Difference between revisions of "1999 IMO Problems/Problem 2"
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Revision as of 18:01, 20 August 2008
Problem
(Marcin Kuczma, Poland)
Let be a fixed integer.
- (a) Find the least constant
such that for all nonnegative real numbers
,
- (b) Determine when equality occurs for this value of
.
Solution
The answer is , and equality holds exactly when two of the
are equal to each other and all the other
are zero. We prove this by induction on the number of nonzero
.
First, suppose that at most two of the , say
and
, are nonzero. Then the left-hand side of the desired inequality becomes
and the right-hand side becomes
. By AM-GM,
with equality exactly when
, as desired.
Now, suppose that our statement holds when at most of the
are equal to zero. Suppose now that
of the
are equal to zero, for
. Without loss of generality, let these be
.
We define
and for convenience, we will denote
. We wish to show that by replacing the
with the
, we increase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.
We note that
If we replace
with
, then
becomes
but none of the other terms change. Since
, it follows that we have strictly increased the right-hand side of the equation, i.e.,
By inductive hypothesis,
and by our choice of
,
Hence the problem's inequality holds by induction, and is strict when there are more than two nonzero
, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.