mateforum.ro

Posted: **Sun Feb 08, 2009 7:46 pm**

Se considera un tetraedru \( OABC \) in care \( OA\perp OB\perp OC\perp OA \). Aratati ca \( \frac{1}{OA}+\frac{1}{OB}+\frac{1}{OC}\ge\sqrt{2}\left(\frac{1}{AB}+\frac{1}{BC}+\frac{1}{CA}\right) \).

Cezar Lupu, lista scurta, 2005

Posted: **Sun Feb 08, 2009 9:54 pm**

Notand OA=a , Ob=b , OC=c inegalitatea devine

\( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \sqrt{2}(\frac{1}{\sqrt{a^2+b^2}}+\frac{1}{\sqrt{b^2+c^2}}+\frac{1}{\sqrt{c^2+a^2}}) \)

mateforum.ro

Inegalitate intre muchiile unui tetraedru tridreptunghic

Inegalitate intre muchiile unui tetraedru tridreptunghic