On Proving $ a^6 + b^6 + c^ 6 − 3a ^2 b^ 2 c^ 2 + 2(a^ 2 + bc)(b ^2 + ca)(c^ 2 + ab) ≥ 0 $ where $a,b,c$ are real numbers.

Question

On this problem, I made some observations and then started to solve.
Here one important note is that the inequality is trivial for $a,b,c\ge0$ and $a,b,c\le0$ and the inequality is symmetric.
Another observation: Working cases when one is negative or one is positive are the same.
Suppose WLOG $a\ge b\ge c$ when one is negative. So we need to prove that$-$ $$ a^6 + b^6 + c^ 6 − 3a ^2 b^ 2 c^ 2 + 2(a^ 2 - bc)(b ^2 - ca)(c^ 2 + ab) ≥ 0 $$ Suppose to the contrary that WLOG $c\ge b\ge a$ when one is positive. So we need to prove that$-$ $$ a^6 + b^6 + c^ 6 − 3a ^2 b^ 2 c^ 2 + 2(a^ 2 - bc)(b ^2 - ca)(c^ 2 + ab) ≥ 0 $$ Which are the same. Working any could do. In both the cases, $a,b,c\ge0$. Say I do the first one. $$ a^6 + b^6 + c^ 6 − 3a ^2 b^ 2 c^ 2 - 2 a^4 b c + 2 a^3 b^3 - 2 a^3 c^3 + 4 a^2 b^2 c^2 - 2 a b^4 c + 2 a b c^4 - 2 b^3 c^3 \ge 0 $$ $$ a^6 + b^6 + c^ 6 + 2 a^3 b^3 + a ^2 b^ 2 c^ 2+ 2 a b c^4 \ge 2 a^4 b c + 2 a^3 c^3 + 2 a b^4 c + 2 b^3 c^3 $$ How to prove this? Thanks!