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Fie \( x,y,z\in(0,\infty) \) astfel incat \( x+y+z=xyz \). Demonstrati ca:

\( \frac{1}{x+y+\sqrt{xy}}+\frac{1}{y+z+\sqrt{yz}}+\frac{1}{z+x+\sqrt{zx}}\le \frac{\sqrt{3}}{3} \)

\( \sum_{cyc}\frac{1}{x+y+\sqrt{xy}}\le\sum_{cyc}\frac{1}{3\sqrt{xy}} \)

Ramane de demonstrat ca \( \sum_{cyc}\frac{1}{sqrt{xy}}\le\sqrt{3} \)

Relatia e echivalenta cu \( \sum_{cyc}\sqrt{x}\le\sqrt{3xyz} \)

Din inegalitatea Cauchy Buniakovsky Schwarz avem \( (\sqrt{x}\cdot 1+\sqrt{y}\cdot 1+\sqrt{z}\cdot 1)^2\le 3(x+y+z)=3xyz\Longrightarrow \sum_{cyc}sqrt{x}\le\sqrt{3xyz} \), adica inegalitatea cautata.

Corect: inegalitatea
Cauchy-Buniakovski-Schwarz