Difference between revisions of "2012 JBMO Problems/Problem 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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== Solution == | == Solution == | ||
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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>. | ||
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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>. | ||
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+ | 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. | ||
+ | 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> | ||
+ | Solving the equations yields that equality holds when <math>\boxed{a=b=c=\frac{1}{3}}</math> | ||
+ | |||
+ | Solution by Someonenumber011 :) | ||
+ | |||
+ | {{JBMO box|year=2012|before=[[2011 JBMO]]|after=[[2013 JBMO]]}} | ||
+ | [[Category: JBMO]] |
Latest revision as of 15:48, 11 January 2021
Problem
Let be positive real numbers such that
. Prove that
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
Let
and similarly for
and
.
The inequality now simplifies to
Note that because
,
, and
are positive real numbers less than
,
,
, and
are always positive real numbers.
Rearranging terms shows that this further simplifies to
which equals
By the trivial inequality we know that this is always true. Finally, we have equality when
and
Solving the equations yields that equality holds when
Solution by Someonenumber011 :)
2012 JBMO (Problems • Resources) | ||
Preceded by 2011 JBMO |
Followed by 2013 JBMO | |
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