Difference between revisions of "1997 USAMO Problems/Problem 5"
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<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>. | <math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>. | ||
− | == Solution == | + | == Solution 1 == |
Because the inequality is homogenous (i.e. <math>(a, b, c)</math> can be replaced with <math>(ka, kb, kc)</math> without changing the inequality other than by a factor of <math>k^n</math> for some <math>n</math>), without loss of generality, let <math>abc = 1</math>. | Because the inequality is homogenous (i.e. <math>(a, b, c)</math> can be replaced with <math>(ka, kb, kc)</math> without changing the inequality other than by a factor of <math>k^n</math> for some <math>n</math>), without loss of generality, let <math>abc = 1</math>. | ||
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<cmath>\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}</cmath> | <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> | <cmath>\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.</cmath> | ||
+ | |||
+ | == Solution 2 == | ||
+ | Rearranging the AM-HM inequality, we get <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}</math>. | ||
==See Also == | ==See Also == |
Revision as of 23:02, 21 August 2020
Contents
Problem
Prove that, for all positive real numbers
.
Solution 1
Because the inequality is homogenous (i.e. can be replaced with without changing the inequality other than by a factor of for some ), without loss of generality, let .
Lemma: Proof: Rearranging gives , which is a simple consequence of and
Thus, by :
Solution 2
Rearranging the AM-HM inequality, we get .
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.