$$\begin{array}{r} \text { Equation (31) } \Leftrightarrow \frac{\sqrt{a+b+c}}{b+c} \geqslant \frac{3 \sqrt{3 a}}{2(a+b+c)} \\ 4(a+b+c)^{3} \geqslant 27 a(b+c)^{2} \end{array}$$

Since
$$\begin{aligned} 4(a+b+c)^{3}= & 4\left(\frac{b+c}{2}+\frac{b+c}{2}+a\right)^{3} \geqslant \\ & 4 \cdot 27 \cdot \frac{b+c}{2} \cdot \frac{b+c}{2} \cdot a= \\ & 27 a(b+c)^{2} \end{aligned}$$

We obtain equation ( ).

Note: From equation (31), we can derive
$$\begin{array}{c} \sum w_{a}^{2} \geqslant \frac{27}{2} R r \quad(\text { Liu Jian's Conjecture }) \\ \sum a(b+c) w_{a}^{2} \geqslant 162(R r)^{2} \\ \sum\left(\sqrt{-a+b+c} \cdot w_{a}\right) \geqslant \frac{3 \sqrt{3 a b c}}{2} \end{array}$$

Here, $w_{a}, w_{b}, w_{c}$ are the angle bisectors of $\triangle A B C$ at $\angle A, \angle B, \angle C$, respectively, and $R$ and $r$ are the circumradius and inradius of $\triangle A B C$, respectively. The equalities in (32), (33), and (34) hold if and only if $\triangle A B C$ is an equilateral triangle.