# Difference between revisions of "2006 Romanian NMO Problems/Grade 8/Problem 4"

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''selected by Mircea Lascu'' | ''selected by Mircea Lascu'' | ||

==Solution== | ==Solution== | ||

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+ | It is easy to see that the function <math>f(a,b,c)=\frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1}</math> is convex in each of the three variables (since each term is linear or of the form <math>\frac{p}{x+q}</math> for each variable <math>x</math>). Thus, its value is maximized at the endpoints. Checking the values of <math>f</math> for all possible values of <math>a,b,c</math> such that <math>a,b,c\in \{\frac{1}{2},1\}</math> yields a maximum of <math>3</math> as desired. | ||

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+ | As for the minimum, we have | ||

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+ | <math>2\le \frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1}</math> | ||

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+ | <math>\Leftrightarrow 5\le \frac{a+b+c+1}{c+1}+\frac{a+b+c+1}{a+1}+\frac{a+b+c+1}{b+1}</math> | ||

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+ | Applying AM-HM to the right hand side yields | ||

+ | |||

+ | <math>9\left(\frac{a+b+c+3}{a+b+c+1}\right)^{-1}\le\frac{a+b+c+1}{c+1}+\frac{a+b+c+1}{a+1}+\frac{a+b+c+1}{b+1}</math> | ||

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+ | <math>\Rightarrow 9\left(1-\frac{2}{a+b+c+3}\right)\le\frac{a+b+c+1}{c+1}+\frac{a+b+c+1}{a+1}+\frac{a+b+c+1}{b+1}</math> | ||

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+ | Obviously, <math>\frac{2}{a+b+c+3}</math> is maximized when <math>a,b,c</math> are minimized. That is, when <math>a=b=c=\frac{1}{2}</math>. Thus, we have that | ||

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+ | <math>5\le 9\left(1-\frac{2}{a+b+c+3}\right)</math> | ||

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+ | as desired. | ||

+ | |||

==See also== | ==See also== | ||

*[[2006 Romanian NMO Problems/Grade 8/Problem 3 | Previous problem]] | *[[2006 Romanian NMO Problems/Grade 8/Problem 3 | Previous problem]] | ||

*[[2006 Romanian NMO Problems]] | *[[2006 Romanian NMO Problems]] | ||

[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |

## Latest revision as of 18:49, 26 August 2008

## Problem

Let . Prove that

*selected by Mircea Lascu*

## Solution

It is easy to see that the function is convex in each of the three variables (since each term is linear or of the form for each variable ). Thus, its value is maximized at the endpoints. Checking the values of for all possible values of such that yields a maximum of as desired.

As for the minimum, we have

Applying AM-HM to the right hand side yields

Obviously, is maximized when are minimized. That is, when . Thus, we have that

as desired.