Difference between revisions of "2012 JBMO Problems/Problem 1"
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The inequality now simplifies to <cmath>A^2+B^2+C^2+12 \geq 4(A+B+C)</cmath> | The inequality now simplifies to <cmath>A^2+B^2+C^2+12 \geq 4(A+B+C)</cmath> | ||
Note that because <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers less than <math>1</math>, <math>A</math>, <math>B</math>, and <math>C</math> are always positive real numbers. | Note that because <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers less than <math>1</math>, <math>A</math>, <math>B</math>, and <math>C</math> are always positive real numbers. | ||
| − | Rearranging terms shows that this further simplifies to <cmath>(A-2)^2+(B-2)^2+(C-2)^2\geq0</cmath> | + | Rearranging terms shows that this further simplifies to <cmath>(A^2-4A+4)+(B^2-4A+4)+(C^2-4C+4)\geq0</cmath> which equals <cmath>(A-2)^2+(B-2)^2+(C-2)^2\geq0</cmath> |
By the [[trivial inequality]] we know that this is always true. Finally, we have equality when <cmath>A=B=C=2</cmath> and <cmath>\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4</cmath> | By the [[trivial inequality]] we know that this is always true. Finally, we have equality when <cmath>A=B=C=2</cmath> and <cmath>\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4</cmath> | ||
Solving the equations yields that equality holds when <math>\boxed{a=b=c=\frac{1}{3}}</math> | Solving the equations yields that equality holds when <math>\boxed{a=b=c=\frac{1}{3}}</math> | ||
Solution by Someonenumber011 :) | Solution by Someonenumber011 :) | ||
Revision as of 19:27, 22 December 2020
Section 1
Let 
be positive real numbers such that 
. Prove that
![\[\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right ).\]](http://latex.artofproblemsolving.com/6/0/d/60d96354dbf897c9576bd8f15779ad2e96d9fc8e.png)
When does equality hold?
Solution
The LHS rearranges to 
. Since 
we have that 
. Therefore, the LHS rearranges again to 
.
Now, distribute the 
on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get ![\[\frac{2-2a}{a} + \frac{2-2b}{b} + \frac{2-2c}{c}+12 \geq 4(\sqrt{\frac{2-2a}{a}} + \sqrt{\frac{2-2b}{b}} + \sqrt{\frac{2-2c}{c}})\]](http://latex.artofproblemsolving.com/e/e/2/ee2a6ededa03919190cd3b0604288c5bc4e27ed2.png)
Let 
and similarly for 
and 
.
The inequality now simplifies to ![\[A^2+B^2+C^2+12 \geq 4(A+B+C)\]](http://latex.artofproblemsolving.com/9/a/e/9aed2014a094c0456ba84fe578db14a99e4d92c4.png)
Note that because 
, 
, and 
are positive real numbers less than 
, 
, , and are always positive real numbers.
Rearranging terms shows that this further simplifies to ![\[(A^2-4A+4)+(B^2-4A+4)+(C^2-4C+4)\geq0\]](http://latex.artofproblemsolving.com/a/c/b/acb82dd66c0132cc18a58cf3a8f3afafa736fe50.png)
which equals ![\[(A-2)^2+(B-2)^2+(C-2)^2\geq0\]](http://latex.artofproblemsolving.com/a/f/c/afc5e9035df13f59ab899bc70434d53ff0441d14.png)
By the trivial inequality we know that this is always true. Finally, we have equality when ![\[A=B=C=2\]](http://latex.artofproblemsolving.com/3/6/b/36bb115f96b93492053450579ffbbd16f1df7315.png)
and ![\[\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4\]](http://latex.artofproblemsolving.com/0/2/a/02ac30320261f93daa3b38aeecff5d7f06d72d8c.png)
Solving the equations yields that equality holds when 
Solution by Someonenumber011 :)
