6. For a triangle with sides of length $a, b, c$, its area is $\frac{1}{4}$, 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
(C)
6. 【Analysis and Solution】 $\because c=2 R \sin C=2 \sin C$, and $\frac{1}{2} a b \sin C=\frac{1}{4}$,
$$
\begin{array}{l} 
\because a b c=1, \text { thus } \\
t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right) \frac{1}{2}\left(\frac{1}{c}+\frac{1}{a}\right) \\
\geqslant \sqrt{\frac{1}{a b}}+\sqrt{\frac{1}{b c}}+\sqrt{\frac{1}{c a}} \\
=\frac{\sqrt{c}+\sqrt{a}+\sqrt{b}}{\sqrt{a b c}} \\
=\sqrt{a}+\sqrt{b}+\sqrt{c}=s .
\end{array}
$$

And the equality cannot be achieved, otherwise $a=b=c=R=1$, which is clearly impossible, hence the answer is $(\mathrm{C})$.