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
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:
![\[\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.\]](http://latex.artofproblemsolving.com/4/0/6/40699bc0defba4331f503edc329983f1d2048e7f.png)
Proof: Rearranging gives 
, which is a simple consequence of 
and
![\[(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.\]](http://latex.artofproblemsolving.com/d/7/3/d73b9c3c44c70799a4efe54c36b8834d174e99eb.png)
Thus, by :
![\[\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}\]](http://latex.artofproblemsolving.com/9/8/e/98ee2820cb96c8bf7f543afcbde3eabaf90751e5.png)
![\[\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.\]](http://latex.artofproblemsolving.com/9/6/5/9658d126aac55adb471e760d4bd7def3ddb71a72.png)
Solution 2
Rearranging the AM-HM inequality, we get 
. Letting 
, 
, and 
, we get ![\[\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc}.\]](http://latex.artofproblemsolving.com/5/8/0/5802891d521ebfc5a35b2721e0ce5d6d205621e2.png)
By AM-GM on 
, 
, and 
, we have ![\[a^{3}+b^{3}+c^{3} \ge 3abc \Rightarrow 2a^{3}+2b^{3}+2c^{3}+3abc \ge 9abc \Rightarrow \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc} \le \frac{1}{abc}.\]](http://latex.artofproblemsolving.com/7/e/4/7e4ccdc00fb116bbd35579093d54b1b16a6c554f.png)
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. 
