Irish Mathematical Olympiad

If we use the abbreviation $u = a^3$ and $v = b^3 + c^3$, the given equation can be written as
$$\frac{u^2}{2v} + \frac{v^2}{2u} + \frac{2uv}{u+v} = u+v.$$
This is the same as
$$\frac{u^3 + v^3}{2uv} = u + v - \frac{2uv}{u+v} = \frac{u^2 + v^2}{u+v},$$
or $(u+v)(u^3+v^3) = 2uv(u^2+v^2)$, which simplifies to $u^4+v^4 = u^3v + uv^3$.
This is easily be seen to be the same as $(u-v)(u^3-v^3) = 0$, which can be written as $(u-v)^2(u^2+uv+v^2) = 0$. Because $u > 0$ and $v > 0$, this equation is only possible if $u = v$. This shows that the original equation is equivalent to $a^3 = b^3 + c^3$. In particular, $b < a$ and $c < a$ and so $b/a < 1$ and $c/a < 1$ and the angle $\angle A$, which is opposite $a$, is the largest internal angle of $\triangle ABC$. We now obtain
$$\left(\frac{b}{a}\right)^2 + \left(\frac{c}{a}\right)^2 > \left(\frac{b}{a}\right)^3 + \left(\frac{c}{a}\right)^3 = 1,$$
hence $b^2+c^2 > a^2$. The Cosine Rule $2bc \cos \angle A = b^2+c^2-a^2 > 0$ shows that $\cos \angle A > 0$ and $\angle A$ must be acute. Therefore, all angles in triangle $ABC$ are acute.