Show that if $u\in C^2(\Omega)$ is subharmonic then $\Delta u\geq 0$ in $\Omega$.
Here is my proposed solution. I read this solution but couldn't follow every step (I am still relatively new to PDE). Please let me know if my proposed solution is correct or if I made a major mistake.
First, let's define $u\in C^2(\Omega)$ as subharmonic if it satisfies
$$u(x_0)\leq \frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x_0)}u(y)dA_{\partial B_r(x_0)}(y)$$
for all $x_0\in\Omega$ where $r\geq 0$ and $\overline{B_r(x_0)}\subset\Omega$.
We will argue by contradiction. Assume that there is an $x_0\in\Omega$ such that $\Delta u(x_0)<0$. For concreteness, assume that $\Delta u(x_0)=-\delta<0$.
As $u\in C^2(\Omega)$, there exists a $r_{\delta} < \mathrm{dist}(x_0,\partial\Omega)$ such that
$$|\Delta u(x_0)- \Delta u(y)| < \frac{\delta}{2}$$
for all $y$ where $|x_0-y|<r_{\delta}$. In particular, $\Delta u(y)< -\delta/2$ in $B_{r_\delta}(x_0)$.
So, let's define $\Psi(r)$ according to
$$\Psi(r)=\frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x_0)}u(y)dA_{\partial B_r(x_0)}(y)$$
Then, for $r<r_\delta$, we have that
$$\Psi'(r)=\frac{1}{\omega_n} \int_{B_1(0)} \Delta u(rz + x_0)dz < -\frac{1}{\omega_n} \int_{B_1(0)} r^2\frac{\delta}{2}dz < 0$$
As $u\in C^2(\Omega)$, we have that $\lim_{r\to 0^+}\Psi(r)=u(x_0)$. Thus, by the above equation, we see that $\Psi(r)<u(x_0)$ for $r\in(0,r_{\delta})$. This is a direct contradiction to the subharmonic property
$$u(x_0)\leq \frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x_0)}u(y)dA_{\partial B_r(x_0)}(y)$$
for all $x_0\in\Omega$ where $r\geq 0$ and $\overline{B_r(x_0)}\subset\Omega$.