2.3.11 * $\star$ 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 $f=a+b-2R-2r$, try to determine the sign of $f$ using the size of angle $C$.
Let $A$, $B$, and $C$ represent the three interior angles of $\triangle ABC$. It is easy to think of the well-known equations in a triangle: $a=2R \sin A, b=2R \sin B, r=4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. Therefore, 
\[ f=2R(\sin A + \sin B - 1 - 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) = 2R \left[2 \sin \frac{B+A}{2} \cdot \cos \frac{B-A}{2} - 1 + 2 \left( \cos \frac{B+A}{2} - \cos \frac{B-A}{2} \right) \sin \frac{C}{2} \right] \]
\[ = 2R \left[2 \cos \frac{B-A}{2} \left( \sin \frac{B+A}{2} - \sin \frac{C}{2} \right) - 1 + 2 \cos \frac{B+A}{2} \sin \frac{C}{2} \right] \]
\[ = 2R \left[2 \cos \frac{B-A}{2} \left( \cos \frac{C}{2} - \sin \frac{C}{2} \right) - 1 + 2 \sin^2 \frac{C}{2} \right] \]
\[ = 2R \left( \cos \frac{C}{2} - \sin \frac{C}{2} \right) \cdot \left( 2 \cos \frac{B-A}{2} - \cos \frac{C}{2} - \sin \frac{C}{2} \right). \]
Since $A \leqslant B \leqslant C$, we have $0 \leqslant B-A < \frac{\pi}{2}$, so $\cos \frac{B-A}{2} > \cos \frac{C}{2}$, and $\cos \frac{B-A}{2} > \cos \frac{B+A}{2} = \cos \frac{\pi - C}{2} = \sin \frac{C}{2}$. Therefore, $2 \cos \frac{B-A}{2} > \cos \frac{C}{2} + \sin \frac{C}{2}$. Thus, 
\[ f > 0 \Leftrightarrow \cos \frac{C}{2} > \sin \frac{C}{2} \Leftrightarrow C < \frac{\pi}{2}. \]