Proving inequality for positive real $a,b,c > 0$: $$ (a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$$
2026-04-03 00:10:45.1775175045
Inequality: $(a^3+3b^2+5)(b^3+3c^2+5)(c^3+3a^2+5) \ge 27(a+b+c)^3$
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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!