Difference between revisions of "1997 USAMO Problems/Problem 5"
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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>. Sum similar expressions to obtain the desired result.) | 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>. Sum similar expressions to obtain the desired result.) | ||
+ | |||
+ | ==Solution 3 (Isolated fudging)== | ||
+ | Without loss of generality, let <math>abc = 1</math>. | ||
+ | |||
+ | Lemma: | ||
+ | <cmath>\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.</cmath> | ||
+ | Proof: Rearranging gives <math>(a^3 + b^3) c + c \ge a + b + c</math>, which is a simple consequence of <math>a^3 + b^3 = (a + b)(a^2 - ab + b^2)</math> and | ||
+ | <cmath>(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.</cmath> | ||
+ | |||
+ | Thus, by <math>abc = 1</math>: | ||
+ | <cmath>\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}</cmath> | ||
+ | <cmath>\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.</cmath> | ||
==See Also == | ==See Also == |
Revision as of 23:42, 29 March 2015
Problem
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. (Alternatively, one sees the required AM-GM is
. Sum similar expressions to obtain the desired result.)
Solution 3 (Isolated fudging)
Without loss of generality, let .
Lemma:
Proof: Rearranging gives
, which is a simple consequence of
and
Thus, by :
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.