$$\leqslant \frac{2 \sqrt{b c s(s-a)}}{2 \sqrt{b c}}=\sqrt{s(s-a)}$$

It can be known that $r_{h}+r_{c} \geqslant 2 w_{c}$,
Similarly, $r_{c}+r_{u} \geqslant 2 w_{b}, r_{u}+r_{h} \geqslant 2 w_{c}$.
Thus, inequalities (3) and (4) can be strengthened to
$$\begin{array}{l} \left(\frac{w_{a}}{r_{h}+r_{c}}\right)^{\lambda}+\left(\frac{w_{b}}{r_{c}+r_{a}}\right)^{\prime}+\left(\frac{w_{c}}{r_{u}+r_{h}}\right)^{\lambda} \\ \leqslant \frac{3}{2^{\lambda}}\left(\lambda \in R^{+}\right) \\ \frac{w_{a}}{r_{h}+r_{c}} \frac{w_{b}}{r_{c}+r_{a}} \frac{w_{c}}{r_{a}+r_{h}} \leqslant \frac{1}{8} \end{array}$$

This paper provides similar and strengthened versions of (1) and (2).
$$\begin{array}{l} \because r_{b}+r_{c} \\ =\sqrt{\frac{s(s-c)(s-a)}{s-b}}+\sqrt{\frac{s(s-a)(s-b)}{s-c}} \\ \geqslant 2 \sqrt{s(s-a)}, \quad\left(s=\frac{a+b+c}{2}\right) \\ h_{a}=\frac{2 \sqrt{s(s-a)(s-b)(s-c)}}{a} \\ =\sqrt{s(s-a)} \cdot \frac{1}{a} \cdot 2 \sqrt{(s-b)(s-c)} \\ \leqslant \frac{\sqrt{s(s-a)}[(s-b)+(s-c)]}{a} \\ =\sqrt{s(s-a)}, \\ \therefore r_{h}+r_{c} \geqslant 2 h_{a} \text {. } \end{array}$$

Similarly, we have
$$r_{c}+r_{u} \geqslant 2 h_{b} . \quad r_{a}+r_{h} \geqslant 2 h_{c} .$$

From this, it is easy to obtain the similar inequalities of (1) and (2):
$$\begin{array}{l} \frac{h_{a}}{r_{b}+r_{c}}+\frac{h_{b}}{r_{c}+r_{a}}+\frac{h_{c}}{r_{a}+r_{b}} \leqslant \frac{3}{2} \\ \frac{h_{a}}{r_{b}+r_{c}} \frac{h_{b}}{r_{c}+r_{a}} \frac{h_{c}}{r_{a}+r_{h}} \leqslant \frac{1}{8} \end{array}$$

Let the angle bisectors of $\triangle A B C$ at angles $A$, $B$, and $C$ be $w_{a}$, $w_{b}$, and $w_{c}$, respectively, where
$$w_{a}=\frac{2 \sqrt{b c s(s-a)}}{b+c}$$
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