Which one is greater in value: $3ab^2$ or $a^3+2b^3$?
Solution 1:
Hint: If you are given that $a,b \ge 0$, then by AM-GM, $$\dfrac{a^3+b^3+b^3}{3} \ge \sqrt[3]{a^3 \cdot b^3 \cdot b^3}.$$ Can you take it from here?
If you aren't given that $a,b \ge 0$, then noticing that $(a^3+2b^3)-3ab^2 = (a-b)(a^2+ab-2b^2)$ is a good first step. Now, you just have to further factor $a^2+ab-2b^2 = (a-b)(a+2b)$ to get $(a^3+2b^3)-3ab^2 = (a-b)^2(a+2b)$. Can you tell when this is positive/negative/zero?
Solution 2:
Hint. $a^2+ab-2b^2=(a-b)(a+2b)$