Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=3\)
\(P=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\)
\(P_{min}=3\) khi \(a=b=c=1\) hay \(x=y=z=1\)