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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.

Maximilian Janisch
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Isaac Tu
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  • Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz Mar 18 '20 at 12:54
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    Hint: RHS $= \frac13( a^3 + b^{3/2} + b^{3/2})$. – achille hui Mar 18 '20 at 13:00
  • Ohh, thanks very much – Isaac Tu Mar 18 '20 at 13:10
  • many thanks!... – Isaac Tu Mar 18 '20 at 13:16
  • Read up on weighted AM-GM inequality $\frac13a^3+\frac23b^{3/2} \geqslant ab$ is a direct application. – Macavity Mar 18 '20 at 13:21
  • this is a verbatim special case of Young's inequality for products – Maximilian Janisch Mar 18 '20 at 13:21

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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.

Sarvesh Ravichandran Iyer
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