We consider the configuration shown in the diagram below. (Our proof uses directed lengths and can be easily modified for different configurations.)

Let $a, b, c$ denote the side lengths of $BC, CA, AB$, and let $A, B, C$ denote $\angle A, \angle B, \angle C$, respectively. Suppose that the incircle touches sides $BC, CA, AB$ at $P, Q, R$, respectively. It is well known that
$$FB = CE = QA = AR = \frac{b+c-a}{2}. \qquad (25)$$
Denote by $E_a, F_a$ the feet of the perpendiculars from $E, F$ to line $BC$. It is clear that $EF \ge E_aF_a$. By (25), we have
$$FE \ge F_aE_a = BC - (BF_a + E_aC) = a - \frac{b+c-a}{2} (\cos B + \cos C).$$
Summing the above inequality and its cyclic analogues yields
$$EF + FD + DE \ge a + b + c - \sum_{\text{cyc}} \frac{b+c-a}{2} (\cos B + \cos C)$$
or
$$EF + FD + DE \ge a + b + c - (a \cos A + b \cos B + c \cos C). \qquad (26)$$
By the sum-to-product formulas, we have
$$\frac{1}{2}(\sin 2A + \sin 2B) = \sin(A+B)\cos(A-B) = \sin C \cos(A-B) \ge \sin C.$$
Summing this inequality and its cyclic analogues yields
$$\sin A + \sin B + \sin C \ge 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C$$
Multiplying both sides by $2R$ and applying the extended Law of Sines, we obtain
$$a + b + c \ge 2a \cos A + 2b \cos B + 2c \cos C. \qquad (27)$$
Substituting (27) into (26) yields the desired
$$EF + FD + DE \ge \frac{a+b+c}{2}.$$

We maintain the notations of the first solution. The result clearly follows from the following two lemmas.

Lemma 1. The sum of the perimeters of triangles $DEF$ and $PQR$ is at least the perimeter of triangle $ABC$.

Proof. By symmetry, it suffices to show that $DE + PQ \ge AB$. Let $M$ and $N$ be the midpoints of segments $CA$ and $CB$, respectively. We want to show that $DE + PQ \ge AB = 2MN$. As in the first solution, we have $CE = AQ$ and $CD = BP$ so that $M$ and $N$ are midpoints of segments $PD$ and $QE$, respectively. Computing using vectors, we obtain
$$\begin{aligned} \overrightarrow{MN} &= \overrightarrow{MC} + \overrightarrow{CN} = \frac{1}{2}(\overrightarrow{QC} + \overrightarrow{EC}) + \frac{1}{2}(\overrightarrow{CP} + \overrightarrow{CD}) \\ &= \frac{1}{2}(\overrightarrow{QC} + \overrightarrow{CP}) + \frac{1}{2}(\overrightarrow{EC} + \overrightarrow{CD}) = \frac{1}{2}(\overrightarrow{QP} + \overrightarrow{ED}). \end{aligned}$$
By the triangle inequality, we have $MN \le \frac{1}{2}(QP + ED)$, which implies
$$PQ + DE \ge 2MN = AB. \quad \square$$

Lemma 2. The perimeter of triangle $ABC$ is at least twice that of triangle $PQR$.

Proof. Note that $PQ = 2CP \sin \frac{C}{2}$, $QR = 2AQ \sin \frac{A}{2}$, and $RP = 2BR \sin \frac{B}{2}$. The perimeter of triangle $PQR$ is equal to
$$S_{PQR} = (b+c-a) \sin \frac{A}{2} + (c+a-b) \sin \frac{B}{2} + (a+b-c) \sin \frac{C}{2}.$$
By symmetry, we may assume that $a \le b \le c$. This yields the orderings
$$A \le B \le C, \quad b+c-a \ge c+a-b \ge a+b-c, \quad \text{and} \quad \sin \frac{A}{2} \le \sin \frac{B}{2} \le \sin \frac{C}{2}.$$
By Chebyshev's inequality, we have
$$S_{PQR} \le \frac{1}{3} \left( (b+c-a) + (c+a-b) + (a+b-c) \right) \left( \sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \right),$$
so it suffices to show that
$$\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2} \le \frac{3}{2}. \qquad (28)$$
But (28) follows from Jensen's inequality for $y = \sin x$ (which is concave for $0 \le x \le \frac{\pi}{2}$). $\square$

Consider the following diagram, which contains several copies of $ABC$ rotated and translated so that $ABC$, $A_2B_2C_2$, $A_4B_4C_4$, and $A_6B_6C_6$ are congruent and $B$, $C = A_2$, $B_2 = C_4$, $A_4 = B_6$, and $C_6$ are collinear. Define $D_i$, $E_i$, and $F_i$ to be the images of $D$, $E$, and $F$ in $A_iB_iC_i$.

Observe that $\triangle ECE_2 \simeq \triangle FBD$, $\triangle D_2B_2D_4 \simeq \triangle EAF$, and $\triangle F_4A_4F_6 \simeq \triangle DAE$. Using these congruences, the triangle inequality, and the fact that $FD \parallel F_6D_6$, we obtain
$$2(DE + EF + FA) \ge FE + EE_2 + E_2D_2 + D_2D_4 + D_4F_4 + F_4F_6 \ge FF_6 = AB + BC + CA.$$