Austria competition problems, 2014

(a) By the cosine theorem, we have $c^2 = a^2 + b^2 - 2ab \cos \gamma$. Since $\cos \gamma > 0 \iff \gamma < 90^\circ$, we see that $a^2 + b^2 > c^2$ is equivalent to $\gamma < 90^\circ$. Since the analogous results hold for the other two inequalities, we see that these all hold exactly for acute angled triangles $ABC$.

(b) Without loss of generality, assume $c \ge b \ge a$. Since $p = \frac{a}{c} \le 1$ and $q = \frac{b}{c} \le 1$, the three inequalities hold iff $p^n + q^n > 1$ holds for all positive integers $n$. If both $p$ and $q$ are less than $1$, there exists sufficiently large values of $n$, such that $p^n < \frac{1}{2}$ and $q^n < \frac{1}{2}$ both hold, and therefore $p^n + q^n < 1$. It therefore follows that $q = 1$ and $p \le 1$ must hold. The triangle must therefore be isosceles with $b = c$ and $a \le c$, i.e. with $\alpha \le 60^\circ$. The inequalities certainly hold for such triangles, and the proof is complete.