2011 IMO Problems/Problem 3
Let be a real-valued function defined on the set of real numbers that satisfies for all real numbers and . Prove that for all .
Solution
Let be the given assertion. Comparing and yields,
Suppose then Now implies that
Then yields a contradiction.
From we get thus we get as desired.
~ZETA_in_olympiad
See Also
2011 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |