Without loss of generality, assume that $a_1 \le a_2 \le \cdots \le a_n.$ We have to prove that if $a_1+a_2 \ge 0,$ then $a_1x_1+a_2x_2+\cdots +a_n x_n \ge a_1x_1^2+a_2x_2^2+\cdots +a_nx_n^2;$ if the inequality $a_1x_1+a_2x_2+\cdots +a_nx_n \ge a_1x_1^2+a_2x_2^2+\cdots +a_nx_n^2$ holds, then $a_1+a_2 \ge 0.$ Firstly, we will prove a. The initial inequality can be written as \[a_1(x_1-x_1^2)+a_2(x_2-x_2^2)+\cdots +a_n(x_n-x_n^2) \ge 0.\] Since $a_2 \le a_3\le \cdots \le a_n$ and $x_2^2+x_3^2+\cdots +x_n^2 \le (x_2+x_3+\cdots +x_n)^2,$ we have \[\begin{aligned} a_2(x_2-x_2^2)+\cdots +a_n(x_n-x_n^2) &\ge a_2(x_2+x_3+\cdots +x_n-x_2^2-x_3^2-\cdots -x_n^2) \\ &\ge a_2\big[(x_2+x_3+\cdots +x_n)-(x_2+x_3+\cdots +x_n)^2\big] \\ &=a_2\big[(1-x_1)-(1-x_1)^2\big]=a_2(x_1-x_1^2).\end{aligned}\] It follows that \[a_1(x_1-x_1^2)+a_2(x_2-x_2^2)+\cdots +a_n(x_n-x_n^2) \ge (a_1+a_2)(x_1-x_1^2) \ge 0.\] For b, we can see that it follows immediately from the inequality by choosing $x_1=x_2=\frac{1}{2}$ and $x_3=x_4=\cdots =x_n=0.$

I am sorry, but I fail to understand what is the converse of the problem. For the statement itself I have a proof. Note that \[\left(\sum_{i=1}^{n}{a_ix_i}\right)\left(\sum_{i=1}^{n} x_i\right)=\sum_{i=1}^{n} a_ix_i^2+\sum_{1\le i<j\le n} (a_i+a_j)x_ix_j\] Give that $a_i+a_j\ge 0$ for any $i\neq j$, also that $x_i\ge 0 $ and $\sum x_i=1$. Thus the result follows immediately.

The converse would be to show the existence of the x's.

The condition is equivalent to \begin{align*} \left(\sum_{i=1}^n a_i x_i\right) \left(\sum_{i=1}^n x_i\right)& \geq \sum_{i=1}^n a_i x_i^2 \\ \sum_{i \neq j} a_i x_i x_j &\geq 0 \\ \sum_{i<j} (a_i+a_j)x_ix_j &\geq 0 \end{align*}which is obvious by the condition. Furthermore, if $a_i + a_j < 0$, let $x_k = 0$ for all $k \neq i, j$, and let $x_i = x_j = \frac 12$. Then the last inequality is violated, which is equivalent to the original. As a result, the converse must also be true.

The other posts already provide solutions given this problem statement.