3. Let $a, b, c \in\left[\frac{1}{2}, 1\right]$, and let $s=\frac{a+b}{1+c}+\frac{b+c}{1+a}+\frac{c+a}{1+b}$, then the range of $s$ is $\qquad$
3. Solution: Since $a+b \geqslant 1$, then $\frac{a+b}{1+c} \geqslant \frac{a+b}{a+b+c}$, thus $\frac{b+c}{1+a} \geqslant \frac{b+c}{a+b+c}, \frac{c+a}{1+b} \geqslant \frac{c+a}{a+b+c}$, adding them up, we get $s \geqslant 2$, which is achieved when $a=b=c=\frac{1}{2}$.
Since $a \leqslant 1$, then $\frac{a}{1+c} \leqslant \frac{a}{a+c}, \frac{c}{1+a} \leqslant \frac{c}{a+c}$, similarly,
$$
\frac{b}{1+c} \leqslant \frac{b}{b+c}, \frac{b}{1+a} \leqslant \frac{b}{a+b} ; \frac{c}{1+b} \leqslant \frac{c}{b+c}, \frac{a}{1+b} \leqslant
$$
$\frac{a}{a+b}$, adding them up, we get $s \leqslant 3$, which is achieved when $a=b=c=1$. $\therefore s \in[2, 3]$.