In that case, assume WLOG that $a_n<0$. Then $a_1+\ldots+a_{n-1}+a_n<a_1+\ldots+a_{n-1}\le \sqrt{(n-1)(a_1^2+\ldots+a_{n-1}^2)}\le \sqrt{(n-1)(a_1^2+\dots+a_{n-1}^2+a_n^2)}$, which contradicts the hypothesis of b).

So, for b) : Assume that for such numbers, we have $a_n < 0$. Then : $a_1 + \cdots + a_{n-1} > a_1 + \cdots a_{n-1} + a_n \geq \sqrt {(n-1)(a_1^2 + \cdots +a_n^2)} > 0$. It follows from (1) that $(n-1)(a_1^2 + \cdots + a_{n-1}^2) \geq (a_1 + \cdots + a_{n-1})^2 > (n-1)(a_1^2 + \cdots +a_n^2)$ so that $(n-1)a_n^2 < 0$, a contradiction. Pierre.

Part A is extremely obvious using any of these theorems: QM-AM, Jensen's inequality, Chebycheff's inequality, Engel's inequality. I think It can also be done by Induction but that's much more than these 3 line solutions