This question is pretty old now, so I would answer the following. Consider a particle inside some box. Its hamiltonian is $$\tag{1} H = \frac{p^2}{2m}. $$ The state of lowest energy is $|E_{0}\rangle$ and you have $$\tag{2} \Delta p \ge \frac{\hbar}{2 \Delta x}, $$ where $\Delta x$ is about the size of the box. Then you have, since $\Delta p^2 \equiv \langle p^2 \rangle - \langle p \rangle^2$ : \begin{align} \langle E \rangle_0 \equiv \langle E_0 | H | E_0 \rangle = \frac{\langle p^2 \rangle}{2m} &\equiv \frac{\Delta p^2}{2 m} + \frac{\langle p \rangle^2}{2m} \\[12pt] &\ge \frac{\hbar^2}{8 m \Delta x^2} + \frac{\langle p \rangle^2}{2m} \ge \frac{\hbar^2}{8 m \Delta x^2}. \tag{3} \end{align} Then obviously there's a minimum value to the energy (the state vector $| E_0 \rangle$ is an eigen-ket of the hamiltonian (1) with minimal energy $E_0$).
For an ultra-relativistic particle, I would use the hamiltonian $H = p c$. Then $H^2 = p^2 c^2$ and \begin{align} E_0^2 \equiv \langle H^2 \rangle_0 = \langle p^2 \rangle c^2 &\equiv \Delta p^2 c^2 + \langle p \rangle^2 c^2 \\[12pt] &\ge \frac{\hbar^2 c^2}{4 \Delta x^2} + \langle p \rangle^2 c^2 \ge \frac{\hbar^2 c^2}{4 \Delta x^2}, \tag{4} \end{align} thus $$\tag{5} E_0 \ge \frac{\hbar c}{2 \Delta x}. $$