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Subiectul IV Gheorghe Lazar 2009 Sibiu
Posted: Sun Mar 22, 2009 9:01 pm
by DrAGos Calinescu
Daca \( ABC \) este un triunghi de laturi \( a,b,c \), iar \( 0<\alpha\le\beta<\frac{\pi}{2} \), atunci
\( b^2+c^2\tan ^2\beta >a^2\sin ^2\alpha \)
Emil C. Popa, Sibiu
Re: Subiectul IV Gheorghe Lazar 2009 Sibiu
Posted: Mon Mar 23, 2009 3:11 am
by Virgil Nicula
DrAGos Calinescu wrote:Daca \( ABC \) este un triunghi si \( 0<\alpha\le\beta<\frac{\pi}{2} \), atunci \( b^2+c^2\tan ^2\beta >a^2\sin ^2\alpha \) .
Dem. Notam
\( \tan\alpha =m \) ,
\( \tan\beta =n \) , unde
\( 0< m\le n \) . Intrucat
\( \sin^2\alpha=\frac {\tan^2\alpha}{1+\tan^2\alpha}=\frac {m^2}{1+m^2} \) ,
inegalitatea propusa devine
\( \left(1+m^2\right)\left(b^2+n^2c^2\right)>a^2m^2 \) . Se observa ca
\( b+c\ >\ a \) si
\( \left(m^2+1\right)\left(b^2+n^2c^2\right)\stackrel{(C.B.S.)}{\ \ \ge\ \ }(mb+nc)^2\ge (mb+mc)^2=m^2(b+c)^2>m^2a^2\ . \)