Difference between revisions of "1997 USAMO Problems/Problem 5"
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4. Cancel like mad. | 4. Cancel like mad. | ||
− | 5. You are left with <math>a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3</math>. Homogenize the inequality by multiplying each term of the LHS by <math> | + | 5. You are left with <math>a^3 + a^3 + b^3 + b^3 + c^3 + c^3 \le a^6b^3 + a^6c^3 + b^6c^3 + b^6a^3 + c^6a^3 + c^6b^3</math>. Homogenize the inequality by multiplying each term of the LHS by <math>a^2b^2c^2</math>. Because <math>(6, 3, 0)</math> ''majorizes'' <math>(5, 2, 2)</math>, this inequality holds true by bunching. |
==See Also == | ==See Also == |
Revision as of 22:59, 22 April 2014
Contents
Problem
Prove that, for all positive real numbers
.
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.