Example 3 Let $x_{i} \geqslant 0(i=1,2, \cdots, 7)$, and satisfy $x_{1}+x_{2}+\cdots+x_{7}=a$ (a constant), denote
$$
A=\max \left\{x_{1}+x_{2}+x_{3}, x_{2}+x_{3}+x_{4}, \cdots, x_{5}+x_{6}+x_{7}\right\} \text {. }
$$

Try to find $A_{\min }$.
Solution: Notice that
$$
\begin{array}{l}
A=\max \left\{x_{1}+x_{2}+x_{3}, x_{2}+x_{3}+x_{4}, \cdots, x_{5}+x_{6}+x_{7}\right\} \\
\geqslant \max \left\{x_{1}+x_{2}+x_{3}, x_{4}+x_{5}+x_{6}, x_{5}+x_{6}+x_{7}\right\} .
\end{array}
$$

Since $x_{i} \geqslant 0(i=1,2, \cdots, 7)$, we have
$$
x_{5}+x_{6}+x_{7} \geqslant x_{7} \text {. }
$$

Therefore, $A \geqslant \max \left\{x_{1}+x_{2}+x_{3}, x_{4}+x_{5}+x_{6}, x_{7}\right\}$
$$
\begin{array}{l}
\geqslant \frac{1}{3}\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}\right) \\
=\frac{a}{3} .
\end{array}
$$

Furthermore, when $x_{1}=x_{4}=x_{7}=\frac{a}{3}, x_{2}=x_{3}=x_{5}=x_{6}=0$, we have $A=\frac{a}{3}$.
Thus, $A_{\min }=\frac{a}{3}$.