Theo giả thiết \(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xz}{y}}+\sqrt{\frac{xy}{z}}=3\)
\(\Rightarrow\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}+2x+2y+2z=9\)
Mặt khác , ta có BĐT phụ : \(\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}\ge x+y+z\)
\(\Rightarrow9\ge3\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z\le3\)
Áp dụng BĐT Cauchy Shwarz \(\Rightarrow\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)\le9\)
\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le3\)
Ta có : \(P=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{2016}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{2007}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\ge2.\sqrt{9}+\frac{2007}{3}=675\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Chúc bạn học tốt !!!
