<s>[INST] How to prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$ Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
 [/INST] $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
Since
$(a-bc)^2\geq 0$,
$(b-ac)^2\geq 0$,
$(c-ab)^2\geq 0$,
then
$a^2+b^2c^2\geq 2abc$,
$b^2+a^2c^2\geq 2abc$,
$c^2+a^2b^2\geq 2abc$.
By of collectted through for through three inequalities last will be obtained
$a^2+b^2c^2+b^2+a^2c^2+c^2+a^2b^2\geq6abc$,
or
$a^2+a^2b^2+b^2+b^2c^2+c^2+a^2c^2\geq6abc$.
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