Question

[Nguyễn Ngọc Minh]

Cho x,y,z > 0 thoả x + y + z = xyz

CMR : \(\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

Cảm ơn m bạn nhìu nha ^^

Answer 1

[Phùng Minh Quân]

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

\(=\frac{xyz}{xy\left(\frac{1}{x}+\frac{1}{y}\right)zx\left(\frac{1}{z}+\frac{1}{x}\right)}=\frac{xyz}{\left(x+y\right)\left(z+x\right)}\)

Tương tự, ta cũng có: \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)}\)\(;\)\(\frac{3z}{1+z^2}=\frac{3xyz}{\left(y+z\right)\left(z+x\right)}\)

\(VT=\frac{xyz}{\left(x+y\right)\left(z+x\right)}+\frac{2xyz}{\left(x+y\right)\left(y+z\right)}+\frac{3xyz}{\left(y+z\right)\left(z+x\right)}\)

\(=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\) ( đpcm )