1974 USAMO Problems/Problem 2
Contents
Problem
Prove that if , , and are positive real numbers, then
Solution 1
Consider the function . for ; therefore, it is a convex function and we can apply Jensen's Inequality:
Apply AM-GM to get
which implies
Rearranging,
Because is an increasing function, we can conclude that:
which simplifies to the desired inequality.
Solution 2
Note that .
So if we can prove that and , then we are done.
WLOG let .
Note that . Since , , , and , it follows that .
Note that . Since , , , and , it follows that .
Thus, , and cube-rooting both sides gives as desired.
Solution 3
WLOG let . Let and , where and .
We want to prove that .
Simplifying and combining terms on each side, we get .
Since , we can divide out to get .
Take the th root of each side and then cube both sides to get .
This simplifies to .
Since and , we only need to prove for our given .
WLOG, let and for . Then our expression becomes
This is clearly true for .
Solution 4
WLOG let . Then sequence majorizes . Thus by Muirhead's Inequality, we have , so .
Solution 5
Let and Then and a straightforward calculation reduces the problem to WLOG, assume Then and Therefore,
J.Z.
Solution 6
Cubing both sides of the given inequality gives If we take as the product of 's, 's, and , we get that
by GM-HM, as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1974 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
- Simple Olympiad Inequality
- Hard inequality
- Inequality
- Some q's on usamo write ups
- ineq
- exponents (generalization)
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.