Theo giả thiết \(\sqrt{\dfrac{yz}{x}}+\sqrt{\dfrac{xz}{y}}+\sqrt{\dfrac{xy}{z}}=3\)
\(\Rightarrow\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}+2x+2y+2z=9\)
Mặt khác, ta có bđt phụ: \(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge x+y+z\)
\(\Rightarrow9\ge3\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z\le3\)
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: \(M=\sqrt{x}+\sqrt{y}+\sqrt{z}+\dfrac{2016}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(=\sqrt{x}+\sqrt{y}+\sqrt{z}+\dfrac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\dfrac{2007}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\ge2\times\sqrt{9}+\dfrac{2007}{3}=675\)
Dấu "=" xảy ra ⇔ x = y = z = 1
