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 ==
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== 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>
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== Solution 2 ==
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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>.
  
 
==See Also ==
 
==See Also ==

Revision as of 23:02, 21 August 2020

Problem

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

$(a^3+b^3+abc)^{-1}+(b^3+c^3+abc)^{-1}+(a^3+c^3+abc)^{-1}\le(abc)^{-1}$.

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 $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}$.

See Also

1997 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions

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