Let us denote the width of the triangle as $ABC$, and the diameters of the hedgehogs as $d_a, d_b, d_c$, respectively. Then suppose $d_a + d_b > 2c$. This means that any point on the side $AB$ of the triangle is inside one of the hedgehogs, and therefore it is impossible to reach this side. Then we have $d_a + d_b \le 2c$, and similarly $d_a + d_c \le 2b$ and $d_b + d_c \le 2a$, which implies that $d_a + d_b + d_c \le a + b + c$.
We will prove that there exists an example where equality is achieved in this inequality. Let $I$ be the center of the inscribed circle of the triangle and let $A_1, B_1, C_1$ be its points of tangency to the sides of the triangle (Fig. 5). Then it suffices to consider the hedgehogs sitting at the vertices and whose corresponding circles have radii $AB_1 = AC_1$, $BA_1 = BC_1$, and $CA_1 = CB_1$. Since $IA_1 \perp BC$, $IB_1 \perp AC$, and $IC_1 \perp AB$, this means that $IA_1, IB_1, IC_1$ are tangents to the hedgehogs sitting at the corresponding vertices, and therefore they will not touch any hedgehogs, which is what we wanted to prove.

Fig. 5