This problem took so much time… Sum both equations, the result is the expansion of cyc (ab - a - c - 1)^2

a_507_bc wrote: Let $a, b, c$ be reals such that $a^2(c^2-2b-1)+b^2(a^2-2c-1)+c^2(b^2-2a-1)=0.$ Show that $$3(a^2+b^2+c^2)+4(a+b+c)+3 \geq 6abc.$$ Equality holds when $ a=b=c=1-\sqrt 2 .$

We will add up $3\sum a^2+4\sum a+3-6abc$ by zero, i.e., $a^2(c^2-2b-1)+b^2(a^2-2c-1)+c^2(b^2-2a-1)$ to obtain $\sum a^2b^2-2\sum a^2b-6abc+2\sum a^2+4\sum a+3\ge 0.$ That is, \[(ab-a-c-1)^2+(bc-b-a-1)^2+(ca-c-b-1)^2\ge 0.\]This completes our proof.