Prueba $\sin \alpha+\sin \beta+\sin \gamma \geq\sin 2\alpha+\sin 2\beta+\sin 2\gamma $

Question

Demostrar que $\sin \alpha+\sin \beta+\sin \gamma \geq\sin 2\alpha+\sin 2\beta+\sin 2\gamma $ donde $\alpha$ $,\beta$ $,\gamma$ son los ángulos de un triángulo

Answer 1

Utilice $$\sin{2A}+\sin{2B}=2\sin{C}\cos{(A-C)}\le 2\sin{C}$$ $$\sin{2B}+\sin{2C}\le 2\sin{A},$$ $$\sin{2C}+\sin{2A}\le 2\sin{B}$$

Answer 2

$\sin A+\sin B+\sin C=\dfrac{a+b+c}{2R}=\dfrac{S}{rR}$

$\sin 2A+\sin 2B+\sin 2C=2(\sin A\cos A+\sin B\cos B+\sin C\cos C)$
$=\dfrac{a\cos A+b\cos B+c\cos C}{R}=\dfrac{2S}{R^{2}}$

$\sin A+\sin B+\sin C\geq \sin 2A+\sin 2B+\sin 2C$
$\Longleftrightarrow \dfrac{S}{rR}\geq \dfrac{2S}{R^{2}}$ $\Longleftrightarrow$ $R\geq 2r$ : Fórmula de Euler