2008 IMO Problems/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 .
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) , that is: Expressing from the first equation and substituting into the second, we get as the sole condition we need to satisfy in rational numbers.
If and for some integers ,,and , they would need to satisfy For to be integer, we would like to divide . Consider the example where divides for any integer . Substituting back, that gives us A simple check shows that are rational and well defined and that and for any integer (even for ).
Moreover, from and for large , we see that infinitely many generate infinitely many different triplets of , , and . That completes the proof of part (ii). --Vbarzov 03:03, 5 September 2008 (UTC)