with $C \geq 0.00127$. An analogous inequality holds for the relativistic case as well: $$\sum_{j=1}^{N} |p_j + \sqrt{\alpha} A (x_j)| \geq \frac{C}{q^{1/3}}
\frac{1}{R} \sum^N_{j=1} N_j (\underbar{X}, R)^{1/3}$$ \[theo:3.3\]

By using the kinetic energy previously saved together with (\[eq:3.8\]) and the previous discussion, we get $$H \geq \sum^N_{j=1} \left\{  N_j (\underbar{X},R)^{2/3}
+  \sqrt{\alpha} \Lambda^{3/2} \frac{1}{\sqrt{N_j
(\underbar{X},R)+1}} \right\} - N \Lambda ~ .$$ By minimizing over $N_j$ the desired estimate is obtained. The upper bound is fairly elementary and is omitted. $\blacksquare$

RELATIVISTIC ENERGY BOUNDS {#relativistic}
==========================

[*Theorem \[theo:4.1\]:*]{} Sketch of Proof.

An upper bound for $E_0$ is easy to obtain, but it is indirect. Note that $$|p + \sqrt{\alpha} A(x) | \leq \varepsilon [ p + \sqrt{\alpha} A (x)
 ]^2 + (4 \varepsilon)^{-1}
\label{eq:4.3}$$ for any $\varepsilon > 0$. Take $\Psi = f (x) \otimes | 0\rangle$ with $|0\rangle$ being the Fock space vacuum. Using (\[eq:4.3\])

$$\begin{aligned}
(\Psi, H \Psi) &\leq& \varepsilon ~\int_{\mathbb{R}^{\rm{3}}} \{ \alpha
\langle 0 | A 
(x)^2 | 0 \rangle | f (x)|^2 + | \nabla f (x)|^2 \} dx +
\varepsilon^{-1} \nonumber \\
&=& \frac{\varepsilon \alpha \Lambda^2}{4 \pi} ~ + \int |\nabla f|^2 +
\frac{1}{4 \varepsilon}~,
\label{eq:4.5a}\end{aligned}$$

since $\langle 0 | A (x)^2 | 0 \rangle = 