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Inegalitate geometrica 4

Posted: Sat Apr 18, 2009 7:36 pm
by Marius Mainea
Daca a,b,c sunt lungimile laturilor unui triunghi , atunci

\( \sqrt{2(b^2+c^2)-a^2}<\sqrt{2(c^2+a^2)-b^2}+\sqrt{2(a^2+b^2)-c^2} \)

Concursul Gh. Titeica

Posted: Mon Apr 20, 2009 11:06 am
by mihai miculita
INDICATIE: \( \mbox{Intrucat: } m_a^2=\frac{2.(b^2+c^2)-a^2}{4};\dots, \mbox{ avem: }
\sqrt{2(b^2+c^2)-a^2}<\sqrt{2(a^2+c^2)-b^2}+\sqrt{2(a^2+b^2)-c^2}\Leftrightarrow m_a<m_b+m_c. \)

Cu alte cuvinte, inegalitatea fiind simetrica in a, b si c problema revine, la a arata ca: "Cu lungimile medianelor unui triunghi putem construi un alt triunghi."