From C-S: $\sum_{i=1}^n a_i^5\sum_{i=1}^n a_i \geq (\sum_{i=1}^n a_i^3)^2$ $5\cdot \sum_{i=1}^n a_i \geq 9$ $\sum_{i=1}^n a_i \geq \frac{9}{5} > \frac{9}{6} = \frac{3}{2}$

By Cauchy-Schwarz, $$(a_1+a_2+\dots +a_n)(a_1^5+a_2^5+\dots a_n^5)\ge (a_1^3+a_2^3+\dots a_n^3)^2$$This is equivalent to $$\sum_{i=1}^n a_i \ge \frac{9}{5}=1.8 > 1.5 =\frac{3}{2}.$$