Cho \(\frac{{xy + 1}}{y} = \frac{{yz + 1}}{z} = \frac{{zx + 1}}{x}\). Chứng minh rằng: x = y = z hoặc x²y²z² = 1.

\(\frac{{xy + 1}}{y} = \frac{{yz + 1}}{z} = \frac{{zx + 1}}{x}\)

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

Do đó:

• x – y = \(\frac{1}{z} - \frac{1}{y} = \frac{{y - z}}{{yz}}\);

• \(y - z = \frac{1}{x} - \frac{1}{z} = \frac{{z - x}}{{xz}}\);

• \(z - x = \frac{1}{y} - \frac{1}{x} = \frac{{x - y}}{{xy}}\).

Suy ra (x – y)(y – z)(z – x) = \(\frac{{(x - y)(y - z)(z - x)}}{{{x^2}{y^2}{z^2}}}\)

\( \Leftrightarrow \frac{{(x - y)(y - z)(z - x).{x^2}{y^2}{z^2} - (x - y)(y - z)(z - x)}}{{{x^2}{y^2}{z^2}}} = 0\)

\( \Leftrightarrow \) (x – y)(y – z)(z – x)(x²y²z² – 1) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x - y = 0\\y - z = 0\\z - x = 0\\{x^2}{y^2}{z^2} - 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = y = z\\{x^2}{y^2}{z^2} = 1\end{array} \right.\)

 Vậy ta có điều phải chứng minh.