Example 2. Let $A, B, C$ be the three interior angles of a triangle. Find the maximum value of $\sin A+\sin B+\sin C$.

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From the trigonometric formula, we have
$$
\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}.
$$

It follows that, under the condition that $C$ is fixed, $\sin A+\sin B+\sin C$ achieves its maximum value when $A=B$. Therefore, it can be conjectured that the maximum value is obtained when $A=B=C$.

By symmetry, we can assume $A \leqslant B \leqslant C$. If $A, B, C$ are not all equal, then $A<\frac{\pi}{3}<C$. Let
$$
A^{\prime}=\frac{\pi}{3}, B^{\prime}=B, C^{\prime}=A+C-\frac{\pi}{3}.
$$

Thus, $\left|A^{\prime}-C^{\prime}\right|<|A-C|$. Therefore,
$$
\begin{aligned}
& \sin A+\sin C=2 \sin \frac{A+C}{2} \cos \frac{A--C}{2} \\
< & 2 \sin \frac{A^{\prime}+C^{\prime}}{2} \cos \frac{A^{\prime}-C^{\prime}}{2} \\
= & \sin A^{\prime}+\sin C^{\prime}.
\end{aligned}
$$

If $C^{\prime} \neq \frac{\pi}{3}$, then since $B^{\prime}+C^{\prime}=\frac{2 \pi}{3}$, we have
$$
\begin{array}{l}
\sin A+\sin B+\sin C \\
<\sin A^{\prime}+\sin B^{\prime}+\sin C^{\prime} \\
=\sin \frac{\pi}{3}+2 \sin \frac{B^{\prime}+C^{\prime}}{2} \cos \frac{B^{\prime}-C^{\prime}}{2} \\
<3 \sin \frac{\pi}{3}=\frac{3 \sqrt{3}}{2}.
\end{array}
$$

This proves that the maximum value of $\sin A+\sin B+\sin C$ is $\frac{3 \sqrt{3}}{2}$, which is achieved if and only if $A=B=C=\frac{\pi}{3}$.