2001 USAMO Problems/Problem 3
Problem
Let and satisfy

Show that

Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Without loss of generality, we assume . From the given equation, we can express
in terms of
and
,

Thus,

From Cauchy,

This completes the proof.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |