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:

\(x^2+1=x^2+xy+yz+zx\)

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

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

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

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

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

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

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

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

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

 Tương tự như thế, ta được

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

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.

biết làm rồi

VẬy bạn giải ra cho mọi người xem được ko?

Lớn hơn hoặc bằng kí hiệu trong Latex là \geq nha!

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

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

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)

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

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

ta có :

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

tương tự ta sẽ có :

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