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>a^2b^2c^2</math>. Because <math>(6, 3, 0)</math> ''majorizes'' <math>(5, 2, 2)</math>, this inequality holds true by bunching. (Alternatively, one sees the required AM-GM is <math>\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2</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. (Alternatively, one sees the required AM-GM is <math>\frac{a^6b^3 + a^6b^3 + a^3c^6}{3} \ge a^5b^2c^2</math>.) |
==See Also == | ==See Also == | ||
Revision as of 22:05, 22 April 2014
Contents
Problem
Prove that, for all positive real numbers 
.
Prove that, for all positive real numbers
.
Solution
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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. (Alternatively, one sees the required AM-GM is 
.)
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. 
