Weak maximum principle for the Helmholtz equation

Question

Suppose one is given the weak maximum principle for subharmonic functions, if $u \in C^2(\Omega)\cap C^0(\overline{\Omega})$ with $\Omega$ open, bounded and $\Delta u \geqslant 0$ in $\Omega$, then $\text{max}_{\overline{\Omega}} \ u = \text{max}_{\delta \Omega} \ u$ . Apparently, a weak maximum principle for the Helmholtz equation should be an easy consequence: if $\Delta u +\lambda u=0$ in $\Omega$, with $\lambda < 0$ and it is known that $\text{max}_{\overline{\Omega}} \ u > 0$, then $\text{max}_{\overline{\Omega}} \ u = \text{max}_{\delta \Omega} \ u$. One should apply the weak maximum priniciple to $u$ in $\Omega_{0}=\left \{ z \in \Omega \mid u(z)>0 \right \}$, where $u$ is subharmonic. However, it would appear to me that $u\equiv 0$ on $\delta \Omega_0$ so I cannot see how that will be of any use (leads to an immediate contradiction). Any thoughts?