Example 3 Let set $A$ be a finite set, and
$$
\begin{array}{l}
|A| \geqslant 2, \\
Q(A)=\left\{\left.\frac{a-b}{c-d} \right\rvert\, a, b, c, d \in A, c \neq d\right\} .
\end{array}
$$

Find the smallest real number $\lambda$ such that $|Q(A)| \leqslant \lambda|A|^{4}$ holds for all sets $A$.
(2015, National Mathematics Competition Problem Workshop Exchange Question)
Solve $\lambda$ with the minimum value being $\frac{1}{2}$.
Define $D(A)=\{a-b \mid a, b \in A, a \geqslant b\}$.
Thus, $|D(A)| \leqslant 1+\frac{1}{2}|A|(|A|-1)$.
Since the set $Q(A)$ can be expressed as
$$
\begin{aligned}
Q(A)= & \{0, \pm 1\} \cup\left\{\left. \pm \frac{x}{y} \right\rvert\, x, y \in D(A),\right. \\
& x \neq y, x y \neq 0\},
\end{aligned}
$$

Therefore, from equations (1) and (2), we have
$$
|Q(A)| \leqslant \frac{1}{2}|A|^{4} \text {. }
$$

We proceed with the inductive construction.
Assume $a_{1}a_{k-1}, a_{k}-a_{k-1}>\max \left\{D\left(A_{k-1}\right)\right\}$;
(2) There do not exist $a, b, c, d \in D\left(A_{k}\right)$ such that
$$
\frac{a}{b}=\frac{c}{d}>1 \text {, and } b \neq d \text {. }
$$

For condition (1), $a_{k}$ can be chosen as a sufficiently large positive integer.
Next, we find $a_{k}$ to satisfy condition (2).
Let $D_{k}=\left\{a_{k}-a_{1}, a_{k}-a_{2}, \cdots, a_{k}-a_{k-1}\right\}$. Then
$D_{k} \cap D\left(A_{k-1}\right)=\varnothing$,
$D\left(A_{k}\right)=D_{k} \cup D\left(A_{k-1}\right)$,
$\min \left\{D_{k}\right\}>\max \left\{D\left(A_{k-1}\right)\right\}$.
Consider $\frac{a}{b}=\frac{c}{d}$ as an equation in $a_{k}$, and discuss the values of $a, b, c, d$.
(i) If $a, b, c, d$ are all taken from $D\left(A_{k-1}\right)$, by assumption, $\frac{a}{b}=\frac{c}{d}$ does not hold;
(ii) If $a, b, c, d$ are partially taken from $D_{k}$, then $\frac{a}{b}=\frac{c}{d}$ is an equation in $a_{k}$, with only a finite number of solutions.

In summary, there are only a finite number of $a_{k}$ such that $\frac{a}{b}=\frac{c}{d}>1$, $b \neq d$ holds, thus condition (2) is proven.

Therefore, when $a_{k}$ is sufficiently large, it can satisfy (1) and (2), completing the construction. At this point,
$D(A)=1+\frac{1}{2}|A|(|A|-1)$,
$Q(A)=\frac{1}{2}|A|^{4}-|A|^{3}-\frac{1}{2}|A|^{2}+|A|+3$.
If $cc|A|^{4}$.
Finally, let's explain the origin of this problem.
Problem 1: Let $A$ be a finite set of positive integers. Define
$B=\left\{\left.\frac{a+b}{c+d} \right\rvert\, a, b, c, d \in A\right\}$.
Prove: $|B| \geqslant 2|A|^{2}-1$.
(2014, China National Training Team Test)
In additive number theory, for a finite set $A, |A| \geqslant 2$, define the quotient set of $A$ as
$Q(A)=\left\{\left.\frac{a-b}{c-d} \right\rvert\, a, b, c, d \in A, c \neq d\right\}$.
Is there a similar result for the quotient set of a set $A$ as in Problem 1?

Since $a, b, c, d$ each have $|A|$ choices, it is easy to see a simple conclusion:
$|Q(A)| \leqslant|A|^{4}$.
We observe the relationship between $|A|$ and $|Q(A)|$ through specific examples.
When $A=\{1,2\}$, $Q(A)=\{0, \pm 1\}$;
When $A=\{1,2,5\}$,
$Q(A)=\left\{0, \pm 1, \pm 4, \pm \frac{1}{3}, \pm \frac{4}{3}, \pm \frac{1}{4}, \pm \frac{3}{4}\right\}$.
This indicates that the coefficient 1 in front of $|A|^{4}$ is not optimal.
Then, does there exist an optimal coefficient? This leads to the problem above.