$4 \cdot 28$. If $a>b>c>d$, and if $x=(a+b)(c+d)$, and $y=(a+c)(b+d)$, and $z=(a+d)(b+c)$. Then the order of magnitude of $x, y, z$ is
(A) $x<z<y$.
(B) $y<z<x$.
(C) $x<y<z$.
(D) $z<y<x$.
(China Beijing High School Mathematics Competition, 1990)
[Solution]
$$
\begin{aligned}
y-x & =(a b+c b+a d+c d)-(a c+b c+a d+b d) \\
& =a(b-c)+d(c-b) \\
& =(a-d)(b-c) .
\end{aligned}
$$

Given $a>b>c>d$, we know $y-x>0$, i.e., $y>x$.
$$
\text { Also, } \begin{aligned}
z-y & =(a b+d b+a c+d c)-(a b+c b+a d+c d) \\
& =a(c-d)+b(d-c) \\
& =(a-b)(c-d),
\end{aligned}
$$
$$
\therefore z-y>0 \text {, i.e., } z>y \text {. }
$$

In summary, $x<y<z$.
Therefore, the answer is $(C)$.