fourier analysis – How is the Cauchy-Schwarz inequality used to bound this derivative?

In “Hardy’s Uncertainty Principle, Convexity and Schrödinger Evolutions” (link) on page 5, the authors state that they are using the Cauchy-Schwarz inequality to bound the derivative of the $L^2(mathbb{R}^n)$ norm of a solution to a certain differential equations, but I am not sure how exactly they applied it.

Some context: Let $v$, $phi$, $V$, $F$ be nice enough functions of $x$ and $t$ so that the following integrals are well-defined, $A>0, Binmathbb{R}$ be constants, and $u=e^{-phi} v$ solve
$$partial_t u = (A+iB)left( Delta u + Vu+Fright).$$
Denote the $L^2$ inner product on $mathbb{R}^n$ between some $f$ and $g$ as $(f, g) = int f g^{dagger} dx$, where $g^dagger$ is the complex conjugate of $g$, and define $f^+=mathrm{max}{f, 0}$.

We know the equality
$$partial_t vert !vert v vert!vert^2_{L^2} = 2,mathrm{Re}left(Sv,vright) + 2,mathrm{Re}left((A+iB)e^phi F, vright),$$
$$mathrm{Re}left(Sv,vright) = -Aint |nabla v|^2 + left(A|nabla phi|^2+partial_t phi right) |v|^2 + 2B ,mathrm{Im}, v^{dagger} nablaphicdotnabla v + left( A,mathrm{Re},V – B,mathrm{Im}, Vright)|v|^2dx,$$
holds true. The authors go on to conclude that the Cauchy-Schwarz inequality implies that
$$partial_t vert !vert v(t) vert!vert^2_{L^2} le 2vert !vert A , left(mathrm{Re},V(t)right)^+ – B,mathrm{Im}, V(t) vert !vert_{infty}vert !vert v(t) vert !vert^2_{L^2} + 2 sqrt{A^2+B^2} vert !vert F e^phi vert !vert_{L^2} vert !vert v(t) vert !vert_{L^2}$$
$$left(A+frac{B^2}{A}right)|nablaphi|^2 + partial_t phile 0, ,,,,mathrm{in}, mathbb{R}_+^{n+1}.$$
However, I am not sure how the authors used the C.S. inequality to arrive at this conclusion, and am especially confused as to where the factor of $B^2/A$ came from, and why we only need the constraint to hold over $mathbb{R}_+^{n+1}$ when we are integrating over all of $mathbb{R}^{n}$, though I understand why we only care about positive time.

Does anyone have any insight here?