Without going into the details of the theorem, we have the following definitions: $\mathbf{B}$ is a non-empty finite set of integers and $\mathbf{T}$ be a set of prime powers.
Suppose for each $t\in\mathbf{T}$, there exists $u(t)$ such that $|\mathbf{B}\pmod t|\le u(t)$.
Define $Z(\mathbf{B};t,r):=|\{ b\in\mathbf{B} : b\equiv r\pmod t \} |$ and hence we have that $|\mathbf{B}| = \sum_{r\pmod t} Z(\mathbf{B};t,r)$.
They then claim that we have $\sum_{r\pmod t} Z(\mathbf{B};t,r) \le u(t)^{1/2}( \sum_{r\pmod t} Z(\mathbf{B};t,r)^2) ^{1/2}$, but I'm not sure how they got this.
Any help clarifying this would be greatly appreciated. Thanks in advance.
I have in fact figured it out. Using the Cauchy-Schwarz inequality with the first series being $Z(\mathbf{B}; t,r)$ and the second just being $1^2$, we get that $$\left(\sum_{r\mod t} Z(\mathbf{B}; t,r)\right)^2 \le \left(\sum_{r\mod t}Z(\mathbf{B};t,r)^2\right)\left(\sum_{r\mod t} 1\right).$$
But we have that $u(t)$ is exactly the number of congruence classes we have mod $t$ (and hence the number of elements we sum over). Thus $(\sum_{r\mod t} 1) = u(t)$ and the result follows.