Example 1 Given positive real numbers $a, b$ and integer $n \geqslant 2$, let
$$
f(x)=(x+a)(x+b) \text {. }
$$

For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find
$$
F=\sum_{1 \leqslant i<j \leqslant n} \min \left\{f\left(x_{i}\right), f\left(x_{j}\right)\right\}
$$

the maximum value.
(2012, China Mathematical Olympiad)
Solve: 
$$
\begin{array}{l}
\min \left\{f\left(x_{i}\right), f\left(x_{j}\right)\right\} \leqslant \sqrt{f\left(x_{i}\right) f\left(x_{j}\right)} \\
=\sqrt{\left(x_{i}+a\right)\left(x_{i}+b\right)\left(x_{j}+a\right)\left(x_{j}+b\right)} \\
\leqslant \frac{\left(x_{i}+a\right)\left(x_{j}+b\right)+\left(x_{j}+a\right)\left(x_{i}+b\right)}{2} \\
=x_{i} x_{j}+\frac{a+b}{2}\left(x_{i}+x_{j}\right)+a b,
\end{array}
$$

Then \( F \leqslant \sum_{1 \leqslant i<j \leqslant n}\left(x_{i} x_{j}+\frac{a+b}{2}\left(x_{i}+x_{j}\right)+a b\right) \) 
\( =\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}+\frac{a+b}{2} \sum_{1 \leqslant i<j \leqslant n}\left(x_{i}+x_{j}\right)+\mathrm{C}_{n}^{2} a b \) 
\( \leqslant \frac{n-1}{2 n}\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\frac{a+b}{2}(n-1) \sum_{i=1}^{n} x_{i}+ \)
$$
\begin{aligned}
& \frac{n(n-1)}{2} a b \\
= & \frac{n-1}{2 n}+\frac{a+b}{2}(n-1)+\frac{n(n-1)}{2} a b \\
= & \frac{n-1}{2 n}\left(1+n(a+b)+n^{2} a b\right) .
\end{aligned}
$$

When \( x_{i}=\frac{1}{n} (1 \leqslant i \leqslant n) \), the equality holds.
Thus, \( F_{\text {max }}=\frac{n-1}{2 n}\left(1+n(a+b)+n^{2} a b\right) \).