Example 7 Let $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$ and $x_{1}+x_{2}+\cdots+x_{n} \geqslant k, k$ be a positive constant, find the maximum value of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{x_{1} \sqrt{\sum_{i=1}^{n} x_{i}}}{\left(\sum_{i=1}^{n-1} x_{i}\right)^{2}+x_{n}}$.
Let $a=x_{1}, b=\sum_{i=2}^{n-1} x_{i}, c=x_{n}$, then $a, b, c \geqslant 0, a+b+c \geqslant k$, up to this point the original function is reduced from $n$ variables to three variables, i.e., $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\frac{a \sqrt{a+b+c}}{(a+b)^{2}+c} \xlongequal{\triangle} g(a, b, c)$.
Further let $s=a+b+c$, then $s \geqslant k$,
$$
g(a, b, c)=\frac{a \sqrt{s}}{(s-c)^{2}+c} \leqslant \frac{(s-c) \sqrt{s}}{(s-c)^{2}+c}=\frac{\sqrt{s}}{(s-c)+\frac{c}{s-c}} .
$$

Notice that $(s-c)+\frac{c}{s-c} \geqslant 2 \sqrt{s}-1 \Leftrightarrow(c-s+\sqrt{s})^{2} \geqslant 0$,
thus we have $g(a, b, c) \leqslant \frac{\sqrt{s}}{2 \sqrt{s}-1}=\frac{1}{2-\frac{1}{\sqrt{s}}} \leqslant \frac{1}{2-\frac{1}{\sqrt{k}}}=\frac{\sqrt{k}}{2 \sqrt{k}-1}$,
when $b=0, c=k-\sqrt{k}, a=\sqrt{k}$, i.e., $x_{1}=\sqrt{k}, x_{2}=x_{3}=\cdots=x_{n-1}=0, x_{n}=k-\sqrt{k}$, the equality holds. Therefore, the maximum value of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is $\frac{\sqrt{k}}{2 \sqrt{k}-1}$.