We use the following

Lemma. For any positive $a$ and $b$ the following inequality is valid
$$(a^3 + b^3)^2 \geq 2ab(a^4 + b^4). \qquad (*)$$
Indeed,
$$(*) \Leftrightarrow a^6 - 2a^5b + 2a^3b^3 - 2ab^5 + b^6 \geq 0 \Leftrightarrow (a-b)^2(a^4 - a^2b^2 + b^4) \geq 0$$
which is true.

Now, using the lemma, we have
$$\sum_{i=1}^{n} \frac{x_i^8}{(x_i^4 + x_{i+1}^4)x_{i+1}} = \sum_{i=1}^{n} \frac{x_i^9}{(x_i^4 + x_{i+1}^4)x_i x_{i+1}} \geq \sum_{i=1}^{n} \frac{2x_i^9}{(x_i^3 + x_{i+1}^3)^2} =$$
$$= 2 \sum_{i=1}^{n} \frac{(x_i^3)^3}{(x_i^3 + x_{i+1}^3)^2} \ge [\text{the H\"older inequality}] \ge 2 \frac{\left(\sum_{i=1}^{n} x_i^3\right)^3}{\left(2 \sum_{i=1}^{n} x_i^3\right)^2} = \frac{1}{2} \sum_{i=1}^{n} x_i^3 \ge \frac{1}{2} n \left(\sqrt[n]{x_1 \cdots x_n}\right)^3 = \frac{n}{2},$$
as was to be proved.