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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.

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    $\begingroup$ What's nice about it?! $\endgroup$
    – user230734
    Jul 5, 2015 at 18:36
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    $\begingroup$ It just follows from the AM-GM inequality, see zeraoulia's answer below. $\endgroup$
    – Jack D'Aurizio
    Jul 5, 2015 at 19:39
  • $\begingroup$ For future reference, if direct manipulations seem to make things more complicated, look for a change of variables to eliminate some clutter. $\endgroup$
    – hardmath
    Jul 6, 2015 at 0:57

1 Answer 1

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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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