<s>[INST] Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$ Prove $a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)\geq 6abc$ 
My attempt: 
$a^{2}(1+b^{2})+b^{2}(1+c^{2})+c^{2}(1+a^2)-6abc\geq 0$ 
$\implies a^{2}+a^{2}b^{2}+b^{2}+b^{2}c^{2}+c^{2}+c^{2}a^{2}-2abc-2abc-2abc\geq 0$ 
$\implies (a-bc)^{2}+(b-ac)^{2}+(c-ab)^{2}\geq 0$ 
Each of these terms must be non-negative, thus the sum is also non-negative. 
I'm new to writing proofs, so I don't know whether this proof is fine.
 [/INST] Alternatively by $AM \ge GM$,
\begin{align*}
  a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2) &=
  a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2 \\
  & \ge 6\sqrt[6]{a^6b^6c^6}  \\
\end{align*}
</s>