Croatian Mathematical Competitions

We have
$$\begin{aligned} \frac{a_1^3}{a_1^2 + a_2 a_3} &= \frac{a_1^3 + a_1 a_2 a_3 - a_1 a_2 a_3}{a_1^2 + a_2 a_3} = a_1 - a_1 a_2 a_3 \cdot \frac{1}{a_1^2 + a_2 a_3} \\ &\geqq \overset{\text{A-G}}{a_1 - a_1 a_2 a_3} \cdot \frac{1}{2a_1\sqrt{a_2 a_3}} = a_1 - \frac{1}{2}\sqrt{a_2 a_3} \overset{\text{A-G}}{\geqq} a_1 - \frac{a_2 + a_3}{4} \end{aligned}$$
Summing those $n$ inequalities gives us:
$$\begin{aligned} & \frac{a_1^3}{a_1^2 + a_2 a_3} + \frac{a_2^3}{a_2^2 + a_3 a_4} + \dots + \frac{a_{n-1}^3}{a_{n-1}^2 + a_n a_1} + \frac{a_n^3}{a_n^2 + a_1 a_2} \\ & \geqslant \left(a_1 - \frac{a_2+a_3}{4}\right) + \left(a_2 - \frac{a_3+a_4}{4}\right) + \dots + \left(a_{n-1} - \frac{a_n+a_1}{4}\right) + \left(a_n - \frac{a_1+a_2}{4}\right) \\ & = \frac{a_1 + a_2 + \dots + a_n}{2} = \frac{1}{2}. \end{aligned}$$