If triangle's angles are $\alpha \leq \beta \leq \gamma$, then respective opposite sides are $a \leq b \leq c$. How to handle obtuse case?

Question

Knowing that $a$, $b$ and $c$ are the lengths of the sides opposite to (respective) angles $\alpha \le \beta \le \gamma$, prove that $a\le b \le c$.

I use the Law of Sines: $$a = 2R\sin(\alpha) $$ $$b = 2R\sin(\beta) $$ $$c= 2R\sin(\gamma)$$
Now, I subsitute these values into the initial inequality and divide by $2R$: $$\sin(\alpha) \le \sin(\beta) \le \sin(\gamma) $$

As for an acute triangle, it works great, but how do I prove that this inequality holds for — for example — an obtuse triangle?

Answer 1

My original answer was inadequate. Try this:

We have the angles $\alpha<\beta<\gamma$. If all are acute, your analysis applies, because the sine is increasing in $[0,\pi/2]$.

If not all are acute, then only $\gamma$ is obtuse, and since $\gamma+\beta<\pi$, we also have $\beta<\pi-\gamma$, both acute. Then, by your argument, we have $\sin\alpha<\sin\beta<\sin(\pi-\gamma)$. Since $\sin\gamma=\sin(\pi-\gamma)$, the result $a<b<c$ follows in this case as well.