6. Let $a, b, c$ be non-negative real numbers. Then the algebraic expression
$$
\begin{aligned}
S= & \sqrt{\frac{a b}{(b+c)(c+a)}}+\sqrt{\frac{b c}{(c+a)(a+b)}} \\
& +\sqrt{\frac{c a}{(a+b)(b+c)}}
\end{aligned}
$$

has a minimum value of $\qquad$ and a maximum value of $\qquad$.
$6.1, \frac{3}{2}$.
Let $\frac{a}{b+c}=u^{2}, \frac{b}{c+a}=v^{2}, \frac{c}{a+b}=w^{2}$, and set $a+b+c=1$, then $a=\frac{u^{2}}{1+u^{2}}, b=\frac{v^{2}}{1+v^{2}}, c=\frac{w^{2}}{1+w^{2}}$.
Thus, $\frac{u^{2}}{1+u^{2}}+\frac{v^{2}}{1+v^{2}}+\frac{w^{2}}{1+w^{2}}=1$,

which means $(w v)^{2}+(v w)^{2}+(w u)^{2}+2(w w)(v w)(w u)=1$.
We can let $w=\cos C, v w=\cos A, w u=\cos B$, and $\angle A+\angle B+\angle C=\pi, \angle A, \angle B, \angle C \in\left[0, \frac{\pi}{2}\right]$.
Therefore, $S=\cos A+\cos B+\cos C$
$=2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}+\cos C$
$\leqslant 2 \sin \frac{C}{2}+\cos C=-2\left(\sin \frac{C}{2}-\frac{1}{2}\right)^{2}+\frac{3}{2}$
$\leqslant \frac{3}{2}$.
Equality holds if and only if $a=b=c$.
Similarly, $\cos A+\cos B+\cos C$
$$
\begin{array}{l}
=2 \sin \frac{C}{2} \cdot \cos \frac{A-B}{2}+\cos C \\
=1+4 \sin \frac{C}{2}\left(\sin \frac{A}{2} \cdot \sin \frac{B}{2}\right) \geqslant 1 .
\end{array}
$$

Equality holds when $a=0$.