By expanding, the condition a) is equivalent to $\sum \vert b_k\vert^2-\vert a_k\vert^2 \ge 2\text{Re}(z\overline{S})$ where $S=\sum b_k-a_k$. So clearly, putting $z=0$ this implies $\sum \vert b_k\vert^2 \ge \sum \vert a_k\vert^2$. Moreover, it can only be true if $S=0$ since otherwise $z\overline{S}$ takes all complex values and in particular arbitrarily large real parts. Conversely, if $S=0$ and $\sum \vert b_k\vert^2 \ge \sum \vert a_k\vert^2$ it is clear that the above inequality holds.