As shown in the figure, $PD \perp BC$, $PE \perp CA$, $PF \perp AB$.

Since $\angle AEP + \angle AFP = 90^\circ + 90^\circ = 180^\circ$, $A, E, P, F$ are concyclic, and $AP = d_A$ is a diameter of this circle, from which we obtain: $\angle EPF = 180^\circ - \angle A = \angle B + \angle C$.

Using the Law of Sines $EF = d_A \sin A$, and the Law of Cosines, we can derive
$$\begin{aligned} (d_A \sin A)^2 &= EF^2 = d_2^2 + d_3^2 - 2d_2 d_3 \cos \angle EPF \\ &= d_2^2 + d_3^2 - 2d_2 d_3 \cos(\angle B + \angle C) \\ &= d_2^2(\cos^2 C + \sin^2 C) + d_3^2(\cos^2 B + \sin^2 B) - 2d_2 d_3(\cos B \cos C - \sin B \sin C) \\ &= (d_2 \cos C - d_3 \cos B)^2 + (d_2 \sin C + d_3 \sin B)^2 \\ &\geq (d_2 \sin C + d_3 \sin B)^2. \end{aligned}$$
Therefore we have $d_A \sin A > d_2 \sin C + d_3 \sin B$, that is, $d_A \geq \dfrac{d_2 \sin C + d_3 \sin B}{\sin A}$.

Similarly we can prove
$$d_B \geq \dfrac{d_3 \sin A + d_1 \sin C}{\sin B}, \quad d_C \geq \dfrac{d_1 \sin B + d_2 \sin A}{\sin C}.$$
Therefore
$$\begin{aligned} d_A + d_B + d_C &\geq \dfrac{d_2 \sin C + d_3 \sin B}{\sin A} + \dfrac{d_3 \sin A + d_1 \sin C}{\sin B} + \dfrac{d_1 \sin B + d_2 \sin A}{\sin C} \\ &= d_1 \left( \dfrac{\sin C}{\sin B} + \dfrac{\sin B}{\sin C} \right) + d_2 \left( \dfrac{\sin A}{\sin C} + \dfrac{\sin C}{\sin A} \right) + d_3 \left( \dfrac{\sin B}{\sin A} + \dfrac{\sin A}{\sin B} \right) \\ &\geq 2(d_1 + d_2 + d_3). \end{aligned}$$