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Given that $x$, $y$, $z$ are real positive numbers such that $xyz(x+y+z)=1$, how can I show that $xyz\leq\frac{\sqrt{\sqrt 3}}{3}$?

We have $(xyz)^{1/3}\leq \frac{x+y+z}{3}$ by inequalities between geometric mean and arithmetic mean. But then what?

Please, can someone help me?

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We have (by reciprocal on AM-GM inequality) $$xyz=\frac{1}{x+y+z}\leq\frac{1}{3(xyz)^{1/3}}\Rightarrow(xyz)^{4/3}\leq\frac{1}{3}\Rightarrow xyz\leq\frac{1}{3^{3/4}}=\frac{\sqrt{\sqrt{3}}}{3}$$

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  • $\begingroup$ Thank u alot for ur great answer :D can we have a communication via exchange or facebook or somthing like that ? $\endgroup$ Commented Mar 13, 2015 at 21:30
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You have $$(xyz)^{1/3}\leq \frac{x+y+z}{3}$$ by AMGM. So then you can substitute in $(x+y+z)=\frac{1}{xyz}$:

$$(xyz)^{1/3}\leq \frac{1}{3xyz}$$ $$(xyz)^{4/3}\leq \frac{1}{3}$$ Then note that $f(x)=x^{4/3}$ is increasing, so we may take roots: $$xyz\leq \frac{1}{3^{3/4}}=\frac{\sqrt{\sqrt{3}}}{3}$$

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