Inegalitate geometrica in sin si cos de unghiuri pe jumatate

[Cezar Lupu]

Sa se demonstreze ca in orice triunghi \( ABC \) are loc inegalitatea:

\( \sum_{cyc}\frac{1}{\sin^2\frac{A}{2}\cos\frac{A}{2}}\geq 4\sqrt{3}\cdot\frac{R}{r} \).

Cezar Lupu

[algeomath]

Hi Cezar my friend, it is along time I haven't met you. Here is my solution:

Let \( I \) be incenter
use \( 2\sin\frac{A}{2}\cos\frac{A}{2}=\sin A=\frac{a}{2R},\sin\frac{A}{2}=\frac{r}{IA} \), it is equivalent to

\( \sum\frac{4RIA}{ra}\ge 4\sqrt{3}\frac{R}{r}\Leftrightarrow \sum\frac{IA}{a}\ge\sqrt{3}
\)

Which is true by \( \sum \frac{PA}{a}\ge\sqrt{3} \) forall \( P \) in the plane.