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Thay \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\) ta có:
\(\left\{{}\begin{matrix}1+x=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}+x=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)\\1+y=\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{y}+\sqrt{x}\right)\\1+z=\left(\sqrt{z}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\end{matrix}\right.\)
\(P=\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\left(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}\right)\)
\(P=\sum\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)=2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)=2\)
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Xét hạng tử: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\)
Thay \(xy+yz+zx=1\); ta có:
\(x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)^2\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}=xy+xz\)
Tượng tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=xy+yz;z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=xz+yz\)
Do đó: \(A=2\left(xy+yz+zx\right)=2.1=2\)
ĐS:...
Thay \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)
\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)
ta có: xy+yz+zx=1
=> \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
c/m tương tự ta đc: \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(y+z\right)\left(z+x\right)\)
thay vào A ta đc:
\(A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(y+z\right)\left(z+x\right)}{\left(x+z\right)\left(x+y\right)}}+y\sqrt{\frac{\left(y+z\right)\left(z+x\right)\left(x+z\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(x+y\right)\left(y+z\right)}{\left(y+z\right)\left(x+z\right)}}\)\(\Rightarrow A=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(\Rightarrow A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(\Rightarrow A=2\left(xy+yz+zx\right)\)
\(\Rightarrow A=2\) vì xy+yz+zx=1
Ta co: \(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=y+z\)
Thê vào ta được
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
Thay \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) ta có
\(1+x=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Tương tự \(1+y=\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\) và \(1+z=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)\)
\(\Rightarrow\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
và \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{z}+\sqrt{x}\right)+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
Do đó P = 2