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Aplicatie C.B.S.
Posted: Tue Dec 22, 2009 7:04 pm
by Marius Mainea
Se considera 100 numere reale \( a_1,a_2,...,a_{100} \) astfel incat
\( a_1^2+a_2^2+...+a_{100}^2+(a_1+a_2+...+a_{100})^2=101 \).
Aratati ca \( |a_k|\le 10 \) pentru orice \( k\in\{1,2,...,100\} \).
Posted: Thu Dec 24, 2009 8:01 pm
by Marius Mainea
Indicatie:
Presupunem prin absurd ca unul dintre numere, sa zicem \( a_1 \), are modulul \( |a_1|>10 \).
Posted: Mon Mar 22, 2010 9:18 pm
by Mateescu Constantin
Presupunem prin absurd ca \( |a_1|\ >\ 10 \) , de unde \( a_1^2\ >\ 100 \) si din relatia data obtinem ca :
\( a_2^2+a_3^2+\ldots +a_{100}^2+(a_1+a_2+\ldots +a_{100})^2\ <\ 1\ (\ast) \) . Pe de alta parte, aplicand inegalitatea C.B.S obtinem :
\( a_1^2=[(a_1+a_2+\ldots +a_{100})-a_2-a_3-\ldots -a_{100}]^2\ \le^{\small CBS}\ 100\cdot [(a_1+a_2+\ldots +a_{100})^2+a_2^2+a_3^2+\ldots +a_{100}^2]\ \stackrel{(\ast)}{<}\ 100 \)
Deci \( a_1^2\ <\ 100 \) , in contradictie cu presupunerea facuta . Prin urmare , \( |a_k|\le 10\ ,\ \forall\ k=\overline{1,100} \) .