Prove that $a b \leq \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$

Question

for any $a,b\in R^+$, how to prove that $$a b \leq \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$$ my try: if I can prove that $\frac{a^2+b^2}{2}\leq \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$then this inequality can be proved but it seems that $\frac{a^2+b^2}{2}> \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$ I had no clue then.

Answer 1

Write $ab = \sqrt[3]{a^3b^3} = \sqrt[3]{a^3b^{\frac 32} b ^ {\frac 32}}$. Apply AM-GM, recombine the $b^{\frac 32}$ terms.