Example 8 (2000 China Mathematical Olympiad) Given that $a, b, c$ are the three sides of $\triangle ABC$, $a \leqslant b \leqslant c$, and $R$ and $r$ are the radii of the circumcircle and incircle of $\triangle ABC$ respectively. Let $f=a+b-2 R-2 r$, try to determine the sign of $f$ using the size of angle $C$.

---

The translation is provided as requested, maintaining the original text's line breaks and format.
Analysis We will unify the expression in terms of angles.
Solution Since $S_{\triangle A B C}=\frac{1}{2} r(a+b+c)$
Therefore, $r=2 \frac{S_{\triangle A B C}}{a+b+c}=\frac{4 R^{2} \sin A \sin B \sin C}{2 R(\sin A+\sin B+\sin C)}$
$$
=2 R \frac{\sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
$$

Thus, $f=2 R\left(\sin A+\sin B-1-4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)$
$$
\begin{array}{l}
=2 R(\sin A+\sin B-\cos A-\cos B-\cos C) \\
=2 R\left[\left(2 \sin \frac{A+B}{2}-2 \cos \frac{A+B}{2}\right) \cos \frac{A-B}{2}-\left(\cos ^{2} \frac{C}{2}-\sin ^{2} \frac{C}{2}\right)\right] \\
=2 R\left(\cos \frac{C}{2}-\sin \frac{C}{2}\right)\left(2 \cos \frac{A-B}{2}-\cos \frac{C}{2}-\cos \frac{A+B}{2}\right)
\end{array}
$$

By $a \leqslant b \leqslant c$ and the monotonicity of the cosine function, we know:
$$
\cos \frac{A-B}{2}>\cos \frac{C}{2}, \cos \frac{A-B}{2}>\cos \frac{A+B}{2}
$$

Therefore, $2 \cos \frac{A-B}{2}-\cos \frac{C}{2}-\cos \frac{A+B}{2}>0$
If $\cos \frac{C}{2}1, C \in\left(\frac{\pi}{2}, \pi\right)$ then $f\sin \frac{C}{2}$, i.e., $\tan \frac{C}{2}0$. Also, since $c \geqslant b \geqslant a$, we have: $C \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$
If $\cos \frac{C}{2}=\sin \frac{C}{2}$, i.e., $\tan \frac{C}{2}=1, C=\frac{\pi}{2}$, then $f=0$

In summary: When $C \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right)$, $f>0$; when $C=\frac{\pi}{2}$, $f=0$; when $C \in\left(\frac{\pi}{2}, \pi\right)$, $f<0$.