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Inegalitate conditionata cu interpretare trigonometrica
Posted: Fri Mar 13, 2009 9:22 am
by zeta
Fie \( x,y,z\in (0,1) \) astfel incat \( \sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1 \). Aratati ca avem \( 8xyz\geq (1-x)(1-y)(1-z) \).
Observatie: incercati o abordare trigonometrica.
Posted: Fri Mar 13, 2009 12:29 pm
by Marius Mainea
Notand \( x=\tan^2\frac{A}{2} \) si analoagele inegalitea se reduce la
\( (1-\cos A)(1-\cos B)(1-\cos C)\ge \cos A\cos B\cos C \)
care este cunoscuta si se deduce de exemplu din identitatea
\( IH^2=2r^2-4R^2\cos A\cos B\cos C \)
Posted: Fri Mar 13, 2009 5:01 pm
by zeta
Avem \( \frac{a^2}{tgA}+\frac{b^2}{tgB}+\frac{c^2}{tgC}=4S\geq\frac{4p^2}{tgAtgBtgC}. \) Dar \( S=p^2tg\frac{A}{2}tg\frac{B}{2}tg\frac{C}{2} \)si atunci tinand cont ca \( tgA=\frac{2tg\frac{A}{2}}{1-tg^2{\frac{A}{2}} \) si ca \( \sum{tg\frac{A}{2}tg\frac{B}{2}}=1 \), notand \( x=tg^2{\frac{A}{2}} \) si analoagele, obtinem inegalitatea din enunt.