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2008 IMO Problems/Problem 2

Revision as of 21:35, 4 September 2008 by Vbarzov (talk | contribs) (→‎Solution)

Problem 2

(i) If $x$, $y$ and $z$ are three real numbers, all different from $1$, such that $xyz = 1$, then prove that $\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1$. (With the $\sum$ sign for cyclic summation, this inequality could be rewritten as $\sum \frac {x^{2}}{\left(x - 1\right)^{2}} \geq 1$.)

(ii) Prove that equality is achieved for infinitely many triples of rational numbers $x$, $y$ and $z$.


Solution

Consider the transormation $f:\mathbb{R}/\{1\} \rightarrow \mathbb{R}/\{-1\}$ defined by $f(u) = \frac{u}{1-u}$ and put $\alpha = f(x), \beta = f(y), \gamma = f(z)$. Since $f$ is also one-to one from $\mathbb{Q}/\{1\}$ to $\mathbb{Q}/\{-1\}$, the problem is equivalent to showing that \[\alpha^2+\beta^2+\gamma^2 \ge 1 \quad (1)\] subject to \[\left(\frac{\alpha}{\alpha+1}\right) \left(\frac{\beta}{\beta+1}\right) \left(\frac{\gamma}{\gamma+1}\right) = 1 \quad (2)\] and that equallity holds for infinitely many triplets of rational $\alpha,\beta,\gamma$.

Now, rewrite (2) as $\alpha\beta\gamma = (1+\alpha)(1+\beta)(1+\gamma)$ and express it as \[0 = 1 + p + q\] where $p=\alpha+\beta+\gamma$ and $q = \alpha\beta+\beta\gamma+\gamma\alpha$. Notice that (1) can be written as \[p^2-2q \ge 1.\] But from $p = -1-q$, we get \[p^2-2q = (1+q)^2-2q = 1 + q^2 \ge 1,\] with equality holding iff $q = 0$. That proves part (i) and points us in the direction of looking for rational $\alpha,\beta,\gamma$ for which $q=0$ and (hence) $p=-1$, that is: \begin{align*} \alpha+\beta+\gamma & = -1\\ \alpha\beta+\beta\gamma+\gamma\alpha & = 0 \end{align*} Expressing $\alpha$ from the first equation and substituting into the second, we get \[\beta\gamma + ( \beta+\gamma ) ( -1 - \beta -\gamma ) = 0\] as the sole condition we need to satisfy in rational numbers.

If $\beta = \frac{b}{m}$ and $\gamma = \frac{c}{m}$ for some integers $b$,$c$,and $m$, they would need to satisfy \[bc = m(b+c)+(b+c)^2 \Leftrightarrow m = \frac{bc}{b+c} - (b+c).\] For $m$ to be integer, we would like $b+c$ to divide $bc$. Consider the example \[b=t, c=t^2-t, m = t-1-t^2,\] where $b+c = t^2$ divides $bc = t(t^2-t)$ for any integer $t \ne 0$. Substituting back, that gives us \[\beta = \frac{t}{t-1-t^2},\quad \gamma = \frac{t^2-t}{t-1-t^2},\quad \alpha = \frac{1-t}{t-1-t^2}.\] A simple check shows that $\alpha,\beta,\gamma$ are rational and well defined and that $p=-1$ and $q=0$ for any integer $t$ (even for $t=0$).

Moreover, from $\lim_{t\rightarrow +\infty} \beta = 0$ and $\beta < 0$ for large $t$, we see that infinitely many $t$ generate infinitely many different triplets of $\alpha$, $\beta$, and $\gamma$. That completes the proof of part (ii).

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