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Difference between revisions of "2008 IMO Problems/Problem 2"

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'''(ii)''' Prove that equality is achieved for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>.
 
'''(ii)''' Prove that equality is achieved for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>.
  
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== Problem 2 ==
 
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'''(i)''' If <math>x</math>, <math>y</math> and <math>z</math> are three real numbers, all different from <math>1</math>, such that <math>xyz = 1</math>, then prove that
 
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<math>\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</math>.
 
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(With the <math>\sum</math> sign for cyclic summation, this inequality could be rewritten as <math>\sum \frac {x^{2}}{\left(x - 1\right)^{2}} \geq 1</math>.)
 
  
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'''(ii)''' Prove that equality is achieved for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>.
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== Solution ==
 
== Solution ==
 
Consider the transormation <math>f:\mathbb{R}/\{1\} \rightarrow \mathbb{R}/\{-1\}</math> defined by <math>f(u) = \frac{u}{1-u}</math> and put <math>\alpha = f(x), \beta = f(y), \gamma = f(z)</math>. Since <math>f</math> is also one-to one from <math>\mathbb{Q}/\{1\}</math> to <math>\mathbb{Q}/\{-1\}</math>, the problem is equivalent to showing that  
 
Consider the transormation <math>f:\mathbb{R}/\{1\} \rightarrow \mathbb{R}/\{-1\}</math> defined by <math>f(u) = \frac{u}{1-u}</math> and put <math>\alpha = f(x), \beta = f(y), \gamma = f(z)</math>. Since <math>f</math> is also one-to one from <math>\mathbb{Q}/\{1\}</math> to <math>\mathbb{Q}/\{-1\}</math>, the problem is equivalent to showing that  

Revision as of 20:57, 4 September 2008

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 equation we need to satisfy in rational numbers.

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