Question 10
Let $a_1,a_2,\ldots,a_n$ be positive real numbers such that $a_1+a_2+\cdots+a_n=1$. Prove that the following inequality holds:
$$\begin{aligned}&\frac{a_1^3}{a_1^2+a_2a_3}+\frac{a_2^3}{a_2^2+a_3a_4}+\cdots\\&\quad+\frac{a_{n-1}^3}{a_{n-1}^2+a_na_1}+\frac{a_n^3}{a_n^2+a_1a_2}\ge\frac12.\end{aligned}$$
$$\begin{aligned}&\frac{a_1^3}{a_1^2+a_2a_3}+\frac{a_2^3}{a_2^2+a_3a_4}+\cdots\\&\quad+\frac{a_{n-1}^3}{a_{n-1}^2+a_na_1}+\frac{a_n^3}{a_n^2+a_1a_2}\ge\frac12.\end{aligned}$$