Given $a \geqslant b \geqslant c$, and $a=c+\alpha+\beta, b=c+\alpha, c \geqslant \sqrt{2 \alpha \beta+2 \beta^{2}}+\beta, \alpha, \beta \in \overline{\mathbf{R}^{-}}$, then
$$\begin{aligned} \text { Equation (11) } \Leftrightarrow \frac{\left(\alpha^{2}+\alpha \beta+\beta^{2}\right) c+\left(2 \alpha \beta^{2}+\beta^{3}\right)}{c(c+\alpha)(c+\alpha+\beta)} & \geqslant \frac{3 \alpha \beta(\alpha+\beta)}{c(c+\alpha)(c+\alpha+\beta)} \Leftrightarrow \\ \left(\alpha^{2}+\alpha \beta+\beta^{2}\right) c+\left(2 \alpha \beta^{2}+\beta^{3}\right) & \geqslant 3 \alpha \beta(\alpha+\beta) \end{aligned}$$

Substituting $c \geqslant \sqrt{2 \alpha \beta+2 \beta^{2}}+\beta$ and simplifying, we get
$$\begin{array}{l} \left(\alpha^{2}+\alpha \beta+\beta^{2}\right) c+\left(2 \alpha \beta^{2}+\beta^{3}\right)-3 \alpha \beta(\alpha+\beta) \geqslant \\ \left(\alpha^{2}+\alpha \beta+\beta^{2}\right)\left(\sqrt{2 \alpha \beta+2 \beta^{2}}+\beta\right)+2 \alpha \beta^{2}+\beta^{3}-3 \alpha \beta(\alpha+\beta)= \\ \left(\alpha^{2}+\alpha \beta+\beta^{2}\right) \sqrt{2 \alpha \beta+2 \beta^{2}}-2 \beta\left(\alpha^{2}-\beta^{2}\right)= \\ \sqrt{2 \alpha \beta+2 \beta^{2}}\left[\left(\alpha^{2}+\alpha \beta+\beta^{2}\right)-\sqrt{2}(\alpha-\beta) \sqrt{\beta(\alpha+\beta)}\right] \end{array}$$

Thus, to prove Equation (11), it suffices to prove
$$\alpha^{2}+\alpha \beta+\beta^{2} \geqslant \sqrt{2}(\alpha-\beta) \sqrt{\beta(\alpha+\beta)}$$

When $\alpha<\beta$, the above inequality is clearly true;
When $\alpha \geqslant \beta$, it suffices to prove the inequality after squaring both sides. In this case,
$$\left(\alpha^{2}+\alpha \beta+\beta^{2}\right)^{2}-2(\alpha-\beta)^{2}\left(\alpha \beta+\beta^{2}\right)=\alpha^{4}+5 \alpha^{2} \beta^{2}+4 \alpha \beta^{3}-\beta^{4} \geqslant 0$$

Note that there is a stronger inequality:
$$1-\frac{2 r}{R} \geqslant \frac{(2 \sqrt{2}+1)|(a-b)(b-c)(a-c)|}{a b c}$$

Equality holds in Equation (12) if and only if $a=b=c$.
From the above proof, we know that in this case,
$$\begin{array}{l} \left(\alpha^{2}+\alpha \beta+\beta^{2}\right) c+2 \alpha \beta^{2}+\beta^{3}-(2 \sqrt{2}+1) \alpha \beta(\alpha+\beta) \geqslant \\ \left(\alpha^{2}+\alpha \beta+\beta^{2}\right)\left(\sqrt{2 \alpha \beta+2 \beta^{2}}+\beta\right)+2 \alpha \beta^{2}+\beta^{3}-(2 \sqrt{2}+1) \alpha \beta(\alpha+\beta)= \\ \left(\alpha^{2}+\alpha \beta+\beta^{2}\right) \sqrt{2 \alpha \beta+2 \beta^{2}}+\beta(\alpha+\beta)^{2}+\beta\left(\alpha \beta+\beta^{2}\right)-(2 \sqrt{2}+1) \alpha \beta(\alpha+\beta)= \\ \left(\alpha \beta+\beta^{2}\right)\left[\left(\frac{\sqrt{2} \alpha^{2}}{\sqrt{\alpha \beta+\beta^{2}}}+\sqrt{2} \cdot \sqrt{\alpha+\beta+\beta^{2}}\right)+\alpha+2 \beta-(2 \sqrt{2}+1) \alpha\right] \geqslant \\ \left(\alpha \beta+\beta^{2}\right)[2 \sqrt{2} \alpha+\alpha+2 \beta-(2 \sqrt{2}+1) \alpha]= \\ 2\left(\alpha \beta+\beta^{2}\right) \beta \geqslant 0 \end{array}$$