1997 USAMO Problems/Problem 5
Problem
Prove that, for all positive real numbers
.
Solution
Solution 2
Outline:
1. Because the inequality is homogenous, scale by an arbitrary factor such that .
2. Replace all with 1. Then, multiply both sides by to clear the denominators.
3. Expand each product of trinomials.
4. Cancel like mad.
5. You are left with . Homogenize the inequality by multiplying each term of the LHS by . Because majorizes , this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is . Sum similar expressions to obtain the desired result.)
Solution 3 (Isolated fudging)
Because the inequality is homogenous (i.e. can be replaced with without changing the inequality other than by a factor of for some ), without loss of generality, let .
Lemma: Proof: Rearranging gives , which is a simple consequence of and
Thus, by :
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |