China Southeastern Mathematical Olympiad

Since
$$\frac{a^k}{a+b} + \frac{1}{4}(a+b) + \underbrace{\frac{1}{2} + \frac{1}{2} + \dots + \frac{1}{2}}_{k-2} \ge k \cdot \sqrt[k]{\frac{a^k}{2^k}} = \frac{k}{2}a,$$
then
$$\frac{a^k}{a+b} \ge \frac{k}{2}a - \frac{1}{4}(a+b) - \frac{k-2}{2}.$$
Similarly,
$$\frac{b^k}{b+c} \ge \frac{k}{2}b - \frac{1}{4}(b+c) - \frac{k-2}{2},$$
$$\frac{c^k}{c+a} \ge \frac{k}{2}c - \frac{1}{4}(c+a) - \frac{k-2}{2}.$$
Adding the three inequalities above, we obtain
$$\begin{aligned} & \frac{a^k}{a+b} + \frac{b^k}{b+c} + \frac{c^k}{c+a} \\ \ge & \frac{k}{2}(a+b+c) - \frac{1}{2}(a+b+c) - \frac{3}{2}(k-2) \\ = & \frac{k-1}{2}(a+b+c) - \frac{3}{2}(k-2) \\ \ge & \frac{3}{2}(k-1) - \frac{3}{2}(k-2) \\ = & \frac{3}{2}, \end{aligned}$$
as desired.