(1) Let $a, b, c$ be the three sides of $\triangle ABC$, $a \leqslant b \leqslant c$, $R$ and $r$ be the circumradius and inradius of $\triangle ABC$, respectively. Let $f=a+b-2 R-2 r$, try to determine the sign of $f$ using the size of $\angle C$.
Let $A, B, C$ represent the three interior angles of $\triangle ABC$. It is well known that
$$
\begin{array}{c}
a=2 R \sin A, b=2 R \sin B \\
r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
\end{array}
$$

Thus,
$$
\begin{aligned}
f= & 2 R\left(\sin A+\sin B-1-4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)= \\
& 2 R\left(2 \sin \frac{B+A}{2} \cos \frac{B-A}{2}-1+\right. \\
& \left.2\left(\cos \frac{B+A}{2}-\cos \frac{B-A}{2}\right) \sin \frac{C}{2}\right)= \\
& 4 R \cos \frac{B-A}{2}\left(\sin \frac{B+A}{2}-\sin \frac{C}{2}\right)- \\
& 2 R+4 R \cos \frac{\pi-C}{2} \sin \frac{C}{2}= \\
& 4 R \cos \frac{B-A}{2}\left(\sin \frac{\pi-C}{2}-\sin \frac{C}{2}\right)-2 R+4 R \sin ^{2} \frac{C}{2}= \\
& 4 R \cos \frac{B-A}{2}\left(\cos \frac{C}{2}-\sin \frac{C}{2}\right)- \\
& 2 R\left(\cos \frac{C}{2}-\sin ^{2} \frac{C}{2}\right)= \\
& 2 R\left(\cos \frac{C}{2}-\sin \frac{C}{2}\right)\left(2 \cos \frac{B-A}{2}-\cos \frac{C}{2}-\sin \frac{C}{2}\right)
\end{aligned}
$$

Let $A \leqslant B \leqslant C$, so
$$
0 \leqslant B-A\cos \frac{C}{2} \\
\cos \frac{B-A}{2}>\cos \frac{B+A}{2}=\cos \frac{\pi-C}{2}=\sin \frac{C}{2}
\end{array}
$$

Thus, $2 \cos \frac{B-A}{2}>\cos \frac{C}{2}+\sin \frac{C}{2}$
Then,
$$
\begin{array}{l}
f>0 \Leftrightarrow \cos \frac{C}{2}>\sin \frac{C}{2} \Leftrightarrow C<\frac{\pi}{2}
\end{array}
$$