\(P=\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\)
\(=\sqrt{x.\left(x+y+z\right)+yz}+\sqrt{y.\left(x+y+z\right)+zx}+\sqrt{z\left(x+y+z\right)+xy}\)
\(=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}\)
\(\le\dfrac{x+y+x+z+y+z+y+x+z+x+z+y}{2}=2.\left(x+y+z\right)=2\)
Dấu "=" xảy ra khi $x=y=z=\dfrac{1}{3}$
\(P=\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}=\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+zx}+\sqrt{z\left(x+y+z\right)+xy}\)
\(=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}\)
Áp dụng hệ quả BĐT Cô-si:
\(P\le\dfrac{x+y+x+z}{2}+\dfrac{x+y+y+z}{2}+\dfrac{x+z+y+z}{2}=\dfrac{4\left(x+y+z\right)}{2}=2\)
Dấu "=" xảy ra khi x=y=z=1/3
Vậy MaxP=2 khi x=y=z=1/3
