Since $0 = (a_1 + a_2 + a_3 + a_4 + a_5)^2 = a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + 2 \sum_{i<j} a_i a_j$, it follows that $\sum_{i<j} (a_i - a_j)^2 = 4(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) - 2 \sum_{i<j} a_i a_j = 5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2)$. On the other side, $\sum_{i<j} (a_i - a_j)^2 \le \sum_{i<j} |a_i - a_j|$, for $|a_i - a_j| \le 1$. Assuming $a_1 \le a_2 \le a_3 \le a_4 \le a_5$, we have $\sum_{i<j} |a_i - a_j| = 4(a_5 - a_1) + 2(a_4 - a_2) \le 4 + 2 = 6$. So $5(a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) \le 6$, as requested to be shown.