Hong Kong competition problems, 2008

a. Let $x = \frac{b+c-a}{2}$, $y = \frac{c+a-b}{2}$ and $z = \frac{a+b-c}{2}$. Then we have
$$\begin{align*} \frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c} &\ge 3 \\ \frac{y+z}{2x} + \frac{z+x}{2y} + \frac{x+y}{2z} &\ge 3 \\ \frac{y}{x} + \frac{z}{x} + \frac{z}{y} + \frac{x}{y} + \frac{x}{z} + \frac{y}{z} &\ge 6. \end{align*}$$
$\Leftrightarrow$
$\Leftrightarrow$
This holds by the AM-GM inequality. Equality holds when $x = y = z$, i.e. $a = b = c$.

b. By the power mean inequality and the triangle inequality, we have
$$a^k + b^k \ge 2 \left(\frac{a+b}{2}\right)^k > 2 \left(\frac{c}{2}\right)^k = \frac{c^k}{2^{k-1}}.$$

c. By the power mean inequality, we have
$$a^k + b^k + c^k \ge 3 \left( \frac{a+b+c}{3} \right)^k = 3 \left( \frac{p}{3} \right)^k = \frac{p^k}{3^{k-1}}.$$
Equality holds when $a = b = c$.