24. Let $a, b, c>0$, and $a+b+c=1$, find the range of $\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b c}{b c+a}}+\sqrt{\frac{c a}{c a+b}}$. (2006 Jiangsu Province Mathematical Olympiad Winter Camp Problem)
$$\begin{array}{l}
\text { 24, from } a+b+c=1 \text { we get } \\
\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b c}{b c+a}}+\sqrt{\frac{-c a}{c a+b}}=\sqrt{\frac{a b}{(a+c)(b+c)}}+\sqrt{\frac{b c}{(a+b)(a+c)}}+ \\
\sqrt{\frac{c a}{(a+b)(b+c)}}
\end{array}$$

It can be proven that
$$\begin{array}{l}
\sqrt{\frac{a b}{(a+c)(b+c)}}>\frac{a b}{a b+b c+c a} \\
\text { Equation (1) } \Leftrightarrow(a b+b c+c a)^{2}>a b(a+c)(b+c)=\left(a^{2}+a c\right)\left(b^{2}+b c\right) \Leftrightarrow \\
b^{2} c^{2}+c^{2} a^{2}+a b c(a+b+c)>0
\end{array}$$

Similarly,
$$\begin{array}{l}
\sqrt{\frac{b c}{(a+b)(a+c)}}>\frac{b c}{a b+b c+c a} \\
\sqrt{\frac{c a}{(a+b)(b+c)}}>\frac{c a}{a b+b c+c a}
\end{array}$$

Adding these inequalities, we get
$$\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b c}{b c+a}}+\sqrt{\frac{c a}{c a+b}}>1$$

By the AM-GM inequality,
$$\sqrt{\frac{a b}{(c+a)(b+c)}}=\sqrt{\frac{a}{c+a}} \sqrt{\frac{b}{b+c}} \leqslant \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{b+c}\right)$$

Similarly,
$$\begin{array}{l}
\sqrt{\frac{b c}{(a+b)(c+a)}} \leqslant \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right) \\
\sqrt{(b+c)(a+b)} \leqslant \frac{c}{2}\left(\frac{c}{b+c}+\frac{a}{a+b}\right)
\end{array}$$

Adding these three inequalities, we get $\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b b c}{b c+a}}+\sqrt{\frac{c a}{c a+b}} \leqslant \frac{3}{2}$. The equality holds when $a=b=c$.
When $c \rightarrow 0$,
$$\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{-b c}{b c+a}}+\sqrt{\frac{c a}{c a+b}} \rightarrow 0$$

Therefore, the range of $\sqrt{\frac{a b}{a b+c}}+\sqrt{\frac{b c}{b c}+a}+\sqrt{\frac{c a}{c a+b}}$ is $\left(0, \frac{3}{2}\right]$.