2000 JBMO Problems/Problem 1
Contents
Problem
Let and be positive reals such that Show that .
Solution 1
After some manipulation we get:
Case 1: , So has to be , so or
Case 2: , So has to be , so or
Thus
(This is wrong :|)
Solution 2
Rearranging the equation yields If in the large equation, then must be a factor of the large equation. Note that we can rewrite the large equation as We can factor the difference of cubes in the first part and factor in the second part, resulting in Finally, we can factor by grouping, which results in By the Zero Product Property, either or However, since and are both positive, can not equal zero, so we have proved that
See Also
2000 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |