Question 75, Given real numbers $a \geq b \geq c \geq d, a+b+c+d=9, a^{2}+b^{2}+c^{2}+d^{2}=21$, find the minimum possible value of $\mathrm{ab}-\mathrm{cd}$.
Let $a+b=x$, then $c+d=9-x$, and $x \geq \frac{9}{2}$. Therefore, we have:
$$
\begin{array}{l}
x^{2}+(9-x)^{2}=(a+b)^{2}+(c+d)^{2} \\
=a^{2}+b^{2}+c^{2}+d^{2}+2(a b+c d)=21+2(a b+c d)
\end{array}
$$

According to the rearrangement inequality, we know: $a b+c d \geq a c+b d \geq a d+b c$, hence
$$
a b+c d \geq \frac{a b+c d+a c+b d+a d+b c}{3}=\frac{(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{6}=10
$$

Substituting into (1), we get
$$
\begin{array}{l}
x^{2}+(9-x)^{2} \geq 21+20=41 \\
\Rightarrow(x-4)(x-5) \geq 0 \\
\Rightarrow x \geq 5
\end{array}
$$

Therefore, we have:
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}+d^{2}+2(a b-c d)=(a+b)^{2}+(c-d)^{2} \geq(a+b)^{2} \geq 25 \\
\Rightarrow a b-c d \geq 2
\end{array}
$$

When $a=3, b=c=d=2$, the equality holds.