<s>[INST] Can we conclude $0^\circMy attempt

Based on the sine rule and the graph of $\sin A = k a$ (where $k$ is a constant) in interval $(0,\pi)$, 
increasing $a$ up to $1/k$ will either


*

*increase $A$ up to $90^\circ$.

*decrease $A$ up to $90^\circ$.


So I cannot conclude that increasing $a$ will increase $A$.
Now I use the cosine rule (it is promising because the cosine is decreasing in the given interval).
\begin{align}
A &= \cos^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)\\
B &= \cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)\\
C &= \cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)\\
\end{align}
It is hard to show that  $0^\circ<A\leq B\leq C<180^\circ$ for any $\triangle ABC$ with $0<a\leq b\leq c$. Could you show it?
It means that I need to show that 
$$
-1<\frac{a^2+b^2-c^2}{2ab}\leq \frac{a^2+c^2-b^2}{2ac} \leq \frac{b^2+c^2-a^2}{2bc}<1
$$
for $0<a\leq b\leq c$.
 [/INST] I would just split the problem in two cases, depending if $c^2$ is greater or smaller than $a^2+b^2$. If $c^2\ge a^2+b^2$, then $C\ge 90^\circ$, so then $A$ and $B$ are less then $90$, therefore less then $C$. After that, apply the law of sines.
</s>