Question

Cho các số thực dương x,y,z thỏa mãn xy+yz+xz=1. CMR:

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

Answer 1

Lời giải:

Vì $xy+yz+xz=1$ nên:

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

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

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

Do đó:

\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)

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

\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)

Answer 2

Lời giải:

Vì $xy+yz+xz=1$ nên:

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

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

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

Do đó:

\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)

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

\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)