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

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

Revision as of 14:25, 12 April 2023

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

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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