$20 \cdot 77$ A triangle with side lengths $a, b, c$ has an area equal to $\frac{1}{4}$, and the radius of its circumcircle is 1. If $s=\sqrt{a}+\sqrt{b}+\sqrt{c}, t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then the relationship between $s$ and $t$ is
(A) $s>t$.
(B) $s=t$.
(C) $s<t$.
(D) Uncertain.
(China High School Mathematics League, 1986)
［Solution］From the given and the formula for the area of a triangle, we have
$$
S=\frac{a b c}{4 R}=\frac{a b c}{4}=\frac{1}{4},
$$

Thus,
$$
a b c=1 .
$$

By the AM-GM inequality, we have $\frac{1}{a}+\frac{1}{b} \geqslant 2 \cdot \sqrt{\frac{1}{a b}}=2 \sqrt{c}$;
Similarly, $\frac{1}{b}+\frac{1}{c} \geqslant 2 \sqrt{a}, \frac{1}{c}+\frac{1}{a} \geqslant 2 \sqrt{b}$.

Adding the above three inequalities, we get $t \geqslant s$, with equality holding if and only if $a=b=c$. In this case, from $a b c=1$, we know
$$
a=b=c=1, \quad S=\frac{\sqrt{3}}{4} \neq \frac{1}{4},
$$

Therefore, we must have $t>s$.
Hence, the correct choice is (C).