Difference between revisions of "1997 USAMO Problems/Problem 5"
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+ | == Problem == | ||
+ | Prove that, for all positive real numbers <math>a, b, c,</math> | ||
+ | <math>\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}</math>. | ||
+ | |||
+ | == 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | == 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>. Letting <math>x=a^{3}+b^{3}+abc</math>, <math>y=b^{3}+c^{3}+abc</math>, and <math>z=c^{3}+a^{3}+abc</math>, we get <cmath>\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc}.</cmath> By AM-GM on <math>a^{3}</math>, <math>b^{3}</math>, and <math>c^{3}</math>, we have <cmath>a^{3}+b^{3}+c^{3} \ge 3abc \Rightarrow 2a^{3}+2b^{3}+2c^{3}+3abc \ge 9abc \Rightarrow \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc} \le \frac{1}{abc}.</cmath> So, <math>\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{1}{abc}</math>. | ||
+ | -Tigerzhang | ||
+ | |||
+ | <math>\textbf{WARNING:}</math> | ||
+ | |||
+ | This solution doesn’t work because <math>(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 9</math>, so <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | If we multiply each side by <math>abc</math>, we get that we must just prove that <cmath> \frac{abc}{a^3+b^3+abc} + \frac{abc}{b^3+c^3+abc} + \frac{abc}{c^3+a^3+abc} \leq 1</cmath> If we divide our LHS equation, we get that <cmath>\frac{1}{\frac{a^3+b^3+abc}{abc}} +\frac{1}{\frac{b^3+c^3+abc}{abc}} + \frac{1}{\frac{c^3+a^3+abc}{abc}}</cmath> <cmath> = \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} +\frac{1}{\frac{b^2}{ac} + \frac{c^2}{ab} + 1} + \frac{1}{\frac{c^2}{ab} + \frac{a^2}{bc} + 1} </cmath> Make the astute observation that by Titu's Lemma, <cmath> \frac{a^2}{bc} + \frac{b^2}{ac} \geq \frac{\left(a+b\right)^2}{bc+ ac}</cmath> <cmath> \sum_{cyc}\frac{\left(a+b\right)^2}{bc+ ac} \leq \sum_{cyc}\frac{a^2}{bc} + \frac{b^2}{ac} </cmath> Therefore: <cmath> \sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} \geq \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1}</cmath> If we expand it out, we get that <cmath> \sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} = \frac{a+b+c}{a+b+c} = 1</cmath> Since our original equation is less than this, we get that <cmath> \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} \leq 1</cmath> <cmath> \sum_{cyc} \frac{abc}{a^3+b^3+abc}\leq 1</cmath> <cmath> \sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \frac{1}{abc}</cmath> | ||
+ | |||
+ | -KEVIN_LIU | ||
+ | |||
+ | == Video Solution (inspired by Solution 1) == | ||
+ | |||
+ | https://youtu.be/6czJm7FMGtk | ||
+ | |||
+ | ==See Also == | ||
+ | {{USAMO newbox|year=1997|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Olympiad Inequality Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:14, 2 September 2024
Contents
[hide]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 . Letting , , and , we get By AM-GM on , , and , we have So, . -Tigerzhang
This solution doesn’t work because , so
Solution 3
If we multiply each side by , we get that we must just prove that If we divide our LHS equation, we get that Make the astute observation that by Titu's Lemma, Therefore: If we expand it out, we get that Since our original equation is less than this, we get that
-KEVIN_LIU
Video Solution (inspired by Solution 1)
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.