Question

Cho x,y,z > 0 và xy + yz + zx = 1

Tính giá trị biểu thức: \(P=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)

Answer 1

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

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

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

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

Do đó : \(P=2\left(xy+yz+zx\right)=2\)