Let $s_i = t_i^2/4$. Since $t_i^3 \le 2t_i^2 = 8s_i$, so for $0 \le s_1, \dots, s_n \le 1$, it suffices to prove
$$\frac{s_1}{s_2+1} + \frac{s_2}{s_3+1} + \dots + \frac{s_n}{s_1+1} \le \frac{n}{2}.$$
Now for any $0 \le x, y \le 1$, we consider:
$$(1+y)(1-y+x) = 2x + (1-y)(1+y-x) \ge 2x$$
which implies:
$$\frac{x}{y+1} \le \frac{1-y+x}{2}.$$

Applying this to the set $\{s_i\}$ cyclically:
$$\sum_{i} \frac{s_i}{s_{i+1} + 1} \le \frac{n}{2} + \frac{1}{2} \sum_{i} (s_i - s_{i+1}) = \frac{n}{2}$$
so the inequality is proved. Equality holds for each $i$ when $s_{i+1} = 1$ or $(s_i, s_{i+1}) = (1, 0)$, i.e., when the sequence $\{t_1, t_2, \dots, t_n, t_1\}$ consists only of 0 and 2, and contains no consecutive 0's.

*Remark.* Refer to the solution of problem E5.