bit more useful is the following.

The following inequality holds in the sense of quadratic forms $$\Pi(x)^2 \geq \frac{1}{4}\frac{1}{(3\pi)^4}\Lambda^6 \frac{1}{A(x)^2}\ .
\label{eq:2.30}$$ \[lemma:2.4\]

For the proof note that $[A_j(x), A_k(y)] = 0$ and compute $$i[\Pi(x)_j, \frac{A_j(x)}{A(x)^2}]
=\frac{1}{(3\pi)^2}\Lambda^3 \left[\frac{1}{A(x)^2} -
2 \left( \frac{A_j(x)}{A(x)^2}\right)^2 \right] \ ,
\label{eq:2.31}$$ and summing over $j$ we obtain that $$i \sum_{j=1}^3[\Pi(x)_j, \frac{A_j(x)}{A(x)^2}]
=\frac{1 }{(3\pi)^2}\Lambda^3 \frac{1}{A(x)^2}
\label{eq:2.32}$$

Our statement follows from the Schwarz inequality.

To prove Theorem \[theo:2.3\] we return to Lemma \[lemma:2.2\] and choose $v_{\lambda}(k)= 
\varepsilon_{\lambda}(k)$ and $w(x)$ any function $\leq 1$. Corollary \[corollary:1\] applied to each of the 3 components of $\Pi(x)$ then yields $$H_f \geq \frac{ 1}{4} \int w(x-y) \Pi (y)^2 {\rm d} y - \Lambda^4 
\frac{3}{8 \pi^2} \int w(y) {\rm d} y \ ,
\label{eq:2.33}$$ for every $x \in {\mathbb R}^3$. By Lemma \[lemma:2.2\] the right side is bounded below by $$\Lambda^6 \int w(x-y) \frac{1}{A(y)^2} {\rm d} y - 
\Lambda^4\int w(y) {\rm d} y \ ,
\label{eq:2.34}$$ and hence $$\begin{aligned}
\langle \Psi, H \Psi \rangle  &\geq&  \frac{1}{2} \int \langle \nabla
\Psi(x), \nabla \Psi(x) 
\rangle {\rm d} x + \frac{\alpha}{2} \int \langle \Psi(x), A(x)^2
\Psi(x) \rangle {\rm d} x     \nonumber  \\
&+&\Lambda^6 \int w(x-y) \langle \Psi(y),
\frac{1}{A(x)^2}\Psi(y) \rangle {\rm d}
y {\rm d} x   