, 2014

First observe that for $u \le 1$ and $v \le 1$ we always have $(1-u)(1-v) \ge 0$, hence $u + v \le 1 + uv$. We apply this now to $u = a_i b_i$ and $v = b_i c_i$. As $b_i = \pm 1$, we have $b_i^2 = 1$ and so $uv = a_i c_i$. This shows that
$$a_i b_i + b_i c_i \le 1 + a_i c_i \quad \text{and so} \quad \sum_{i=1}^n a_i b_i + \sum_{i=1}^n b_i c_i \le n + \sum_{i=1}^n a_i c_i$$
for all $n \ge 1$. Multiplying this inequality by $3$ and taking into account that $a_i^2 = b_i^2 = c_i^2 = 1$, and so $\sum_{i=1}^n a_i^2 = \sum_{i=1}^n b_i^2 = \sum_{i=1}^n c_i^2 = n$, we obtain
$$3 \sum_{i=1}^n a_i b_i + 3 \sum_{i=1}^n b_i c_i \le \sum_{i=1}^n a_i^2 + \sum_{i=1}^n b_i^2 + \sum_{i=1}^n c_i^2 + 3 \sum_{i=1}^n a_i c_i,$$
which gives the desired inequality for $n = 2014$.