In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$. The following was a question in the final of the Flanders Mathematics Olympiad 2018:

In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$.

In this competition, points are assigned for formulating a rigorous and mathematically sound proof.
I proved the above by contradiction. Let $\alpha = \angle BAC, \beta = \angle CBA, \gamma = \angle ACB$. Suppose $\gamma \le 60^\circ$:
$$\gamma \le 60^\circ \iff \sin(\gamma) \leq \frac{\sqrt{3}}{2}$$
Applying the sine rule:
$$\frac{\sin(\alpha)}{|BC|} = \frac{\sin(\beta)}{|AC|} = \frac{\sin(\gamma)}{|AB|}$$
$$\iff\frac{|AC|^3}{|AB|^3} + \frac{|BC|^3}{|AB|^3} = \frac{\sin^3(\alpha) + \sin^3(\beta)}{\sin^3(\gamma)} = 1$$
$$\iff \sin^3(\alpha) + \sin^3(\beta) = \sin^3(\gamma) \le \left( \frac{\sqrt{3}}{2} \right)^3$$
$$\iff\sin(\alpha) \le \frac{\sqrt(3)}{2}, \, \sin(\beta) \le \frac{\sqrt(3)}{2}\tag{1}$$
We also know that:
$$\alpha + \beta = 180^\circ - \gamma \ge 120^\circ\tag{2}$$
$$\alpha + \beta < 180^\circ\tag{3}$$
Without loss of generality, assume $\alpha \ge \beta$. From $(1)$, $(2)$ and $(3)$, it then follows that:
$$a \ge 120^\circ, \, \beta \le 60^\circ$$
$$\implies |BC| > |AB| \qquad \unicode{x21af}$$
Is this answer adequate enough? Can the notation be improved? Are there any alternative approaches to solve this problem?

In the standard notation we need to prove that
$$\frac{a^2+b^2-c^2}{2ab}<\cos60^{\circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$\sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$\measuredangle ACB>60^{\circ}$$ and we are done!
