IMO Hk TST

By the Cauchy-Schwarz inequality, we have
$$((q+r) + (r+p) + (p+q)) \left( \frac{1}{q+r} + \frac{1}{r+p} + \frac{1}{p+q} \right) \ge (1+1+1)^2 = 9.$$
Dividing both sides by $2(p+q+r)$, we obtain
$$\frac{1}{q+r} + \frac{1}{r+p} + \frac{1}{p+q} \ge \frac{9}{2(p+q+r)}.$$

Next, WLOG assume $p \ge q \ge r$. Then we have $\frac{1}{q+r} \ge \frac{1}{r+p} \ge \frac{1}{p+q}$ and $p^m \ge q^m \ge r^m$. By Chebyshev's inequality, we obtain
$$\frac{p^m}{q+r} + \frac{q^m}{r+p} + \frac{r^m}{p+q} \ge \frac{p^m+q^m+r^m}{3} \cdot \left( \frac{1}{q+r} + \frac{1}{r+p} + \frac{1}{p+q} \right).$$

By the power mean inequality, since $m > 1$, we have
$$\frac{p^m + q^m + r^m}{3} \geq \left( \frac{p+q+r}{3} \right)^m .$$
Combining these and the result of the first part, we obtain
$$\frac{p^m}{q+r} + \frac{q^m}{r+p} + \frac{r^m}{p+q} \geq \left(\frac{p+q+r}{3}\right)^m \cdot \frac{9}{2(p+q+r)} = \frac{(p+q+r)^{m-1}}{2 \cdot 3^{m-2}}.$$