In orice tetraedru ABCD are loc inegalitatea :
\( \frac{1}{aria (ABC)}+\frac{1}{aria (BCD)}+\frac{1}{aria (CDA)}+\frac{1}{aria (DAB)} \geq \frac{16r}{3vol (ABCD)} \) unde r este raza sferei inscrise in tetraedru
Mihaly Bencze, Brasov , RMT nr 1 /2008
inegalitate in tetraedru
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Bogdan Posa
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inegalitate in tetraedru
Gradul de cultură al unei ţări se măsoară astăzi, prin nivelul matematic al locuitorilor ţării (André Lichnerowicz)
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Beniamin Bogosel
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Folosim inegalitatea dintre media armonica si aritmetica:
\( \frac{4}{\frac{1}{\sigma(ABC)}+\frac{1}{\sigma(ABD)}+\frac{1}{\sigma(ACD)}+\frac{1}{\sigma(BCD)}}\leq
\frac{\sigma(ABC)+\sigma(ABD)+\sigma(ACD)+\sigma(BCD)}{4}=\frac{A_l}{4}=\frac{\mathcal 3V_{ABCD}}{4r} \).
Inversam, inmultim cu 4 si obtinem ceea ce trebuia demonstrat.
\( \frac{4}{\frac{1}{\sigma(ABC)}+\frac{1}{\sigma(ABD)}+\frac{1}{\sigma(ACD)}+\frac{1}{\sigma(BCD)}}\leq
\frac{\sigma(ABC)+\sigma(ABD)+\sigma(ACD)+\sigma(BCD)}{4}=\frac{A_l}{4}=\frac{\mathcal 3V_{ABCD}}{4r} \).
Inversam, inmultim cu 4 si obtinem ceea ce trebuia demonstrat.