Difference between revisions of "1974 USAMO Problems/Problem 2"
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J.Z. | J.Z. | ||
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+ | ==Solution 6== | ||
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+ | Cubing both sides of the given inequality gives <cmath>a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}</cmath> | ||
+ | If we take <math>a^{3a}b^{3b}c^{3c}</math> as the product of <math>3a</math> <math>a</math>'s, <math>3b</math> <math>b</math>'s, and <math>3c</math> <math>c's</math>, we get that | ||
+ | |||
+ | <center> | ||
+ | <math>\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}</math> | ||
+ | </center> | ||
+ | |||
+ | by GM-HM, as desired. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 19:04, 14 March 2019
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.