Difference between revisions of "2006 Romanian NMO Problems/Grade 8/Problem 4"
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==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 | ||
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| + | <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. | ||
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==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 19:49, 26 August 2008
Problem
Let ![$a,b,c \in \left[ \frac 12, 1 \right]$](http://latex.artofproblemsolving.com/6/1/5/615213db0ab5e339395bf928b755a587570364f8.png)
. 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.
