Áp dụng bđt \(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)ta có:
\(A=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge\sqrt{3\left(\frac{xy\cdot yz}{xz}+\frac{yz\cdot xz}{xy}+\frac{xy\cdot xz}{z\cdot y}\right)}\)
\(=\sqrt{3\left(x^2+y^2+z^2\right)}=\sqrt{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)