Difference between revisions of "1974 USAMO Problems/Problem 2"
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==Solution 4== | ==Solution 4== | ||
WLOG let <math>a\ge b\ge c</math>. Then sequence <math>(a,b,c)</math> majorizes <math>(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})</math>. Thus by Muirhead's Inequality, we have <math>\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}</math>, so <math>a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}</math>. | WLOG let <math>a\ge b\ge c</math>. Then sequence <math>(a,b,c)</math> majorizes <math>(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})</math>. Thus by Muirhead's Inequality, we have <math>\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}</math>, so <math>a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>x=\frac{a}{\sqrt[3]{abc}},</math> <math>y=\frac{b}{\sqrt[3]{abc}}</math> and <math>z=\frac{c}{\sqrt[3]{abc}}.</math> Then <math>xyz=1</math> and a straightforward calculation reduces the problem to | ||
+ | <cmath>x^xy^yz^z \ge 1.</cmath> | ||
+ | WLOG, assume <math>x\ge y\ge z.</math> Then <math>x\ge 1,</math> <math>z\le 1</math> and <math>xy=\frac{1}{z} \ge 1.</math> Therefore, | ||
+ | <cmath> x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.</cmath> | ||
+ | |||
+ | J.Z. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 13:57, 16 May 2018
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.
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.