Example 3 In $\triangle A B C$, $a, b, c$ are the sides opposite to $A, B, C$ respectively, then $\frac{a A+b B+c C}{a+b+c}-\frac{\pi}{3}$ (fill in $\left.\geq 、 \leq 、=\right)$.

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Analysis: The numerator $a A+b B+c C$ of the left side of this problem is easily reminiscent of the rearrangement inequality, and when $a \leq b \leq c$, we have $A \leq B \leq C$, so $a A+b B+c C$ is the sum in the same order. If we notice that $\pi=A+B+C$, the problem becomes easy to solve.

Solution: Without loss of generality, assume $a \leq b \leq c$, thus $A \leq B \leq C$. By the rearrangement inequality, we get $a A+b B+c C \geq b A+c B+a C$,
$a A+b B+c C \geq c A+a B+b C$.
Also, $a A+b B+c C=a A+b B+c C$,
Adding the three inequalities, we get $3(a A+b B+c C) \geq(a+b+c)(A+B+C)$
$=\pi(a+b+c)$, which means $\frac{a A+b B+c C}{a+b+c} \geq \frac{\pi}{3}$. Therefore, the answer is $\geq$.