Prove Acute Triangle Inequality: $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$

[anemone]

Let $\alpha,\,\beta$ and $\gamma$ be the interior angles of an acute triangle.

Prove that if $\alpha \lt \beta \lt \gamma$, then $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

 

[kaliprasad]

Let $\alpha,\,\beta$ and $\gamma$ be the interior angles of an acute triangle.

Prove that if $\alpha \lt \beta \lt \gamma$, then $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

There are 2 cases

Spoiler

no angle is less than $\frac{\pi}{4}$ and from given condition no angle is greater than or equal to $\frac{\pi}{2}$
so we have

$\frac{\pi}{4} \le \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$

hence $\frac{\pi}{2} \le 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$

as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

Spoiler

case 2 :
one angle $\alpha \lt \frac{\pi}{4}$ and hence $\frac{\pi}{2} - \alpha <\beta$ otherwise $\gamma \ge \frac{\pi}{2}$
$\frac{\pi}{4} \le \frac{\pi}{2}- \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$
hence $\frac{\pi}{2} \le \pi- 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$
as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin (\pi-2\alpha) \gt \sin 2\beta \gt \sin 2\gamma$.
or $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

 

[anemone]

There are 2 cases

Spoiler

no angle is less than $\frac{\pi}{4}$ and from given condition no angle is greater than or equal to $\frac{\pi}{2}$
so we have

$\frac{\pi}{4} \le \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$

hence $\frac{\pi}{2} \le 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$

as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

Spoiler

case 2 :
one angle $\alpha \lt \frac{\pi}{4}$ and hence $\frac{\pi}{2} - \alpha <\beta$ otherwise $\gamma \ge \frac{\pi}{2}$
$\frac{\pi}{4} \le \frac{\pi}{2}- \alpha \lt \beta \lt \gamma \lt \frac{\pi}{2}$
hence $\frac{\pi}{2} \le \pi- 2\alpha \lt 2\beta \lt 2\gamma \lt \pi$
as $\sin $ is monotonically decreasing from $\frac{\pi}{2}$ to $\pi$ hence $\sin (\pi-2\alpha) \gt \sin 2\beta \gt \sin 2\gamma$.
or $\sin 2\alpha \gt \sin 2\beta \gt \sin 2\gamma$.

Very well done kaliprasad!(Cool)