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Posted: **Sat May 09, 2009 10:31 am**

Sa se arate ca in orice triunghi \( ABC \) avem inegalitatea:

\( \left\(\frac{1}{b}+\frac{1}{c}\right\)\tan\frac{A}{2}+\left\(\frac{1}{c}+\frac{1}{a}\right\)\tan\frac{B}{2}+\left\(\frac{1}{a}+\frac{1}{b}\right\)\tan\frac{C}{2}\geq \frac{2}{R}. \)

Gh. Szollosy, GM 4/2009

Posted: **Tue May 12, 2009 10:50 pm**

\( LHS=\sum {\frac{1}{2R}\frac{\sin B+\sin C}{\sin B\sin C}\tan \frac{A}{2}}=\frac{1}{2R}\sum {\frac{\cos B+\cos C}{\sin B\sin C}}=\frac{1}{2R}\frac{\sum \sin A}{\prod \sin A}=\frac{1}{2R}\frac{1}{2\prod \sin \frac{A}{2}}\ge \frac{1}{2R}\frac{1}{2\cdot\frac{1}{8}}=RHS \)

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