TITLE: Prove $\alpha+\beta+\gamma \ge \alpha\frac{\sin \beta}{\sin \alpha}+\beta\frac{\sin \gamma}{\sin \beta}+\gamma\frac{\sin \alpha}{\sin \gamma}$
QUESTION [5 upvotes]: Prove the inequality
$$\alpha+\beta+\gamma \ge \alpha\cdot\frac{\sin \beta}{\sin \alpha}+\beta\cdot\frac{\sin \gamma}{\sin \beta}+\gamma\cdot\frac{\sin \alpha}{\sin \gamma}$$
for any three numbers $\alpha, \beta, \gamma \in \left(0;\frac{\pi}2\right)$
My attempt:
I plan to use Rearrangement inequality. But I do not know how to start.

REPLY [3 votes]: Let $f(t) = \sin t - t \cos t, 0 \leq t \leq \pi/2.$ 
We have $f(0) = 0$ and $f'(t) = \cos t - \cos t + t \sin t = t \sin t \geq 0$ if $0 \leq t \leq \pi/2.$
This implies $f(t) \geq 0$ for $0 \leq t \leq \pi/2.$
If $g(t) = \dfrac{t}{\sin t}$  then $g'(t) = \dfrac{f(t)}{\sin^2 t} \geq 0.$ So $\dfrac{t}{\sin t}$ increases on $(0,\dfrac{\pi}{2}).$
Let $ 0 < \alpha \leq \beta \leq \gamma < \pi/2$ then $ 0 < \dfrac{\alpha}{\sin \alpha} \leq \dfrac{\beta}{\sin \beta} \leq \dfrac{\gamma}{\sin \gamma}$ and $0 < \sin \alpha \leq \sin \beta \leq \sin \gamma.$
By the rearrangement inequality $ \dfrac{\alpha}{\sin \alpha} \sin \beta + \dfrac{\beta}{\sin \beta} \sin \gamma + \dfrac{\gamma}{\sin \gamma} \sin \alpha \leq \dfrac{\alpha}{\sin \alpha} \sin \alpha + \dfrac{\beta}{\sin \beta} \sin \beta + \dfrac{\gamma}{\sin \gamma} \sin \gamma = \alpha + \beta + \gamma.$