$1 \cdot 47$ If $x y=a, x z=b, y z=c$, and none of them is zero, then $x^{2}+y^{2}+z^{2}$ equals
(A) $\frac{a b+a c+b c}{a b c}$.
(B) $\frac{a^{2}+b^{2}+c^{2}}{a b c}$.
(C) $\frac{(a+b+c)^{2}}{a b c}$.
(D) $\frac{(a b+a c+b c)^{2}}{a b c}$.
(E) $\frac{(a b)^{2}+(a c)^{2}+(b c)^{2}}{a b c}$.
(34th American High School Mathematics Examination, 1983)
［Solution］From the given conditions, we have

Therefore,
$$
\begin{array}{c}
a b c=x^{2} y^{2} z^{2}=x^{2} c^{2}=a^{2} z^{2}=y^{2} b^{2}, \\
x^{2}=\frac{a b}{c}, \quad y^{2}=\frac{a c}{b}, \quad z^{2}=\frac{b c}{a} .
\end{array}
$$

Adding them up, we get
$$
x^{2}+y^{2}+z^{2}=\frac{a b}{c}+\frac{a c}{b}+\frac{b c}{a}=\frac{(a b)^{2}+(a c)^{2}+(b c)^{2}}{a b c} \text {. }
$$

Therefore, the correct choice is $(E)$.