Prove that $(a^2+b^2)(b^2+c^2)(c^2+a^2) \geq 8a^2b^2c^2$ This is my proof. I'm not sure if it is true but perhaps someone would tell me yes or (no and why). 
Since $(a^2+b^2), (b^2+c^2), (c^2+a^2)$ are greater than or equal to zero, then $(a^2+b^2)(b^2+c^2)(c^2+a^2) \geq 0$. I'll spare you the distribution and say this, $(a^2+b^2)(b^2+c^2)(c^2+a^2) + 8a^2b^2c^2 \geq 8a^2b^2c^2$
Done.

rewriting what andres said : $(a^2+b^2)\geq2ab,(b^2+c^2)\geq2bc,(c^2+a^2)\geq2ca$.
therefore $(a^2+b^2)(b^2+c^2)(c^2+a^2) \geq (2ab)(2bc)(2ca) = 8a^2 b^2 c^2$
