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

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<cmath>p^2-2q \ge 1.</cmath>
 
<cmath>p^2-2q \ge 1.</cmath>
 
But from <math>p = -1-q</math>, we get  
 
But from <math>p = -1-q</math>, we get  
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<cmath>p^2-2q = (1+q)^2-2q = 1 + q^2 \ ge 1,</cmath>
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<cmath>p^2-2q = (1+q)^2-2q = 1 + q^2 \ge 1,</cmath>
−
with equality holding iff <math>q = 0</math>. That proves part '''(i)''' and points us in the direction of looking for rational <math>\alpha,\beta,\gamma</math> for which <math>q=0</math> and (hence) <math>p=-1)</math>.
+
with equality holding iff <math>q = 0</math>. That proves part '''(i)''' and points us in the direction of looking for rational <math>\alpha,\beta,\gamma</math> for which <math>q=0</math> and (hence) <math>p=-1</math>, that is:
 +
<cmath>\begin{align*}
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\alpha+\beta+\gamma & = 1\\
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\alpha\beta+\beta\gamma+\gamma\alpha & = 0
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\end{align*}</cmath>

Revision as of 21:52, 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 $z$.

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*}

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