Difference between revisions of "2001 IMO Shortlist Problems/A6"

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Prove that for all positive real numbers <math>a,b,c</math>,
 
Prove that for all positive real numbers <math>a,b,c</math>,
 
<center><math>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.</math></center>
 
<center><math>\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.</math></center>
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=== Generalization ===
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The leader of the Bulgarian team had come up with the following generalization to the inequality:
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<center><math>\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.</math></center>
  
 
== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
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== Resources ==
 
== Resources ==
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[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
The leader of the Bulgarian team had come up with a generalization of this with putting k in place of 8 and replacing 1 in the RHS by 3/sq root(1+k)
 

Revision as of 12:50, 13 September 2012

Problem

Prove that for all positive real numbers $a,b,c$,

$\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$

Generalization

The leader of the Bulgarian team had come up with the following generalization to the inequality:

$\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.$

Solution

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Resources