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).