Example 6 Given a real number $a>25$. For all positive real number tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ satisfying the condition $\sum_{i=1}^{5} x_{i} \cdot \sum_{i=1}^{5} \frac{1}{x_{i}}=a$, try to find the extremum of $\frac{\max \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}}{\min \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}}$.
Solution: By symmetry, let $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$. By homogeneity, let $x_{1}=1, x_{5}=u, x_{2}, x_{3}, x_{4} \in[1, u], a$ $=f\left(x_{2}, x_{3}, x_{4}\right)=\left(1+x_{2}+x_{3}+x_{4}+u\right)$ $\left(\frac{1}{u}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}+1\right) \geqslant$
$$
\left(\frac{1}{\sqrt{u}}+1+1+1+\sqrt{u}\right)^{2}
$$
$$
\begin{array}{l}
\therefore \sqrt{u}+\frac{1}{\sqrt{u}} \leqslant \sqrt{a}-3, \\
u-(\sqrt{a}-3) \sqrt{u}+1 \leqslant 0, \\
\text { thus } \sqrt{u} \leqslant \frac{\sqrt{a}-3+\sqrt{a+5-6 \sqrt{a}}}{2} .
\end{array}
$$

On the other hand, considering $x_{3}, x_{4}$ as constants,
$$
a=f\left(x_{2}, x_{3}, x_{4}\right)=\alpha \cdot x_{2}+\frac{\beta}{x_{2}}+\gamma(\alpha, \beta, \gamma>0) \text {. }
$$
When $x_{2}>0$, $f$ is a convex function, achieving its maximum value at $x_{2}=1$ or $x_{2}=u$.
Similarly, $f$ achieves its maximum value when $x_{3}, x_{4}=1$ or $u$.
Assume that when $f$ achieves its maximum value, $x_{2}, x_{3}, x_{4}$ have $k$ values of $u$ and $3-k$ values of $1, k=0,1,2$.
At this time, $f=(k u+3-k+1+u)$
$$
\begin{array}{l}
\left(\frac{k}{u}+3-k+1+\frac{1}{u}\right)=-\frac{(u-1)^{2}}{u} k^{2}+ \\
\frac{3(u-1)^{2}}{u} k+\frac{(u+4)(4 u+1)}{u}
\end{array}
$$
$f$ is a parabola opening downwards, with the axis of symmetry at $k=\frac{3}{2}$, so $f$ achieves its maximum value when $k=1$ or 2.
$$
\begin{array}{c}
\therefore a=f\left(x_{2}, x_{3}, x_{4}\right) \leqslant(2 u+3)\left(\frac{2}{u}+3\right)= \\
6\left(u+\frac{1}{u}\right)+13=6\left(\sqrt{u}+\frac{1}{\sqrt{u}}\right)^{2}+1 \\
\therefore \sqrt{u}+\frac{1}{\sqrt{u}} \geqslant \sqrt{\frac{a-1}{6}}, \\
\sqrt{u} \geqslant \frac{\sqrt{a-1}+\sqrt{a-25}}{2 \sqrt{6}} . \\
\therefore u=\frac{\max \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}}{\min \left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}} \in \\
{\left[\left(\frac{\sqrt{a-1}+\sqrt{a-25}}{2 \sqrt{6}}\right)^{2},\left(\frac{\sqrt{a}-3+\sqrt{a+5-6 \sqrt{a}}}{2}\right)^{2}\right] .}
\end{array}
$$