Question 74, Given $a \geq b \geq c \geq d>0, a^{2}+b^{2}+c^{2}+d^{2}=\frac{(a+b+c+d)^{2}}{3}$, find the maximum value of $\frac{a+c}{b+d}$.
Question 74, Solution: According to homogeneity, we may assume $a+b+c+d=3$, then $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+\mathrm{d}^{2}=3$. Let $a+c=x$, then $b+d=3-x$. Therefore, we have
$$
x^{2}+(3-x)^{2}=(a+c)^{2}+(b+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(a c+b d)=3+2(a c+b d)
$$

According to the rearrangement inequality, we know:
$$
\begin{array}{l}
a c+b d \leq a b+c d \\
\Rightarrow 2(a c+b d) \leq a c+b d+a b+c d=(a+d)(b+c) \leq \frac{[(a+d)+(b+c)]^{2}}{4}=\frac{9}{4}
\end{array}
$$

Thus, we have:
$$
\begin{array}{l}
x^{2}+(3-x)^{2} \leq 3+\frac{9}{4}=\frac{21}{4} \\
\Rightarrow x \leq \frac{6+\sqrt{6}}{4}
\end{array}
$$

Therefore, $\frac{a+c}{b+d}=\frac{x}{3-x} \leq \frac{\frac{6+\sqrt{6}}{4}}{3-\frac{6+\sqrt{6}}{4}}=\frac{7+2 \sqrt{6}}{5}$. Equality holds when $a=\frac{3+\sqrt{6}}{4}, b=c=\frac{3}{4}, d=\frac{3-\sqrt{6}}{4}$.