4・40 Let $a>b>c>d>0$, and $X=\sqrt{a b}+\sqrt{c d}, Y=\sqrt{a c}+$ $\sqrt{b d}, Z=\sqrt{a d}+\sqrt{b c}$. Then the size relationship of $X, Y, Z$ is
(A) $X<Z<Y$.
(B) $Y<Z<X$.
(C) $X<Y<Z$.
(D) $Z<Y<X$.
(China Beijing Junior High School Mathematics Competition, 1990)
［Solution］From the given, we have
$$
\begin{aligned}
X-Y & =\sqrt{a}(\sqrt{b}-\sqrt{c})+\sqrt{d}(\sqrt{c}-\sqrt{b}) \\
& =(\sqrt{a}-\sqrt{d})(\sqrt{b}-\sqrt{c})>0, \\
\text { Also, } \quad Y-Z & =\sqrt{a}(\sqrt{c}-\sqrt{d})+\sqrt{b}(\sqrt{d}-\sqrt{c}) \\
& =(\sqrt{a}-\sqrt{b})(\sqrt{c}-\sqrt{d})>0 .
\end{aligned}
$$

Thus, $X>Y>Z$.
Therefore, the answer is $(D)$.