It is well known that $AD = \frac{2bc}{b+c} \cos \frac{A}{2} = \frac{bc\sqrt{3}}{b+c}$. Let $h = AM$ be the length of the altitude from $A$ in the triangle $ABC$. From the theorem of the internal bisector we get
$$BD = \frac{ac}{b+c}, \quad CD = \frac{ab}{b+c}.$$
Let $p_{ABD} = \frac{BD + DA + AB}{2}$ be the half perimeter of triangle $ABD$. If we denote $[ABD]$ the surface of the triangle $ABD$, then
$$r_B = \frac{[ABD]}{p_{ABD}} = \frac{h \cdot BD}{2 \cdot p_{ABD}} = \frac{h \cdot BD}{BD + AD + AB} = \frac{\frac{h \cdot ac}{b+c}}{\frac{ac}{b+c} + \frac{bc\sqrt{3}}{b+c} + c} = \frac{ah}{2p + \sqrt{3} \cdot b}.$$
Analogously we get $r_C = \frac{ah}{2p + \sqrt{3} \cdot c}$, where $p = \frac{a+b+c}{2}$. From this we have
$$\frac{1}{r_B} + \frac{1}{r_C} = \frac{2p + b\sqrt{3}}{ah} + \frac{2p + c\sqrt{3}}{ah} = \frac{4p + (b+c)\sqrt{3}}{ah}.$$
But $r = \frac{[ABC]}{p} = \frac{ah}{2p}$, so $ah = 2pr$. Thus,
$$\frac{1}{r_B} + \frac{1}{r_C} = \frac{4p + (b+c)\sqrt{3}}{2pr}.$$
Now, $b = AC$, $c = AB$, and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so $[ABC] = \frac{1}{2}bc\sin 60^\circ = \frac{bc\sqrt{3}}{4}$, and $r = \frac{[ABC]}{p} = \frac{bc\sqrt{3}}{4p}$, so $\frac{1}{r} = \frac{4p}{bc\sqrt{3}}$.

Also,
$$\frac{1}{b} + \frac{1}{c} = \frac{b + c}{bc}.$$
Therefore,
$$2\left(\frac{1}{r} + \frac{1}{b} + \frac{1}{c}\right) = 2\left(\frac{4p}{bc\sqrt{3}} + \frac{b + c}{bc}\right) = \frac{8p}{bc\sqrt{3}} + \frac{2(b + c)}{bc}.$$
But $ah = 2pr$, so $ah = 2p \cdot \frac{bc\sqrt{3}}{4p} = \frac{bc\sqrt{3}}{2}$.

Now, substitute $ah$ into the earlier formula:
$$\frac{1}{r_B} + \frac{1}{r_C} = \frac{4p + (b+c)\sqrt{3}}{ah} = \frac{4p + (b+c)\sqrt{3}}{\frac{bc\sqrt{3}}{2}} = \frac{2[4p + (b+c)\sqrt{3}]}{bc\sqrt{3}} = \frac{8p}{bc\sqrt{3}} + \frac{2(b+c)}{bc}.$$
This matches the previous expression, so
$$\frac{1}{r_B} + \frac{1}{r_C} = 2\left(\frac{1}{r} + \frac{1}{b} + \frac{1}{c}\right).$$