 Solution

True. Continuity and compactness give a maximum of $\varphi$ on $\overline{\Omega }$. At an interior maximum the Hessian matrix is negative semidefinite, so its trace satisfies $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. The maximum must therefore occur on $\partial \Omega$. This is the strict form of the maximum principle for subharmonic functions.