Example 1 (2002 Shanghai High School Mathematics Competition Question) Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2}=5$. Then the maximum value of $(a-b)^{2}+$ $(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}$ is $\qquad$.
Solution: Fill in 20. Reason: First, notice the matching expression: $4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a+b+c+d)^{2}+(a-b)^{2}+(a-$ $c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}$.

Substitute the given $a^{2}+b^{2}+c^{2}+d^{2}=5$ into the above expression, and we immediately get: $20=4 \times 5=4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a+b+c+d)^{2}+$ $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}$, noting that $(a+b+c+d)^{2} \geqslant 0$, hence the maximum value of $(a-b)^{2}+$ $(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}+(c-d)^{2}$ is 20.