1982 IMO Problems/Problem 3
Contents
Problem
Consider infinite sequences of positive reals such that and .
a) Prove that for every such sequence there is an such that:
b) Find such a sequence such that for all :
Solution 1
By Cauchy, the LHS is at least:
It is clear that and
We thus have that the LHS is at least:
Applying 2-variable AM-GM to and the LHS is at least
We then simply choose such that .
Part b) is likely to occur only if both the AM-GM and Cauchy equality conditions are close to met, or if and , respectively. This points to the geometric series with common ratio as the equality case.
This solution was posted and copyrighted by n^4 4. The original thread for this problem can be found here: [1]
Solution 2
b) is immediate by taking .
For a), we prove , and taking will yield the result. (Here gives the empty sum, so , which is trivial).
The proof is not hard. Let be the sum of the first terms, so . This in turn is . By repeatedly applying AM-GM to the last two terms of the sum, collecting powers of two, and cancelling variables, we eventually obtain
, as desired.
This solution was posted and copyrighted by tastymath75025. The original thread for this problem can be found here: [2]
See Also
1982 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |