China Southeastern Mathematical Olympiad

Since $$\frac{a^k}{a+b}+\frac{1}{4}(a+b)+\underset{k-2}{\underbrace{\frac{1}{2}+\frac{1}{2}+\cdots +\frac{1}{2}}}\ge k\cdot \sqrt[k]{\frac{a^k}{2^k}}=\frac{k}{2}a,$$ 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 $$\begin{array}{rl}& \frac{a^k}{a+b}+\frac{b^k}{b+c}+\frac{c^k}{c+a}\\ \ge & \frac{k}{2}(a+b+c)-\frac{1}{2}(a+b+c)-\frac{3}{2}(k-2)\\ =& \frac{k-1}{2}(a+b+c)-\frac{3}{2}(k-2)\\ \ge & \frac{3}{2}(k-1)-\frac{3}{2}(k-2)\\ =& \frac{3}{2},\end{array}$$ as desired.