1. 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) $z<y<x$
$-1 . D$.
From $y^{2}-x^{2}=a c+b d-(a b+c d)=(c-b)(a-$
$d)0$, we have $y<x$. Similarly, $z<y$. Therefore, $z<y<$