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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 \(P=\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
\(x+y+z+\sqrt{xyz}=4\)
\(\Leftrightarrow xyz=\left(4-x-y-z\right)^2\)
\(\Leftrightarrow xyz=16+x^2+y^2+z^2-8x-8y-8z+2xy+2xz+yz\)
\(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left(16-4y-4z+yz\right)}=\sqrt{16x-4xy-4xz+xyz}\)
\(=\sqrt{16x-4xy-4xz+16+x^2+y^2+z^2-8x-8y-8z+2xy+2yz+2xz}\)
\(=\sqrt{8x-2xy-2xz+2yz+x^2+y^2+z^2-8y-8z+16}\)
\(=\sqrt{\left(-x+y+z-4\right)^2}=\left|y+z-x-4\right|=\left|y+z-x-\left(x+y+z+\sqrt{xyz}\right)\right|\)
\(=\left|-2x-\sqrt{xyz}\right|=2x+\sqrt{xyz}\) (Vì x > 0)
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 \(B=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
Ta có
\(\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x\left[4\left(4-y-z\right)+yz\right]}\)
\(=\sqrt{x\left(4\left(x+\sqrt{xyz}\right)+yz\right)}\)
\(=\sqrt{4x^2+4x\sqrt{xyz}+xyz}\)
\(=2x+\sqrt{xyz}\)
Khi đó \(T=2\left(x+y+z\right)+3\sqrt{xyz}-\sqrt{xyz}=2.4=8\)
Ta có \(4x+4y+4z+4\sqrt{xyz}=16\Rightarrow4x+4\sqrt{xyz}+yz=yz-4y-4z+16\)
=> \(\left(2\sqrt{x}+\sqrt{yz}\right)^2=\left(4-y\right)\left(4-z\right)\Rightarrow\sqrt{\left(4-y\right)\left(4-z\right)}=2\sqrt{x}+\sqrt{yz}\)
=> \(\sqrt{x}\sqrt{\left(4-y\right)\left(4-z\right)}=\sqrt{x}\left(2\sqrt{x}+\sqrt{yz}\right)=2x+\sqrt{xyz}\)
Tương tự, rồi cộng lại, ta có
\(S=2\left(x+y+z\right)+3\sqrt{xyz}-\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=8\)
Vậy S=8
^_^
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