Let the radii of the circles $c_1$, $c_2$, $c_3$ be $r_1$, $r_2$, $r_3$, and the radii of the circles $k_1$, $k_2$, $k_3$ be $R_1$, $R_2$, $R_3$, respectively (Fig. 28). By assumptions,
$$\begin{aligned} R_1 + R_2 &= |K_1K_2| = \frac{1}{2}|C_1C_2|, \\ R_2 + R_3 &= |K_2K_3| = \frac{1}{2}|C_2C_3|, \\ R_3 + R_1 &= |K_3K_1| = \frac{1}{2}|C_3C_1|, \end{aligned}$$
which sum up to $2R_1 + 2R_2 + 2R_3 = \frac{1}{2}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$. The assumptions also imply inequalities
$$\begin{aligned} r_1 + R_3 &\le \frac{1}{2}|C_1C_2|, & R_3 + r_2 &\le \frac{1}{2}|C_1C_2|, \\ r_2 + R_1 &\le \frac{1}{2}|C_2C_3|, & R_1 + r_3 &\le \frac{1}{2}|C_2C_3|, \\ r_3 + R_2 &\le \frac{1}{2}|C_3C_1|, & R_2 + r_1 &\le \frac{1}{2}|C_3C_1|, \end{aligned}$$
which sum up to $2r_1 + 2r_2 + 2r_3 + 2R_1 + 2R_2 + 2R_3 \le |C_1C_2| + |C_2C_3| + |C_3C_1|$.

i) Altogether, we obtain the inequality $2r_1 + 2r_2 + 2r_3 \le \frac{1}{2}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$, which implies $r_1 + r_2 + r_3 \le \frac{1}{4}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$ as desired.

Figure 28

ii) Suppose that $r_1 + r_2 + r_3 = \frac{1}{4}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$. Then all inequalities above must hold as equalities. The equalities $r_1 + R_3 = \frac{1}{2}|C_1C_2| = r_2 + R_3$ imply $r_1 = r_2$, analogously $r_1 = r_3$. Denoting $r = r_1 = r_2 = r_3$, we get
$$r + R_3 = \frac{1}{2}|C_1C_2| = R_1 + R_2,$$
$$r + R_2 = \frac{1}{2}|C_1C_3| = R_1 + R_3,$$
where summing side-by-side gives $2r + R_2 + R_3 = 2R_1 + R_2 + R_3$, i.e., $R_1 = r$. Analogously, $R_2 = R_3 = r$. Thus all sides of the triangle $C_1C_2C_3$ have length $4r$.