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Inegalitate de la Internet Mathematics Olympiad for Students

 
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Laurian Filip
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Joined: 25 Nov 2007
Posts: 471
Location: Bucuresti

PostPosted: Fri Jun 07, 2013 5:12 pm    Post subject: Inegalitate de la Internet Mathematics Olympiad for Students Reply with quote

Urmatoarea problema a fost una dintre problemele de la editia a 9-a a Internet Mathematics Olympiad for Students.
Dintre cei aproape 300 de participanti niciunul nu are reusit sa ia nici macar un punct pe aceasta problema.



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Theodor Munteanu
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Joined: 06 May 2008
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Location: Sighetu Marmatiei/Bucharest

PostPosted: Tue Jun 25, 2013 9:32 pm    Post subject: derivate partiale+hessiana Reply with quote

fie f(a,b,c)=\sqrt{a^2+b^2+c^2}+2\sqrt{ab+bc+ca}-( \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+ab})si observam ca f(a,b,c)=f(\sigma(a),\sigma(b),\sigma(c))deci putem presupune ca a\leq b\leq c

acum \frac{\part f}{\part x}=\frac{a}{\sqrt{a^2+b^2+c^2}}+\frac{b+c}{\sqrt{ab+bc+ca}}-\frac{a}{\sqrt{a^2+2bc}}-\frac{b}{\sqrt{b^2+ac}}-\frac{c}{\sqrt{c^2+ab}}=a\left(\frac{1}{\sqrt{a^2+b^2+c^2}}-\frac{1}{\sqrt{a^2+2bc}}\right)+\frac{b+c}{\sqrt{ab+bc+ac}}-\frac{b}{\sqrt{b^2+2ac}}-\frac{c}{\sqrt{c^2+2ab}}
Observam ca a\cdot \left(\frac{\sqrt{a^2+2bc}-\sqrt{a^2+b^2+c^2}}{P}\right)\leq 0(acestia sunt primii 2 termeni)
de asemenea \frac{b+c}{\sqrt{ab+bc+ca}}-\frac{b}{\sqrt{b^2+2ac}}-\frac{c}{\sqrt{c^2+2ab}}\leq 0deoarece
\frac{b}{\sqrt{ab+bc+ca}}-\frac{b}{\sqrt{b^2+2ac}}\leq 0si \frac{c}{\sqrt{ab+bc+ca}}-\frac{c}{\sqrt{c^2+2ab}}\leq 0
Demonstrez de exemplu primul:
acesta e echivalent cu ab+bc+ca\geq b^2+2ac \Leftrightarrow b(a+c)\geq b^2+ac\Leftrightarrow b(a+c-b)\geq ac
rescriem inegalitatea ca \frac{a+c-b}{c}\geq \frac{a}{b}sau \frac{a-b}{c}+1\geq \frac{a}{b}echivalent cu \frac{a-b}{c}\geq \frac{a-b}{b}care e evident din cauza ca a-b\leq 0 si c\geq b
analog se face si cealalta inegalitate .
Deci obtinem ca \frac{\part f}{\part x}\leq 0cu egalitate daca si numai daca b=c
Deci punctele stationare ale functiei f sunt de forma (a,a,a).
Mai ramane sa vedem daca ele sunt puncte de minim.
Calculam in acest sens hessiana lui f.
\frac{\part^2 f}{\part x^2}(a,b,c)>0 iar hessianul e pozitiv dupa calcule faraonice deci conform criteriului lui Sylvester (a,a,a) e punct de minim pentru f si cum f(a,a,a)=0rezulta concluzia.
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