$13 \cdot 19$ If real numbers $a>b>c>0$. Also, let $l_{1}=\sqrt{(a+c)^{2}+b^{2}}$, $l_{2}=\sqrt{a^{2}+(b+c)^{2}}, \quad l_{3}=\sqrt{(a+b)^{2}+c^{2}}$. Then the smallest of the products $l_{1} l_{2}, l_{1} l_{3}$, $l_{2} l_{3}, l_{2}^{2}, l_{3}^{2}$ is
(A) $l_{1} l_{2}$.
(B) $l_{1} l_{3}$.
(C) $l_{2} l_{3}$.
(D) $l_{2}^{2}$.
(E) $l_{3}^{2}$.
(China Beijing Junior High School Mathematics Competition, 1983)
[Solution 1] Given $a>b>c>0$, we know $a b>a c>b c$, so
$$
\begin{aligned}
a^{2}+2 a b+b^{2}+c^{2}>a^{2}+2 a c+c^{2}+b^{2} \\
>a^{2}+b^{2}+2 b c+c^{2},
\end{aligned}
$$

which means $(a+b)^{2}+c^{2}>(a+c)^{2}+b^{2}>a^{2}+(b+c)^{2}>0$, hence $\sqrt{(a+b)^{2}+c^{2}}>\sqrt{(a+c)^{2}+b^{2}}>\sqrt{a^{2}+(b+c)^{2}}>0$, or $l_{3}>l_{1}>l_{2}>0$. Therefore, $l_{3}^{2}>l_{1} l_{3}>l_{2} l_{3}>l_{1} l_{2}>l_{2}^{2}$. Thus, the answer is $(D)$.
[Solution 2] Let $a=3, b=2, c=1$. It is easy to calculate
$$
l_{1}=\sqrt{20}, \quad l_{2}=\sqrt{18}, \quad l_{3}=\sqrt{26},
$$

Clearly, $l_{2}^{2}$ is the smallest among the products.
Thus, the answer is $(D)$.