 Solution

The difference $u_1-u_2$ is subharmonic. If $M={\max }_{\overline{\Omega }}(u_1-u_2)>0$, its maximum set lies in $\Omega$. At any point of this set, comparison with the harmonic replacement on a small ball and the Strong maximum principle for harmonic functions show that the whole ball belongs to the maximum set. The set is therefore both open and closed in the connected domain $\Omega$, so it is all of $\Omega$, contradicting the boundary inequality. Thus $u_1\le u_2$ throughout $\Omega$. This is the comparison principle for subharmonic and superharmonic functions.