Example 1 Let $0<p \leqslant a, b, c, d, e \leqslant q$, find the maximum value of $F=(a+b+c+d+$
e) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right)$.
Since $F$ is continuous on a closed domain, it must have a maximum value. Fix $a, b, c, d$, and let $a + b + c + d = A$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = B$, then $A, B$ are constants, and $F = (A + e)(B + \frac{1}{e}) = 1 + AB + eB + \frac{A}{e}$.

Consider $f(e) = eB + \frac{A}{e}$. It is easy to see that when $e \leq \sqrt{\frac{A}{B}}$, $f(e)$ is monotonically decreasing, and when $e \geq \sqrt{\frac{A}{B}}$, $f(e)$ is monotonically increasing. Thus, $\sqrt{\frac{A}{B}}$ is the minimum point of $f(e)$. Noting that $p \leq e \leq q$, the maximum value of $f(e)$ can only be achieved at its endpoints: $e = p$ or $e = q$. By symmetry, $F$ can only achieve its maximum value when $a, b, c, d, e \in \{p, q\}$. Therefore, suppose $F$ achieves its maximum value when $a, b, c, d, e$ have $k$ values of $p$ and $5-k$ values of $q$, then the maximum value of $F$ is the maximum value of $F(k)$ of the following form:
$$\begin{aligned}
F(k) & = [kp + (5-k)q]\left(\frac{k}{p} + \frac{5-k}{q}\right) \\
& = k^2 + (5-k)^2 + (5k - k^2)\left(\frac{p}{q} + \frac{q}{p}\right) \\
& = \left[2 - \left(\frac{p}{q} + \frac{q}{p}\right)\right]k^2 + \left[5\left(\frac{p}{q} + \frac{q}{p}\right) - 10\right]k + 25
\end{aligned}$$

In the above quadratic function $F(k)$, the coefficient of the quadratic term $2 - \left(\frac{p}{q} + \frac{q}{p}\right) < 0$, and the x-coordinate of the vertex is $\frac{5}{2}$, but $k \in \mathbb{N}$, so $F(k) < F(2) = F(3) = 13 + 6\left(\frac{p}{q} + \frac{q}{p}\right)$. Also,
$$a = b = c = p, c = d = q \text{ when, } F = 13 + 6\left(\frac{p}{q} + \frac{q}{p}\right),$$

Therefore, $F_{\text{max}} = 13 + 6\left(\frac{p}{q} + \frac{q}{p}\right) = 25 + 6\left(\sqrt{\frac{p}{q}} - \sqrt{\frac{q}{p}}\right)^2$.