\(x^2+4xy+y^2\ge6xy\Leftrightarrow4\left(x^2+xy+y^2\right)\ge3\left(x+y\right)^2\Leftrightarrow\sqrt{x^2+xy+y^2}\ge\dfrac{\sqrt{3}}{2}\left(x+y\right)\)
\(\Rightarrow P=\sum\sqrt{x^2+xy+y^2}\ge\dfrac{\sqrt{3}}{2}\sum\left(x+y\right)=\sqrt{3}\left(x+y+z\right)=3\)
\(\Rightarrow P_{min}=3\) khi \(x=y=z=\dfrac{\sqrt{3}}{3}\)
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