TITLE: Mean-value theorem for subharmonic functions:
QUESTION [6 upvotes]: In this problem, we will consider subharmonic functions. If $U$ is a bounded, open set, we say that $v\in C^2(U)\cap C(\overline{U})$ is subharmonic if 
  $$-\Delta v \leq 0 \ \ \ \text{in} \ U$$
  a.) Prove that if $v$ is subharmonic and $B(x,r)\subseteq U$, then 
  $$v(x) \leq ⨍_{\partial B(x,r)}v(y)dS_y$$

Attempted proof - Suppose $v$ is subharmonic and $B(x,r)\subseteq U$ where $U$ is a bounded, open set. Then by definition $v\in C^2(U)\cap C(\overline{U})$ and $-\Delta v \leq 0$ in $U$ or equivalently $\Delta v \geq 0$ in $U$. Then we have 
$$v(x) = ⨍_{B(x,r)}v(y)S_y$$
I am having some problem with this proof, I don't see where to go. Any suggestions are greatly appreciated.
Note:
I will now try another attempt at proving this. (for consistency with comments: $\oint$ was used as a substitute for $⨍$.)
Attempted proof 1 - Let 
$$f(r) = ⨍_{\partial B(x,r)}v(y)d S_y = ⨍_{|y-x| = r}v(y)d S_y = \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}v(y) dS_y = \frac{1}{n\alpha(n)}\int_{|z| = 1}u(x+rz) d S_z$$
where $y = x + rz$ and $\alpha(n) = \int_{|x|\leq 1}dx = $ volume of unit ball in $n$ dimensions. The outer unit normal at $y\in\partial B(x,r)$ is $\nu(y) = \frac{y-z}{r}$, thus 
\begin{align*}
f'(r) = \frac{1}{n\alpha(n)}\int_{|z|=1}z \nabla v(x+rz) dS_z &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y) \frac{y-x}{r}d S_y\\
&= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y)\nu(y)d S_y
\end{align*}
Since $-\Delta v\leq 0$ in $U$, we have 
$$0\leq \int_{B(x,r)}\Delta v(y) dy = \int_{\partial B(x,r)}v_{\nu}d S_y$$
thus $f'(r)\geq 0$ for $r > 0$. Then we get for $r > 0$
$$v(x) = f(0) \leq f(r) = ⨍_{\partial B(x,r)}v(y)d S_y$$
Therefore we have the result.
I am sure this proof is quite messy and contains some errors. Please give me some comments or suggestions on the latter.

REPLY [8 votes]: Your proof is correct, with just minor errors. Here is revised version of it. 
Let 
$$ f(r) = ⨍_{\partial B(x,r)}v(y)d S_y = ⨍_{|y-x| = r}v(y)d S_y = \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}v(y) dS_y =\\= \frac{1}{n\alpha(n)}\int_{|z| = 1}v(x+rz) d S_z$$
where $y = x + rz$ and $\alpha(n) = \int_{|x|\leq 1}dx = $ volume of unit ball in $n$ dimensions. The outer unit normal at $y\in\partial B(x,r)$ is $\nu(y) = \frac{y-x}{r}$, thus 
\begin{align*}
f'(r) = \frac{1}{n\alpha(n)}\int_{|z|=1}z. \nabla v(x+rz) dS_z &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y). \frac{y-x}{r}d S_y\\
&= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y).\nu(y)d S_y
\end{align*}
By Green-Stokes Theorem we know that 
$$ \int_{|y-x| = r}\nabla v(y).\nu(y)d S_y =\int_{\partial B(x,r)}\nabla v(y).\nu(y)d S_y = \int_{B(x,r)}\Delta v(y) dy $$
Since $-\Delta v\leq 0$ in $U$, we have 
$$0\leq \int_{B(x,r)}\Delta v(y) dy = \int_{|y-x| = r}\nabla v(y).\nu(y)d S_y $$
thus $f'(r)\geq 0$ for $r > 0$. Then we get for $r > 0$
$$v(x) = f(0) \leq f(r) = ⨍_{\partial B(x,r)}v(y)d S_y$$
Therefore we have the result.