Ta có \(2x+yz=\left(x+y+z\right)x+yz=xy+xz+x^2+yz=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{2x+yz}=\sqrt{\left(x+z\right)\left(x+y\right)}\le\dfrac{2x+y+z}{2}\left(Cauchy\right)\)
Tuơng tự \(\sqrt{2y+xz}\le\dfrac{2y+x+z}{2}\)
\(\sqrt{2z+xy}\le\dfrac{2z+x+y}{2}\)
Do đó
\(P\le\dfrac{3}{2}\left(x+y+z\right)=3\)