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Difference between revisions of "2012 JBMO Problems/Problem 1"

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== Section 1 ==
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== Problem ==
 
Let <math>a,b,c</math> be positive real numbers such that <math>a+b+c=1</math>. Prove that
 
Let <math>a,b,c</math> be positive real numbers such that <math>a+b+c=1</math>. Prove that
 
<cmath>\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 ).</cmath>
 
<cmath>\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 ).</cmath>
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The LHS rearranges to <math>\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6</math>. Since <math>b+c=1-a</math> we have that <math>\frac{b+c}{a}=\frac{1-a}{a}</math>. Therefore, the LHS rearranges again to <math>\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6</math>.
 
The LHS rearranges to <math>\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6</math>. Since <math>b+c=1-a</math> we have that <math>\frac{b+c}{a}=\frac{1-a}{a}</math>. Therefore, the LHS rearranges again to <math>\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6</math>.
  
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Now, multiply the <math>\sqrt{2}</math> on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get <cmath>\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}})</cmath> Let <math>\sqrt{\frac{2-2a}{a}} = A</math> and similarly for <math>B</math> and <math>C</math>.
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Now, distribute the <math>\sqrt{2}</math> on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get <cmath>\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}})</cmath> Let <math>\sqrt{\frac{2-2a}{a}} = A</math> and similarly for <math>B</math> and <math>C</math>.
  
 
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.
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Rearranging terms shows that this further simplifies to <cmath>(A-2)^2+(B-2)^2+(C-2)^2\geq0</cmath>
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Rearranging terms shows that this further simplifies to <cmath>(A^2-4A+4)+(B^2-4B+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 :)
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{{JBMO box|year=2012|before=[[2011 JBMO]]|after=[[2013 JBMO]]}}
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[[Category: JBMO]]

Latest revision as of 14:48, 11 January 2021

Problem

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that \[\frac {a}{b} + \frac {a}{c} + \frac {c}{b} + \frac {c}{a} + \frac {b}{c} + \frac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right ).\] When does equality hold?

Solution

The LHS rearranges to $\frac{b+c}{a} + \frac{a+c}{b} + \frac{a+b}{c} + 6$. Since $b+c=1-a$ we have that $\frac{b+c}{a}=\frac{1-a}{a}$. Therefore, the LHS rearranges again to $\frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}+6$.

Now, distribute the $\sqrt{2}$ on the RHS into the parenthesis and multiply the LHS and RHS by 2 to get \[\frac{2-2a}{a} + \frac{2-2b}{b} + \frac{2-2c}{c}+12 \geq 4(\sqrt{\frac{2-2a}{a}} + \sqrt{\frac{2-2b}{b}} + \sqrt{\frac{2-2c}{c}})\] Let $\sqrt{\frac{2-2a}{a}} = A$ and similarly for $B$ and $C$.

The inequality now simplifies to \[A^2+B^2+C^2+12 \geq 4(A+B+C)\] Note that because $a$, $b$, and $c$ are positive real numbers less than $1$, $A$, $B$, and $C$ are always positive real numbers. Rearranging terms shows that this further simplifies to \[(A^2-4A+4)+(B^2-4B+4)+(C^2-4C+4)\geq0\] which equals \[(A-2)^2+(B-2)^2+(C-2)^2\geq0\] By the trivial inequality we know that this is always true. Finally, we have equality when \[A=B=C=2\] and \[\frac{2-2a}{a}=\frac{2-2b}{b}=\frac{2-2c}{c}=4\] Solving the equations yields that equality holds when $\boxed{a=b=c=\frac{1}{3}}$

Solution by Someonenumber011 :)

2012 JBMO (Problems • Resources)
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