The sentence you underlined in red abbreviates these steps:
$$(\mathbf{B}^{1/2} \mathbf{b} \cdot \mathbf{B}^{-1/2} \mathbf{d})^2 \leq (\mathbf{B}^{1/2} \mathbf{b} \cdot \mathbf{B}^{1/2} \mathbf{b})(\mathbf{B}^{-1/2} \mathbf{d} \cdot \mathbf{B}^{-1/2} \mathbf{d}) $$
This is the Cauchy-Schwarz inequality applied as requested. Now we use the transpose-based definition of the dot product $(\mathbf{a} \cdot \mathbf{b}) = \mathbf{a}^T \mathbf{b}$:
$$(\mathbf{b}^T (\mathbf{B}^{1/2})^T \mathbf{B}^{-1/2} \mathbf{d})^2 \leq (\mathbf{b}^T (\mathbf{B}^{1/2})^T \mathbf{B}^{1/2} \mathbf{b})
(\mathbf{d}^T (\mathbf{B}^{-1/2})^T \mathbf{B}^{-1/2} \mathbf{d})$$
Finally we take advantage of the the following observations:
- $\mathbf{B}$ is positive definite and therefore symmetric
- $\mathbf{B}^{1/2}$ is also positive definite and symmetric
- The inverse of a positive definite matrix is positive definite
As a result:
$$(\mathbf{B}^{1/2})^T \mathbf{B}^{-1/2} = \mathbf{B}^{1/2} \mathbf{B}^{-1/2} = \mathbf{I}$$
$$(\mathbf{B}^{1/2})^T \mathbf{B}^{1/2} = \mathbf{B}^{1/2} \mathbf{B}^{1/2} = \mathbf{B}$$
$$(\mathbf{B}^{-1/2})^T \mathbf{B}^{-1/2} = \mathbf{B}^{-1/2} \mathbf{B}^{-1/2} = \mathbf{B}^{-1}$$
Completing the proof.
You can see how this relates to the regular Cauchy-Schwarz inequality (for the dot product) by considering what it states when $\mathbf{B} = \mathbf{I}$ (Since $\mathbf{I}$ is a positive definite matrix). In this case we have
$$(\mathbf{b} \cdot \mathbf{d})^2 \leq (\mathbf{b} \cdot \mathbf{b}) (\mathbf{d} \cdot \mathbf{d})$$
Which is precisely the usual Cauchy-Schwarz inequality for the dot product.