Ta có:
\(P=\dfrac{x}{\sqrt{y+3z}}+\dfrac{y}{\sqrt{z+3x}}+\dfrac{z}{\sqrt{x+3y}}=\dfrac{x^2}{x\sqrt{y+3z}}+\dfrac{y^2}{y\sqrt{z+3x}}+\dfrac{z^2}{z\sqrt{x+3y}}\)
\(P\ge\dfrac{\left(x+y+z\right)^2}{x\sqrt{y+3z}+y\sqrt{z+3x}+z\sqrt{x+3y}}\)
Mà:
\(\sqrt{x}.\sqrt{xy+3xz}+\sqrt{y}.\sqrt{yz+3xy}+\sqrt{z}.\sqrt{xz+3yz}\le\sqrt{\left(x+y+z\right)\left(4xy+4yz+4zx\right)}\)
\(\le\sqrt{\left(x+y+z\right).\dfrac{4}{3}\left(x+y+z\right)^2}=6\)
\(\Rightarrow P\ge\dfrac{3^2}{6}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Hoặc có thể dùng BĐT Holder:
Đặt \(P=\dfrac{x}{\sqrt{y+3z}}+\dfrac{y}{\sqrt{z+3x}}+\dfrac{z}{\sqrt{x+3y}}\)
\(P^2\left[x.\left(y+3z\right)+y.\left(z+3x\right)+z.\left(x+3y\right)\right]\ge\left(x+y+z\right)^3\)
\(\Rightarrow P^2\ge\dfrac{\left(x+y+z\right)^3}{4\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^3}{\dfrac{4}{3}\left(x+y+z\right)^2}=\dfrac{9}{4}\)
\(\Rightarrow P\ge\dfrac{3}{2}\)
