SOLution. We use the mixing all variables method, similarly as in the preceding problem. We can assume that $a \geq b \geq c=0$. In this case, we obtain
$$\begin{aligned} F\left((a-b)^{2},(b-c)^{2},(c-a)^{2}\right) & =a^{6}+b^{6}+(a-b)^{6}+3 a^{2} b^{2}(a-b)^{2}-\left(a^{2}+b^{2}\right)(a-b)^{4} \\ & -a^{2} b^{2}\left(a^{2}+b^{2}\right)-(a-b)^{2}\left(a^{4}+b^{4}\right) \\ & =(a-b)^{2}\left(4 a b\left(a^{2}+b^{2}\right)-\left(a^{2}-b^{2}\right)^{2}+(a-b)^{4}+a^{2} b^{2}\right) \\ & =8 a^{2} b^{2}(a-b)^{2} \end{aligned}$$

Moreover, because $F\left(a^{2}, b^{2}, c^{2}\right)=9(a-b)^{2}\left(a^{2}+b^{2}\right)(a+b)^{2}$, it remains to prove that
$$9\left(a^{2}+b^{2}\right)(a+b)^{2} \geq 72 a^{2} b^{2}$$
which is obvious because $a^{2}+b^{2} \geq 2 a b$ and $(a+b)^{2} \geq 4 a b$. The proof is finished and the equality holds for $a=b=c$ and $a=b, c=0$ up to permutation.