6. For a triangle with side lengths $a$, $b$, and $c$, its area equals $\frac{1}{4}$, and the radius of the circumscribed circle is 1. If $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$ and $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then the relationship between $s$ and $t$ is
A. $s>t$
B. $s=t$
C. $s<t$
D. Uncertain
$$
R=1 \Rightarrow c=2 \sin C \Rightarrow \sin C=\frac{c}{2}, \quad S_{\triangle}=\frac{1}{2} a b \sin C=\frac{1}{4} a b c=\frac{1}{4} \Rightarrow a b c=1 \text {. }
$$

Thus, $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \sqrt{\frac{1}{a b}}+\sqrt{\frac{1}{b c}}+\sqrt{\frac{1}{c a}}=\sqrt{a}+\sqrt{b}+\sqrt{c}=s$.
Note that $a b c=1 \Rightarrow \sin A \sin B \sin C=\frac{1}{8} \Rightarrow a, b, c$ are not all equal, otherwise $A=B=C$ $=60^{\circ}$, which is a contradiction. Therefore, $t>s$, so the answer is $C$.