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 $0<a\leq b\leq c$.

Consider two cases:
Case 1 -- $C \leq 90^\circ$:
Then also $A$ and $B$ are less than $90^\circ$ so we are in the monotonic increasing range of sine, and 
$$
A \leq B \leq C \implies \sin A \leq \sin B \leq \sin C \implies a \leq b \leq c
$$
where the second implication uses the law of sines.
Case 2 -- $C > 90^\circ$:
First  we need to show is that $a \leq b$.
It is trivial that $A, B < 90^\circ$ because the sum of the angles is only $180^\circ$. So for just $A$ and $B$ we are again in the monotonic region of sine, thus
$$
A \leq B  \implies \sin A \leq \sin B  \implies a \leq b 
$$
Now we need to show that $b < c$.  Here the law of cosines is useful, and $\cos C < 0$.
$$
a^2 + b^2 - 2ab \cos C = c^2 \\
a^2 + b^2  = c^2 + 2ab \cos C < c^2\\
0 < a^2 < c^2 - b^2 \\
(c+b)(c-b) > 0 \\
c > b
$$
Thus $c>b$ and $b\geq a$ so 
$$
a \leq b < c
$$
