Ta có:
\(VT=\sqrt{x+z}\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}}+\sqrt{x+y}\sqrt{\dfrac{y}{\left(x+y\right)\left(y+z\right)}}+\sqrt{y+z}\sqrt{\dfrac{z}{\left(x+z\right)\left(y+z\right)}}\)
\(\Rightarrow VT^2\le\left(x+z+x+y+y+z\right)\left(\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(x+y\right)\left(y+z\right)}+\dfrac{z}{\left(x+z\right)\left(y+z\right)}\right)\)
\(\Rightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác ta có:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(\Rightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+zx\right)}{\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)}=\dfrac{9}{2}\)
\(\Rightarrow VT\le\dfrac{3\sqrt{2}}{2}\) (đpcm)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z\)