1.53 Let three real numbers $x, y, z$ satisfy the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2, \\ \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=1 .\end{array}\right.$ Then the value of $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}$ is
(A) 1 .
(B) $\sqrt{2}$.
(C) $\frac{3}{2}$.
(D) $\sqrt{3}$.
(Nanchang City, Jiangxi Province, Junior High School Mathematics Competition, 1991)
[Solution] $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, thus $\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2}=4$.

That is $\square$
$$
\therefore \quad \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{x y}+\frac{2}{y z}+\frac{2}{x z}=4 \text {, }
$$

That is $\square$
$$
\begin{aligned}
& 1+2\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{x z}\right)=4, \\
\therefore & \frac{1}{x y}+\frac{1}{y z}+\frac{1}{x z}=\frac{3}{2} .
\end{aligned}
$$

Therefore, the answer is $(C)$.
Note: The original condition does not hold in the real number range (the original assumption is that $x, y, z$ are real numbers). The reason is: according to the problem,
$$
\begin{array}{c}
4\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2}, \\
\text { i.e., }\left(\frac{1}{x}-\frac{1}{y}\right)^{2}+\left(\frac{1}{y}-\frac{1}{z}\right)^{2}+\left(\frac{1}{z}-\frac{1}{x}\right)^{2}+\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=0,
\end{array}
$$

which is impossible.
If considering in the real number range, one should note the inequality
$$
3\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right) \geqslant\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2}
$$

and the equality condition. For example, set
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \\
\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}=2 .
\end{array}\right.
$$