality shouldn't be too hard to prove.
A: 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 $a=b=c=1$.
