2) Cho x,y,z > 0 thỏa mãn [tex]x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=3\sqrt{xyz},[/tex]
Tính giá trị biểu thức
[tex]A=(1+\sqrt{\frac{x}{y}})(1+\sqrt{\frac{y}{z}})(1+\sqrt{\frac{z}{x}})[/tex]
$x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=3\sqrt{xyz}$
$\Leftrightarrow (\sqrt{x}+\sqrt{y})^{3}-3\sqrt{xy}(\sqrt{x}+\sqrt{y})+z\sqrt{z}=3\sqrt{xyz}$
$\Leftrightarrow (\sqrt{x}+\sqrt{y}+\sqrt{z})^{3}-3\sqrt{xy}(\sqrt{x}+\sqrt{y}+(\sqrt{z})-3\sqrt{z}(\sqrt{x}+\sqrt{y})(\sqrt{x}+\sqrt{y}+\sqrt{z})=0$
$\Leftrightarrow (\sqrt{x}+\sqrt{y}+\sqrt{z})\left [ (\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}-3\sqrt{xy}-3\sqrt{z}(\sqrt{x}+\sqrt{y}) \right ]=0$
$\Leftrightarrow (\sqrt{x}+\sqrt{y}+\sqrt{z})\left ( x+y+z-\sqrt{xy}-\sqrt{yz}-\sqrt{zx} \right )=0$
$\Leftrightarrow (\sqrt{x}+\sqrt{y}+\sqrt{z})\frac{1}{2}\left [ \left ( \sqrt{x}-\sqrt{y} \right )^{2}+\left ( \sqrt{y}-\sqrt{z} \right )^{2}+\left ( \sqrt{z}-\sqrt{x} \right )^{2} \right ]=0$
[tex]\Leftrightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}=0[/tex] hoặc [tex]\sqrt{x}=\sqrt{y}=\sqrt{z}[/tex]
TH1: $\sqrt{x}+\sqrt{y}+\sqrt{z}=0$ (*)
Vì [tex]x;y;z>o\Rightarrow \sqrt{x};\sqrt{y};\sqrt{z}>0\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}>0\Rightarrow (*)[/tex] vô lý
TH2: [tex]\sqrt{x}=\sqrt{y}=\sqrt{z}[/tex] $\Rightarrow x=y=z$
$ \Rightarrow A=(1+\sqrt{\frac{x}{y}})(1+\sqrt{\frac{y}{z}})(1+\sqrt{\frac{z}{x}})=(1+1)(1+1)(1+1)=8$