IMO 2008 · Problem 2

(a) Prove that \[ \frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\geq 1 \] for all real numbers \(x,y,z\), each different from \(1\), and satisfying \(xyz=1\).

(b) Prove that equality holds above for infinitely many triples of rational numbers \(x,y,z\), each different from \(1\), and satisfying \(xyz=1\).

Official text of the problem. Official problem paper (PDF)

Solution

(a) Set \[ a=\frac{x}{x-1},\qquad b=\frac{y}{y-1},\qquad c=\frac{z}{z-1}. \] These are well-defined by the hypotheses. Since \[ a-1=\frac1{x-1},\qquad b-1=\frac1{y-1},\qquad c-1=\frac1{z-1}, \] the condition \(xyz=1\) gives \[ abc=(a-1)(b-1)(c-1). \] Expanding and cancelling \(abc\), we obtain \[ ab+bc+ca=a+b+c-1. \] Consequently, \[ \begin{aligned} \frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2} &=a^2+b^2+c^2\\ &=(a+b+c)^2-2(ab+bc+ca)\\ &=(a+b+c-1)^2+1\\ &\geq 1. \end{aligned} \] This proves the inequality for every admissible real triple. It also shows that equality holds exactly when \[ \frac{x}{x-1}+\frac{y}{y-1}+\frac{z}{z-1}=1. \]

(b) For each positive integer \(n\), take \[ x=-n(n+1),\qquad y=-\frac{n+1}{n^2},\qquad z=\frac{n}{(n+1)^2}. \] All three numbers are rational. Moreover, \(x,y<0\) and \(0<z<1\), so none equals \(1\), and \[ xyz= \bigl(-n(n+1)\bigr) \left(-\frac{n+1}{n^2}\right) \frac{n}{(n+1)^2}=1. \] Writing \(D=n^2+n+1\), direct calculation gives \[ \frac{x}{x-1}=\frac{n(n+1)}D,\qquad \frac{y}{y-1}=\frac{n+1}D,\qquad \frac{z}{z-1}=-\frac nD. \] Their sum is \[ \frac{n(n+1)+(n+1)-n}{D} =\frac{n^2+n+1}{D}=1. \] Thus equality holds by part (a). Finally, the first coordinates \(-n(n+1)\) are distinct for distinct positive integers \(n\), so these give infinitely many distinct triples.

The solution was checked against the official IMO shortlist solutions.