Example 1.39 (Cai Jianxing) For non-negative real numbers $a, b, c$ where not two of them are zero simultaneously, determine the constant $k$ such that the following inequality always holds:
$$\sqrt{S+k} \leqslant \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \leqslant \sqrt{S+k+1}$$

where $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=S$.
Proof: Let $a=b=c$, we have $2 \leqslant k \leqslant 3$, and let $a=0, b=c$, we have $1 \leqslant k \leqslant 2$, hence $k=2$. We first give two identities:
$$\begin{array}{c}
\sum(a+b)(a+c)=(a+b+c)^{2}+a b+b c+c a \\
\sum a(a+b)(a+c)=\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)+3 a b c
\end{array}$$

For the right side of the inequality, transform it into the following form:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \leqslant \sqrt{(a+b+c)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)}$$

Let $s=\sqrt{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}$, divide both sides by $s$ and use identity (1) $\Leftrightarrow$
$$\sum \sqrt{\frac{a^{2}(a+b+c)+a b c}{(a+b+c)^{2}+a b+b c+c a}} \leqslant \sqrt{a+b+c}$$

Since the inequality is symmetric, we can assume $a+b+c=1$, thus we only need to prove:
$$\sum \sqrt{a^{2}+a b c} \leqslant \sqrt{1+a b+b c+c a}$$

Using the Cauchy inequality:
$$\begin{aligned}
\left(\sum \sqrt{a^{2}+a b c}\right)^{2}= & \left(\sum \sqrt{a} \cdot \sqrt{a+b c}\right)^{2} \leqslant \\
& (a+b+c)(a+b+c+a b+b c+c a) \leqslant \\
& 1+a b+b c+c a
\end{aligned}$$

Thus, the right side of the inequality is proved. Equality holds if and only if $a=b=c$.
The left side of the inequality can be written in the following form:
$$\sqrt{\frac{2}{3}(a+b+c) \sum \frac{1}{a+b}+\frac{1}{3} S} \leqslant \sum \sqrt{\frac{a}{b+c}}$$

Divide both sides by $s$ and use identities (1) and (2), we get:
$$\begin{array}{l}
\sqrt{\frac{2}{3}(a+b+c)+\frac{a^{2}(a+b+c)+a b c}{3\left[(a+b+c)^{2}+a b+b c+c a\right]}} \leqslant \\
\sum \sqrt{\frac{a^{2}(a+b+c)+a b c}{(a+b+c)^{2}+a b+b c+c a}}
\end{array}$$

Similarly, let $a+b+c=1$, then the above is equivalent to:
$$\begin{array}{l}
\sqrt{\frac{2}{3}+\frac{a^{2}+b^{2}+c^{2}+3 a b c}{3(1+a b+b c+c a)}} \leqslant \sum \sqrt{\frac{a^{2}+a b c}{1+a b+b c+c a}} \Leftrightarrow \\
\sqrt{1+a b c} \leqslant \sqrt{a^{2}+a b c}+\sqrt{b^{2}+a b c}+\sqrt{c^{2}+a b c}
\end{array}$$

Square both sides, we only need to prove:
$$a^{2}+b^{2}+c^{2}+2 \sum \sqrt{\left(a^{2}+a b c\right)\left(b^{2}+a b c\right)}+3 a b c \geqslant 1+a b c$$

By the Cauchy inequality:
$$\begin{array}{l}
a^{2}+b^{2}+c^{2}+2 \sum \sqrt{\left(a^{2}+a b c\right)\left(b^{2}+a b c\right)}+3 a b c \geqslant \\
a^{2}+b^{2}+c^{2}+3 a b c+2 \sum(a b+a b c)= \\
(a+b+c)^{2}+9 a b c=1+9 a b c \geqslant 1+a b c
\end{array}$$

Thus, the right side of the original inequality holds, equality holds if and only if $a=0, b=c$ or its cyclic permutations.