We know that
$$
\begin{aligned}
r_1 & =\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b} \text { and } r_3=\frac{\Delta}{s-c} . \\
\therefore \quad & r_1+r_2+r_3=\frac{\Delta}{s-a}+\frac{\Delta}{s-b}+\frac{\Delta}{s-c} \\
& =\Delta\left[\frac{1}{s-a}+\frac{1}{s-b}\right]+\frac{\Delta}{s-c}-\frac{\Delta}{s}+\frac{\Delta}{s} \\
& =\Delta\left[\frac{s-b+s-a}{(s-a)(s-b)}\right]+\frac{\Delta(s-s+c)}{s(s-c)}+\frac{\Delta}{s} \\
& =\frac{\Delta c}{(s-a)(s-b)}+\frac{\Delta c}{s(s-c)}+\frac{\Delta}{s} \\
& =\Delta c\left[\frac{s^2-c s+s^2-a s-b s+a b}{s(s-a)(s-b)(s-c)}\right]+\frac{\Delta}{s} \\
& =\left[\frac{\Delta c\left[2 s^2-s(a+b+c)+a b\right.}{s(s-a)(s-b)(s-c)}\right]+\frac{\Delta}{s} \\
& =\frac{\Delta c\left[2 s^2-2 s^2+a b\right]}{\Delta^2}+\frac{\Delta}{s}
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{a b c}{\Delta}+\frac{\Delta}{s} \\
& =4 R+r \quad\left[\because r=\frac{\Delta}{s} \text { and } R=\frac{a b c}{4 \Delta}\right]
\end{aligned}
$$