<s>[INST] How do I prove this nice inequality $x+3\sqrt[3]{xy^2}\geq4\sqrt{xy} $? 
Let $x,y\geq0$. Prove that:
  $$
  x+3\sqrt[3]{xy^2}\geq4\sqrt{xy}
$$

Note: It's seems easy but when I tried to show it I went to complicated formula.
 [/INST] Hint :$x+3\sqrt[3]{xy^2}=x+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}\ge 4\sqrt[4]{x\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}}=4\sqrt{xy}$
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