TITLE: Inequality: $(a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$
QUESTION [6 upvotes]: Proving inequality for positive real $a,b,c > 0$: $$ (a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$$

REPLY [7 votes]: Use AM-GM inequality we have $$a^3+2=a^3+1+1\ge 3a$$
so
$$a^3+3b^2+5\ge 3a+3b^2+3=3(a+b^2+1)$$
so
$$LHS\ge 27(a+b^2+1)(1+b+c^2)(a^2+1+c)$$
Use Holder inequality we have
$$(a+b^2+1)(1+b+c^2)(a^2+1+c)\ge (a+b+c)^3$$
By done!