Difference between revisions of "1997 USAMO Problems/Problem 5"
(→Problem) |
(→Solution 3) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 18: | Line 18: | ||
== 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> | ||
| − | - | + | == Solution 3 == |
| + | 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> | ||
| + | |||
| + | -KEVIN_LIU | ||
| + | |||
| + | == Video Solution (inspired by Solution 1) == | ||
| + | |||
| + | https://youtu.be/6czJm7FMGtk | ||
==See Also == | ==See Also == | ||
Latest revision as of 22:14, 2 September 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 
Solution 3
If we multiply each side by 
, we get that we must just prove that ![\[\frac{abc}{a^3+b^3+abc} + \frac{abc}{b^3+c^3+abc} + \frac{abc}{c^3+a^3+abc} \leq 1\]](http://latex.artofproblemsolving.com/3/9/4/3941571a34486320f956fcf94c5f776d5ff63a9c.png)
If we divide our LHS equation, we get that ![\[\frac{1}{\frac{a^3+b^3+abc}{abc}} +\frac{1}{\frac{b^3+c^3+abc}{abc}} + \frac{1}{\frac{c^3+a^3+abc}{abc}}\]](http://latex.artofproblemsolving.com/f/d/b/fdb96a5afe575c18d6342deef9d0b591b14cfcae.png)
![\[= \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} +\frac{1}{\frac{b^2}{ac} + \frac{c^2}{ab} + 1} + \frac{1}{\frac{c^2}{ab} + \frac{a^2}{bc} + 1}\]](http://latex.artofproblemsolving.com/0/d/0/0d067e90da78b56d725a528ee7ee8fa175cb675d.png)
Make the astute observation that by Titu's Lemma, ![\[\frac{a^2}{bc} + \frac{b^2}{ac} \geq \frac{\left(a+b\right)^2}{bc+ ac}\]](http://latex.artofproblemsolving.com/e/a/3/ea30a01c594c7b9fd495f51385265db7025e143e.png)
![\[\sum_{cyc}\frac{\left(a+b\right)^2}{bc+ ac} \leq \sum_{cyc}\frac{a^2}{bc} + \frac{b^2}{ac}\]](http://latex.artofproblemsolving.com/7/9/7/797d01ba992e112e82550ac2c26067ce761c8955.png)
Therefore: ![\[\sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} \geq \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1}\]](http://latex.artofproblemsolving.com/4/9/d/49d1ee9de1fe9bf2dead81a77cbfcc8b490550b1.png)
If we expand it out, we get that ![\[\sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} = \frac{a+b+c}{a+b+c} = 1\]](http://latex.artofproblemsolving.com/a/f/d/afd89247d8f1235f2e467656eaeb03557bed6f7d.png)
Since our original equation is less than this, we get that ![\[\sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} \leq 1\]](http://latex.artofproblemsolving.com/e/8/7/e87129d2dd14eba242c0261f06e09c505d1a5c93.png)
![\[\sum_{cyc} \frac{abc}{a^3+b^3+abc}\leq 1\]](http://latex.artofproblemsolving.com/1/0/b/10bc2e55ce8d956cb131aebfea447c2d373770ff.png)
![\[\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \frac{1}{abc}\]](http://latex.artofproblemsolving.com/9/2/d/92d5e335d006689d3e32f07c9702b5791e6d599c.png)
-KEVIN_LIU
Video Solution (inspired by Solution 1)
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. 
