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Difference between revisions of "1997 USAMO Problems/Problem 5"

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Prove that, for all positive real numbers <math>a, b, c,</math>
 
Prove that, for all positive real numbers <math>a, b, c,</math>
  
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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>.
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<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>.
  
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Prove that, for all positive real numbers <math>a, b, c,</math>
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== Solution 1 ==
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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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<math>(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}</math>.
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Lemma:
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<cmath>\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.</cmath>
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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
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<cmath>(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.</cmath>
  
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== Solution ==
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Thus, by <math>abc = 1</math>:
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[[File:USAMO97(5-solution).jpg]]
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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>
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<cmath>\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.</cmath>
  
 
== Solution 2 ==
 
== Solution 2 ==
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'''Outline:'''
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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>.
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-Tigerzhang
  
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1. Because the inequality is homogenous, scale <math>a, b, c</math> by an arbitrary factor such that <math>abc = 1</math>.
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<math>\textbf{WARNING:}</math>
  
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2. Replace all <math>abc</math> with 1. Then, multiply both sides by <math>(a^3 + b^3 + 1)(b^3 + c^3 + 1)(a^3 + c^3 + 1)</math> to clear the denominators.
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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>
  
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3. Expand each product of trinomials.
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== Solution 3 ==
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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>
  
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4. Cancel like mad.
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-KEVIN_LIU
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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>.
 
  
 
==See Also ==
 
==See Also ==

Revision as of 12:55, 25 March 2024

Problem

Prove that, for all positive real numbers $a, b, c,$

$\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}$.

Solution 1

Because the inequality is homogenous (i.e. $(a, b, c)$ can be replaced with $(ka, kb, kc)$ without changing the inequality other than by a factor of $k^n$ for some $n$), without loss of generality, let $abc = 1$.

Lemma: \[\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.\] Proof: Rearranging gives $(a^3 + b^3) c + c \ge a + b + c$, which is a simple consequence of $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and \[(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.\]

Thus, by $abc = 1$: \[\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}\] \[\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.\]

Solution 2

Rearranging the AM-HM inequality, we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}$. Letting $x=a^{3}+b^{3}+abc$, $y=b^{3}+c^{3}+abc$, and $z=c^{3}+a^{3}+abc$, we get \[\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc}.\] By AM-GM on $a^{3}$, $b^{3}$, and $c^{3}$, we have \[a^{3}+b^{3}+c^{3} \ge 3abc \Rightarrow 2a^{3}+2b^{3}+2c^{3}+3abc \ge 9abc \Rightarrow \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc} \le \frac{1}{abc}.\] So, $\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{1}{abc}$. -Tigerzhang

$\textbf{WARNING:}$

This solution doesn’t work because $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 9$, so $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

Solution 3

If we multiply each side by $abc$, we get that we must just prove that \[\frac{abc}{a^3+b^3+abc} + \frac{abc}{b^3+c^3+abc} + \frac{abc}{c^3+a^3+abc} \leq 1\] If we divide our LHS equation, we get that \[\frac{1}{\frac{a^3+b^3+abc}{abc}} +\frac{1}{\frac{b^3+c^3+abc}{abc}} + \frac{1}{\frac{c^3+a^3+abc}{abc}}\] \[= \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} +\frac{1}{\frac{b^2}{ac} + \frac{c^2}{ab} + 1} + \frac{1}{\frac{c^2}{ab} + \frac{a^2}{bc} + 1}\] Make the astute observation that by Titu's Lemma, \[\frac{a^2}{bc} + \frac{b^2}{ac} \geq \frac{\left(a+b\right)^2}{bc+ ac}\] \[\sum_{cyc}\frac{\left(a+b\right)^2}{bc+ ac} \leq \sum_{cyc}\frac{a^2}{bc} + \frac{b^2}{ac}\] Therefore: \[\sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} \geq \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1}\] If we expand it out, we get that \[\sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} = \frac{a+b+c}{a+b+c} = 1\] Since our original equation is less than this, we get that \[\sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} \leq 1\] \[\sum_{cyc} \frac{abc}{a^3+b^3+abc}\leq 1\] \[\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \frac{1}{abc}\]

-KEVIN_LIU

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

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