# 2008 IMO Problems/Problem 2

## 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) , 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)