Consider the following problem:
Let $\Omega \subset \mathbb R^N$ be a domain and $u \in C(\Omega)$. Show that $u$ is subharmonic in the Mean Value Inequality sense, that is, $$ u(x) \leq \frac{1}{\omega_N r^N} \int_{B_r(x)} u (y) \ dy, \quad \forall x \in \Omega, \forall 0 < r < d(x, \partial \Omega). $$ if and only if for every $x_0 \in \Omega$ and every $\phi \in C^2(\Omega)$ such that $u - \phi$ attains a local maximum at $x_0$ it holds that $$ - \Delta \phi(x_0) \leq 0, $$ that is, $u$ is a viscosity subsolution of the Laplace equation.
For the $\Rightarrow$ implication, I managed to show that $\phi$ is subharmonic in the sense of the (volumetric) Mean Value Inequality. Since it is $C^2$, it would suffice to show that it implies that $\phi$ is subharmonic in the classic sense. How to do this, if true?
For the $\Leftarrow$ implication I have no ideas, and any hints are welcome.
Thanks in advance.
EDIT
For the $\Rightarrow$ implication, it almost suffices to show that $\phi$ is subharmonic in the Mean Value Property sense. Indeed, suppose that $- \Delta \phi(x_0) > 0$. Then we would have the opposite inequality.
There are lots of ways to prove this, depending on what you know about PDEs and viscosity solutions. Here is one hint: Fix your ball $B_r(x)$ and solve Laplace's equation $\Delta v = 0$ on $B_r(x)$ with boundary conditions $v=u$ on $\partial B_r(x)$ (I assume you know how to construct the solution with Poisson's Integral Formula). Then $v$ is $C^2$, and the viscosity subsolution property for $u$ can be used to show that $u\leq v$ on $B_r(x)$ (this does not require the full comparison principle machinery of viscosity solutions, and is easy to prove by hand; I can add a comment on this if needed). Then since $v$ is harmonic and smooth, use the mean value property for $v$.