Difference between revisions of "2008 IMO Problems/Problem 2"
(New page: == Problem 2 == '''(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 <m...) |
(→Solution) |
||
Line 6: | Line 6: | ||
'''(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>. | ||
+ | == Problem 2 == | ||
+ | '''(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 | ||
+ | <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>. | ||
+ | (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>.) | ||
+ | |||
+ | '''(ii)''' Prove that equality is achieved for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>. | ||
== 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> | + | 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 |
<cmath>\alpha^2+\beta^2+\gamma^2 \ge 1 \quad (1)</cmath> | <cmath>\alpha^2+\beta^2+\gamma^2 \ge 1 \quad (1)</cmath> | ||
− | + | subject to | |
− | and that | + | <cmath>\left(\frac{\alpha}{\alpha+1}\right) |
+ | \left(\frac{\beta}{\beta+1}\right) | ||
+ | \left(\frac{\gamma}{\gamma+1}\right) = 1 \quad (2)</cmath> | ||
+ | and that equallity holds for infinitely many triplets of ''rational'' <math>\alpha,\beta,\gamma</math>. | ||
+ | |||
+ | Now, rewrite (2) as <math>\alpha\beta\gamma = (1+\alpha)(1+\beta)(1+\gamma)</math> and express it as | ||
+ | <cmath>0 = 1 + p + q</cmath> | ||
+ | where <math>p=\alpha+\beta+\gamma</math> and <math>q = \alpha\beta+\beta\gamma+\gamma\alpha</math>. Notice that (1) can be written as | ||
+ | <cmath>p^2-2q \ge 1.</cmath> | ||
+ | But from <math>p = -1-q</math>, we get | ||
+ | <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>. |
Revision as of 20:31, 4 September 2008
Problem 2
(i) If , and are three real numbers, all different from , such that , then prove that . (With the sign for cyclic summation, this inequality could be rewritten as .)
(ii) Prove that equality is achieved for infinitely many triples of rational numbers , and .
Problem 2
(i) If , and are three real numbers, all different from , such that , then prove that . (With the sign for cyclic summation, this inequality could be rewritten as .)
(ii) Prove that equality is achieved for infinitely many triples of rational numbers , and .
Solution
Consider the transormation defined by and put . Since is also one-to one from to , the problem is equivalent to showing that subject to and that equallity holds for infinitely many triplets of rational .
Now, rewrite (2) as and express it as where and . Notice that (1) can be written as But from , we get with equality holding iff . That proves part (i) and points us in the direction of looking for rational for which and (hence) .