Why is the magnetic Schrödinger operator positive?

Let $\mathcal{H}$ denote the Hilbert space of wave functions $\psi:\mathbb{R}^3\to\mathbb{C}$. Recall that $$\langle u|v\rangle =\iiint_{\mathbb{R}^3}\bar{u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz$$ for all $u,v\in \mathcal{H}$. Write $h$ for the operator $(-i\nabla -A)^2$. Observe that \begin{align}\langle hu|v\rangle &=\iiint_{\mathbb{R}^3}\overline{hu}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz \\&=\iiint_{\mathbb{R}^3}(i\nabla -A)^2\bar{u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz \\&=\iiint_{\mathbb{R}^3}(i\nabla -A)\cdot \overline{\Phi u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz, \end{align} where $\Phi=-i\nabla-A$. That is, \begin{align}\langle hu|v\rangle &=i\iiint_{\mathbb{R}^3}\nabla\cdot \overline{\Phi u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz-\iiint_{\mathbb{R}^3} \overline{\Phi u}(x,y,z)\cdot Av(x,y,z)\ dx\ dy\ dz \\&=-i\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot \nabla v(x,y,z)\ dx\ dy\ dz -\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot Av(x,y,z)\ dx\ dy\ dz,\end{align} where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large. That is, \begin{align}\langle hu|v\rangle &=\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot(-i\nabla-A)v(x,y,z)\ dx\ dy\ dz \\&=\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot\Phi v(x,y,z)\ dx\ dy\ dz. \end{align} In particular, \begin{align}\langle hu|u\rangle &=\int_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot\Phi u(x,y,z)\ dx\ dy\ dz\\&=\int_{\mathbb{R}^3}\big\Vert\Phi u(x,y,z)\big\Vert^2\ dx\ dy\ dz\geq 0\end{align}

If $X$ is a vector with Hermitian components, $$\langle\psi |X\cdot X|\psi\rangle=\sum_i\langle\psi |X_i^2| \psi\rangle=\sum_i\langle\psi |X_i^TX_i| \psi\rangle=\sum_i\Vert X_i|\psi\rangle\Vert^2\ge 0.$$