Irish Mathematical Olympiad

Notice that
$$\begin{align*} & (a + b + c)(a^{p-1} + b^{p-1} + c^{p-1}) - 2(a^p + b^p + c^p) \\ &= (b + c)a^{p-1} + (c + a)b^{p-1} + (a + b)c^{p-1} - a^p - b^p - c^p \\ &= (b + c - a)a^{p-1} + (c + a - b)b^{p-1} + (a + b - c)c^{p-1} > 0, \end{align*}$$
by the triangle inequality, since $a$, $b$, $c$ are positive. Hence the left inequality holds.

Clearly, the right inequality holds if $p = 1$. So, suppose $p > 1$ and apply Hölder's inequality to $a+b+c$, with exponents $p$, $q$, where $q = p/(p-1)$, to get
$$a+b+c \le (a^p + b^p + c^p)^{1/p} \cdot 3^{1/q}.$$
Next, apply Hölder's inequality to $a^{p-1} + b^{p-1} + c^{p-1}$, with exponents $q$, $p$, to get
$$a^{p-1} + b^{p-1} + c^{p-1} \le (a^{q(p-1)} + b^{q(p-1)} + c^{q(p-1)})^{1/q} 3^{1/p} = (a^p + b^p + c^p)^{1/q} 3^{1/p}.$$
Combining these inequalities, the right inequality follows.

Here is an alternative proof of the second inequality. Observe that
$$\begin{aligned} & 3(a^p + b^p + c^p) - (a+b+c)(a^{p-1} + b^{p-1} + c^{p-1}) \\ &= 2(a^p + b^p + c^p) - (b+c)a^{p-1} - (c+a)b^{p-1} - (a+b)c^{p-1} \\ &= (a^p - a^{p-1}b - ab^{p-1} + b^p) + (b^p - b^{p-1}c - bc^{p-1} + c^p) \\ &\quad + (c^p - c^{p-1}a - ca^{p-1} + a^p) \\ &= (a^{p-1} - b^{p-1})(a-b) + (b^{p-1} - c^{p-1})(b-c) + (c^{p-1} - a^{p-1})(c-a) \\ &\ge 0, \end{aligned}$$
since $a$, $b$, $c$ are positive, and, for $r \ge 0$, the function $t \mapsto t^r$ is increasing on $(0, \infty)$, so that $(x^{p-1} - y^{p-1})(x - y) \ge 0$ for all positive $x$, $y$.