show that $f(r) = \frac{1}{\mathrm{area}(S_r)} \int_{S_r} u^2 dS$ is increasing, when $u$ is harmonic Heading ##Let $u: \mathbb{R}^3 \to \mathbb{R}$ be a harmonic function. For every $r>0$, I denote:
$$ f(r) = \frac{1}{\mathrm{area}(S_r)} \int_{S_r} u^2 dS $$
when $S_r = \{(x,y,z)\mid x^2+y^2+z^2 = r^2 \}$.
I want to show that $f$ monotonically increasing for $r \in (0, \infty)$.
My problem here is that while $u$ is harmonic, $u^2$ is not necessarily, since $\Delta(u^2) = 2\|\nabla u\|^2$.
So I can't use here the mean-value property. I tried also somehow use the divergence theorem, but I always get a very ugly and non computable integral.
I also tried just calculating $f(r_1)- f(r_2)$, but since those integrals are on different surfaces, I don't see how to do it.
Help would be appreciated.
 A: I think what you can do is to take the derivative of $f$ with respect to r and show that it is always non-negative and conclude from this your desired statement. I happy to provide more details if needed. 
edit: (details).
First notice that 
$$
\frac{1}{\text{area}(S_r)} \int_{S_r} u^2(x) dS(x) =
\frac{1}{\text{area}(S_1)r^2} \int_{S_1} r^2u^2(rx) dS(x) 
=
\frac{1}{\text{area}(S_1)} \int_{S_1} u^2(rx) dS(x).
$$
Now compute $\frac{d}{dr} u^2(rx)$ which gives $x\cdot\nabla u^2(rx)$, where
$\cdot$ denotes the scalar product. Now, argue why you can swap differentiation
and integration to get
$$
\frac{d}{dr} f(r) = \frac{1}{S_1} \int_{S_1} x \cdot \nabla u^2(rx) dS(x) 
= \frac{1}{S_r} \int_{S_r} \frac{x}{r} \cdot \nabla u^2(x) dS(x)  
= \frac{1}{S_r} \int_{B_r} \Delta u^2(x) dx, 
$$
where in the last step the divergence theorem was used.
Can you conclude from here?
A: You have that $u^2$ is subharmonic, so it satisfies the mean value inequality 
$$u^2(x) \leq \frac{1}{area(S_r)}\int_{S_r} u^2 dS$$
for each $x$ in its domain within the sphere.
