thay xyz=(4-x-y-z)²vào

Ta có \(x+y+z+\sqrt{xyz}=4\Rightarrow4x+4y+4z+4\sqrt{xyz}=16\)

Ta lại có \(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{x\left(4x+4\sqrt{xyz}+yz\right)}=\sqrt{4x^2+4x\sqrt{xyz}+xyz}=\sqrt{\left(2x+\sqrt{xyz}\right)^2}=2x+\sqrt{xyz}\)

Tương tự \(\sqrt{y\left(4-z\right)\left(4-x\right)}=2y+\sqrt{xyz}\)

\(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)

Suy ra \(P=\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)

Ta có: \(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{z\left[4\left(4-y-z\right)+yz\right]}\)

\(=\sqrt{x\left[4\left(x+\sqrt{xyz}\right)+yz\right]}=\sqrt{4x^2+4x\sqrt{xyz}+xyz}=2x+\sqrt{xyz}\)

Tương tự ta có: \(\sqrt{y\left(4-z\right)\left(4-z\right)}=2y+\sqrt{xyz}\)

Và: \(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)

Từ trên:

\(\Rightarrow T=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}\)

\(=2\left(x+y+z+\sqrt{xyz}\right)\)

\(=8\)

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Bạn tự tham khảo nhé

\(x+y+z=4-\sqrt{xyz}\)

\(\Leftrightarrow x+y+z+\sqrt{xyz}=4\)

\(\Leftrightarrow4\left(x+y+z\right)+4\sqrt{xyz}=16\)

Ta có: \(x\left(4-y\right)\left(4-z\right)\)

\(=x\left[16-4\left(y+z\right)+yz\right]\)

\(=x\left[4\left(x+y+z\right)+4\sqrt{xyz}-4\left(y+z\right)+yz\right]\)

\(=x\left(4x+4\sqrt{xyz}+yz\right)\)

\(=x\left(2\sqrt{x}+\sqrt{yz}\right)^2\)

\(\Rightarrow\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x}\left(2\sqrt{x}+\sqrt{yz}\right)\)

\(=2x+\sqrt{xyz}\)

Tương tự: \(\sqrt{y\left(4-z\right)\left(4-z\right)}=2y+\sqrt{xyz}\)

\(\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\)

Cộng vế theo vế các đẳng thức vừa chứng minh ta được:

\(P=2\left(x+y+z\right)+3\sqrt{xyz}=2\left(4-\sqrt{xyz}\right)+3\sqrt{xyz}=8+\sqrt{xyz}\)