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.
Solution 7
Replacing with , with , and with , for some positive real we get: This means that this inequality is homogeneous since both sides have the same power of as a factor. Since the inequality is homogeneous, we can scale so that their product is , i.e. . This makes the inequality turn into something much more nicer to deal with. Now we have to prove: given that . Note that even if . Therefore , , and . Multiplying these together we get: . This proves the desired result. Equality holds when .
Solution 8 (Rearrangement)
Let , and , and WLOG . Then we have both and . By the rearrangement inequality, and . Summing, Adding ,we get . Now we substitute back for to get: . Raising to the power of each side, we get
Solution 9
Assume without loss of generality that . Then, we have It follows that We can "distribute" the exponents, which gives Notice that in the in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives We can move around terms on the left to get Now we combine the exponents in the numerators. This gives We multiply both sides by which gives We then multiply both sides by Multiplying both sides by then simplifying both sides with exponent laws, gives Finally, we multiply both sides by Combining exponents one last time gives the desired --MenuThreeOne
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.