Let $x,y,z$ be real numbers ( $x \ne y$, $y\ne z$, $x\ne z$) different from $0$. If $\frac{x^2-yz}{x(1-yz)}=\frac{y^2-xz}{y(1-xz)}$, prove that the following relation holds: $$x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$
Problem
Source: https://artofproblemsolving.com/community/c6h1870126p12682434
Tags: algebra
GorgonMathDota
07.07.2019 05:57
We have \[ (x^2 - yz)(1- xz)y = x(1 - yz)(y^2 - xz) \]\[ (x^2 - yz - x^3 z + xyz^2) y = x(y^2 - y^3 z - xz + xyz^2 ) \] \[ x^2 y - y^2 z - x^3 yz + xy^2 z^2 = xy^2 - xy^3 z - x^2 z + x^2 y z^2 \]\[ xy(x-y) - z(y^2 - x^2) - xyz(x^2 - y^2) + xyz^2 (y - x) = 0 \]\[ (x - y)(xy + xz + yz - xyz(x+y+z) ) = 0 \]Since $x \not= y$, we rhen have $xy + xz + yz = xyz(x+y+z)$, which is equivalent to $x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$.
sqing
07.07.2019 11:32
GorgonMathDota wrote:
We have
\[ (x^2 - yz)(1- xz)y = x(1 - yz)(y^2 - xz) \]\[ (x^2 - yz - x^3 z + xyz^2) y = x(y^2 - y^3 z - xz + xyz^2 ) \]
\[ x^2 y - y^2 z - x^3 yz + xy^2 z^2 = xy^2 - xy^3 z - x^2 z + x^2 y z^2 \]\[ xy(x-y) - z(y^2 - x^2) - xyz(x^2 - y^2) + xyz^2 (y - x) = 0 \]\[ (x - y)(xy + xz + yz - xyz(x+y+z) ) = 0 \]Since $x \not= y$, we rhen have $xy + xz + yz = xyz(x+y+z)$, which is equivalent to $x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$.