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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 \(P=\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-z\right)\left(4-x\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}=2x+\sqrt{xyz}+2y+\sqrt{xyz}+2z+\sqrt{xyz}-\sqrt{xyz}=2x+2y+2z+2\sqrt{xyz}=2\left(x+y+z+\sqrt{xyz}\right)=2.4=8\)
\(x+y+z=4-\sqrt{xyz}\)
\(\Leftrightarrow x+y+z+\sqrt{xyz}=4\)
\(\Leftrightarrow4\left(x+y+z\right)+4\sqrt{xyz}=16\)
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:
\(P=2\left(x+y+z\right)+3\sqrt{xyz}=2\left(4-\sqrt{xyz}\right)+3\sqrt{xyz}=8+\sqrt{xyz}\)
Bạn ơi, P = 8 thôi nhé! Bạn quên là P còn trừ cho \(\sqrt{xyz}\) nữa nhé!! :3
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
^_^
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
https://olm.vn/hoi-dap/tim-kiem?id=199649&subject=1&q=+++++++++++Cho+x,y,z%3E0.+Th%E1%BB%8Fa+m%C3%A3n:+x+y+z+%E2%88%9Axyz=4++T%C3%ADnh+Gi%C3%A1+tr%E1%BB%8B+c%E1%BB%A7a+bi%E1%BB%83u+th%E1%BB%A9c:A=%E2%88%9Ax(4%E2%88%92y)(4%E2%88%92z)+%E2%88%9Ay(4%E2%88%92z)(4%E2%88%92x)+%E2%88%9Az(4%E2%88%92x)(4%E2%88%92y)%E2%88%92%E2%88%9Axyz++++++++++
Bạn tự tham khảo nhé