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 \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)

\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)

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) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)

\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)

1111111111111111111

\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)

Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)

Là xong.

Ta có \(x^2+1=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)

\(y^2+1=\left(y+z\right)\left(y+x\right)\)

\(z^2+1=\left(z+x\right)\left(z+y\right)\)

Khi đó

\(S=x.\sqrt{\left(y+z\right)^2}+y.\sqrt{\left(x+z\right)^2}+z.\sqrt{\left(x+y\right)^2}=2\left(xy+yz+xz\right)=2\)

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:...

Ta có: 

\(1+x^2=xy+yz+xz+x^2=\left(x+y\right)\left(x+z\right)\)

\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)

\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)

Thay vào T ta được:

\(T=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)

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

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=xy+xz+xy+yz+xz+zy\)

\(=2\left(xy+yz+xz\right)=2\left(xy+yz+xz=1\right)\)

Ta có \(1+x^2=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\).

Tương tự ta cũng có \(1+y^2=\left(x+y\right)\left(y+z\right)\) và \(1+z^2=\left(z+x\right)\left(y+z\right)\).

Thu gọn được \(T=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)

\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\frac{\left(xy+yz+xz+y^2\right)\left(xy+yz+xz+z^2\right)}{xy+yz+xz+x^2}}=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left(y+z\right)\)

tương tự ta có

\(y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}=y\left(x+z\right)\) và \(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\left(x+y\right)\)

do đó \(A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2.1=2\)

vậy A=2