To prove in $\triangle ABC$, $\frac{r}{R}=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$, to prove $\frac{a b c}{\sqrt{2\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \geqslant \frac{r}{2 R}$, it suffices to prove
$$\frac{a b c}{\sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \geqslant 2 \sqrt{2} \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$

Considering the symmetry, it suffices to prove
$$\frac{a}{\sqrt{b^{2}+c^{2}}} \geqslant \sqrt{2} \sin \frac{A}{2}$$

It suffices to prove $\frac{a^{2}}{b^{2}+c^{2}} \geqslant 1-\cos A$, which is equivalent to proving
$$a^{2}-\left(b^{2}+c^{2}\right)(1-\cos A) \geqslant 0$$

By the cosine rule,
$$\begin{aligned} a^{2}-\left(b^{2}+c^{2}\right)(1+\cos A)= & b^{2}+c^{2}-2 b c \cos A-\left(b^{2}+c^{2}\right)(1-\cos A)= \\ & \left(b^{2}+c^{2}-2 b c\right) \cos A= \\ & (b-c)^{2} \cos A \end{aligned}$$

Since $(b-c)^{2} \geqslant 0, \cos A>0$, it follows that $a^{2}-\left(b^{2}+c^{2}\right)(1-\cos A) \geqslant 0$, thus $\frac{a^{2}}{b^{2}+c^{2}} \geqslant 1-\cos A$

That is, $\frac{a}{\sqrt{b^{2}+c^{2}}} \geqslant \sqrt{2} \sin \frac{A}{2}$, similarly,
$$\begin{array}{l} \frac{b}{\sqrt{c^{2}+a^{2}}} \geqslant \sqrt{2} \sin \frac{B}{2} \\ \frac{c}{\sqrt{c^{2}+a^{2}}} \geqslant \sqrt{2} \sin \frac{C}{2} \end{array}$$

Multiplying the above three inequalities, we get $\frac{a b c}{\sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \geqslant 2 \sqrt{2} \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$, thus the original inequality holds.