Prove that the inequality is equivalent to
$$P=\sum_{c y c} \sqrt{\frac{a}{(a+c)(a+b)} \cdot \frac{a b}{(b+c)(c+a)}} \leq \frac{3 \sqrt{3}}{4}$$

Without loss of generality, assume $a=\min (a, b, c)$, and consider the following cases:
( i ) If $a \leq b \leq c$, then we have
$$\begin{array}{l} \frac{a b}{(b+c)(c+a)} \leq \frac{c a}{(a+b)(b+c)} \leq \frac{b c}{(c+a)(a+b)} \\ \frac{a}{(a+c)(a+b)} \leq \frac{b}{(a+b)(b+c)} \leq \frac{c}{(b+c)(c+a)} \end{array}$$

Therefore, by the rearrangement inequality, we have
$$\begin{array}{l} P \leq \sqrt{\frac{a^{2} b}{(a+b)(a+c)^{2}(b+c)}}+\sqrt{\frac{a b c}{(a+b)^{2}(b+c)^{2}}}+\sqrt{\frac{b c^{2}}{(c+a)^{2}(a+b)(b+c)}} \\ =\sqrt{\frac{a b c}{(a+b)^{2}(b+c)^{2}}}+\frac{b}{(a+b)(b+c)}\left(\frac{a}{a+c}+\frac{c}{c+a}\right) \end{array}$$
$$\begin{array}{l} \leq \sqrt{3\left[\frac{a b c}{(a+b)^{2}(b+c)^{2}}+2 \cdot \frac{1}{4} \cdot \frac{b}{(a+b)(b+c)}\right]} \text { (since } \sqrt{x}+ \\ \frac{a b c}{(a+b)^{2}(b+c)^{2}}+\frac{1}{2} \cdot \frac{b}{(a+b)(b+c)} \geq \frac{9}{16} \Leftrightarrow(3 a c-b)^{2} \geq 0 \\ \text { i) If } a \leq c \leq b \text {, we have } \end{array}$$

Thus, it suffices to prove
$$\begin{array}{l} \frac{c a}{(a+b)(b+c)} \leq \frac{a b}{(b+c)(c+a)} \leq \frac{b c}{(c+a)(a+b)} \\ \frac{a}{(a+c)(a+b)} \leq \frac{c}{(b+c)(c+a)} \leq \frac{b}{(a+b)(b+c)} \end{array}$$

Therefore, by the rearrangement inequality, we have
$$\begin{array}{l} P \leq \sqrt{\frac{a^{2} c}{(a+c)(a+b)^{2}(b+c)}}+\sqrt{\frac{a b c}{(a+c)^{2}(b+c)^{2}}}+\sqrt{\frac{b^{2} c}{(a+c)(a+b)^{2}(b+c)}} \\ \leq \sqrt{3\left[\frac{a b c}{(a+c)^{2}(b+c)^{2}}+2 \cdot \frac{1}{4} \cdot \frac{c}{(b+c)(c+a)}\right]} \end{array}$$

Thus, it suffices to prove
$$\frac{a b c}{(a+c)^{2}(b+c)^{2}}+\frac{1}{2} \cdot \frac{c}{(b+c)(c+a)} \leq \frac{9}{16} \Leftrightarrow(3 a b-c)^{2} \geq 0$$

Equality holds when $a=b=c=\frac{1}{3}$