Difference between revisions of "1997 USAMO Problems/Problem 5"
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== Solution 2 == | == 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>. | 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> | ||
==See Also == | ==See Also == |
Revision as of 11:09, 14 February 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
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.