Example 7 Given any $n$ points in the plane, any 3 of which form a triangle, each triangle has an area. Let the ratio of the maximum area to the minimum area be $\mu_{n}$. Find the minimum value of $\mu_{5}$.
Let any 5 points in the plane be $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, where no 3 points are collinear.
(1) If the convex hull of the 5 points is not a convex pentagon, then there must be one point inside a certain triangle. Without loss of generality, let $A_{4}$ be inside $\triangle A_{1} A_{2} A_{3}$, then
$$
\mu_{5} \geqslant \frac{S_{\triangle A_{1} A_{2} A_{3}}}{\min \left\{S_{\triangle A_{1} A_{2} A_{4}}, S_{\triangle A_{2} A_{3} A_{4}}, S_{\triangle A_{3} A_{1} A_{4}}\right\}} \geqslant 3 .
$$
(2) If the convex hull of the 5 points is a convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, as shown in Figure 6-36, draw $M N / / A_{3} A_{4}$ intersecting $A_{1} A_{3}$ and $A_{1} A_{4}$ at $M$ and $N$ respectively, and such that
$$
\frac{A_{1} M}{M A_{3}}=\frac{A_{1} N}{N A_{4}}=\frac{\sqrt{5}-1}{2} \text {. }
$$
( i ) If one of $A_{2}, A_{5}$, for example $A_{2}$, is on the same side of the line $M N$ as $A_{3}, A_{4}$ (as shown in Figure 6-36), then
$$
\mu_{5} \geqslant \frac{S_{\triangle A_{1} A_{3} A_{4}}}{S_{\triangle A_{2} A_{3} A_{4}}} \geqslant \frac{A_{1} A_{3}}{M A_{3}}=1+\frac{A_{1} M}{M A_{3}}=1+\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2} .
$$
(ii) If $A_{2}, A_{5}$ are on the same side of the line $M N$ as $A_{1}$ (as shown in Figure 6-37), let $A_{2} A_{5}$ intersect $A_{1} A_{3}$ at $O$, then $A_{1} O \leqslant A M$, thus
$$
\mu_{5} \geqslant \frac{S_{\triangle A_{2} A_{3} A_{5}}}{S_{\triangle_{A_{1} A_{2} A_{5}}}}=\frac{S_{\triangle A_{2} A_{3} O}}{S_{\triangle A_{1} A_{2} O}}=\frac{O A_{3}}{O A_{1}} \geqslant \frac{M A_{3}}{M A_{1}}=\frac{1}{\frac{\sqrt{5}-1}{2}}=\frac{\sqrt{5}+1}{2} .
$$

Notice that $3>\frac{\sqrt{5}+1}{2}$, so we always have $\mu_{5} \geqslant \frac{\sqrt{5}+1}{2}$. And when $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are the 5 vertices of a regular pentagon with side length $a$, we have
$$
\mu_{5}=\frac{S_{\triangle A_{1} A_{3} A_{4}}}{S_{\triangle A_{1} A_{2} A_{3}}}=\frac{\frac{1}{2} A_{1} A_{3} \cdot A_{1} A_{4} \sin 36^{\circ}}{\frac{1}{2} A_{1} A_{2} \cdot A_{1} A_{3} \sin 36^{\circ}}=\frac{A_{1} A_{4}}{A_{1} A_{2}}=\frac{1}{2 \sin 18^{\circ}}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2} \text {. }
$$

In summary, the minimum value of $\mu_{5}$ is $\frac{\sqrt{5}+1}{2}$.