O inegalitate, cred eu, tare intr-un triunghi oarecare.
Posted: Sun Aug 10, 2008 1:42 am
\( \triangle ABC\ \Longrightarrow\ \sum\ \frac {(a+b)^2}{a+b-c}\ \le\ \frac {2R}{r}\cdot (a+b+c) \) (notatii standard).
Folosind formulele trigonometrice avem
\( \sum {\frac{(a+b)^2}{a+b-c}}=\sum {\frac{16R^2\cos^2\frac{C}{2}\cos^2\frac{A-B}{2}}{8R\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}}}\le\sum {2R\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}}=2R\frac{\sum {\sin\frac{A}{2}\cos\frac{A}{2}}}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=R\frac{\sin A+\sin B+\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=\frac{2R}{r}\cdot(a+b+c) \)
deoarece \( \cos^2\frac{A-B}{2}\le1 \) si analoagele.