1997 USAMO Problems/Problem 5: Difference between revisions
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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 22:42, 29 March 2015
Problem
Prove that, for all positive real numbers 
.
Solution
.jpg)
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:
![\[\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.\]](http://latex.artofproblemsolving.com/4/0/6/40699bc0defba4331f503edc329983f1d2048e7f.png)
Proof: Rearranging gives 
, which is a simple consequence of 
and
![\[(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.\]](http://latex.artofproblemsolving.com/d/7/3/d73b9c3c44c70799a4efe54c36b8834d174e99eb.png)
Thus, by :
![\[\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}\]](http://latex.artofproblemsolving.com/9/8/e/98ee2820cb96c8bf7f543afcbde3eabaf90751e5.png)
![\[\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.\]](http://latex.artofproblemsolving.com/9/6/5/9658d126aac55adb471e760d4bd7def3ddb71a72.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
