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Inegalitate conditionata cu suma de patrate (1)
Posted: Tue Dec 22, 2009 6:22 pm
by Marius Mainea
Daca \( a_1,a_2,a_3,a_4,a_5 \) sunt cinci numere reale astfel incat \( a_1^2+a_2^2+a_3^2+a_4^2+a_5^2=1 \) sa se arate ca
\( \begin{array}{cc}\mbox{min}\\1\le i<j\le 5\end{array} (a_i-a_j)^2\le\frac{1}{10} \)
Posted: Wed Dec 30, 2009 4:30 pm
by Mateescu Constantin
Sa presupunem ca \( (a_i-a_j)^2>\frac1{10}\ ,\ \forall\ i\ne j \) . De asemenea presupunem, fara a restrange generalitatea ca
\( a_1<a_2<a_3<a_4<a_5 \) si astfel relatia devine: \( a_{i+1}-a_i>\frac{1}{\sqrt{10}}\ ,\ \forall\ i\in\{1,2,3,4\} \) si deci \( a_j-a_i>\frac{j-i}{\sqrt{10}}\ ,\ 1\le i<j\le 5 \)
Ridicand la patrat si sumand dupa toti indicii \( i \) si \( j \) obtinem :
\( 4\sum a_i^2 - 2\sum a_ia_j>\frac{(5-4)^2+(5-3)^2+(5-2)^2+(5-1)^2+(4-3)^2+(4-2)^2+(4-1)^2+(3-2)^2+(3-1)^2+(2-1)^2}{10}=5 \)
Deoarece \( \sum_{i=1}^5 a_i^2=1 \) ultima relatie devine : \( 4-2\sum a_ia_j>5\ \Longleftrightarrow\ 2\sum a_ia_j+1<0 \)
\( \Longleftrightarrow\ 2\sum a_ia_j+\sum a_i^2<0\ \Longleftrightarrow\ \(\sum a_i^2\)<0 \) , fals. Deci presupunerea facuta este falsa si astfel rezulta concluzia .