An identity $ \int_{\partial B(0,1)}u(x_0+aw)u(x_0+cw)$ with a harmonic function $u$ This is Question 2.18 from Gilbarg and Trudinger, chapter 2.
We are given that $\Omega$ is open bounded smooth boundary. Now fix $x_0\in \Omega$ and a constant $c>0$ such that $B(x_0,c)\subset\subset \Omega$. Next, given $u$ such that $\Delta u=0$ in $\Omega$ and another $2$ positive constants $a$ and $b$ such that $a<b<c$ and $b^2=ac$. Then, the question asks us to prove the following simple and beautiful equation.
$$ \int_{\partial B(0,1)}u(x_0+aw)u(x_0+cw)\,dSw=\int_{\partial B(0,1)}u^2(x_0+bw)\,dSw $$
My try:
Define
$$ v(w):=u(x_0+aw)u(x_0+cw)- u^2(x_0+bw)$$
If I can prove $v$ is harmonic, then by Mean Value Theorem I would be done, since
$$ 0=v(0)=\int_{\partial B(0,1)} v(w)\,dSw$$
Hence, by $v\in C^2$, I just compute $\Delta v$ and obtain that, after simplification,
$$ \Delta v(w)=2b^2(\nabla u(x_0+aw)\cdot\nabla u(x_0+cw)-\nabla u(x_0+bw)\cdot\nabla u(x_0+bw)) $$
I can not go further from here. Any hint would be very welcome!
 A: Let $B$ be the unit ball. Applying  Green's second identity to $\phi(w)=u(aw)$ and $\psi(w)=u(cw)$ yields
$$
\int_B \psi \frac{\partial \phi}{\partial n}= \int_B \phi \frac{\partial \psi}{\partial n}\tag{1}
$$
hence (by the Chain Rule)
$$
a \int_B u(cw) u_r(aw)= c \int_B u(aw)u_r(cw) \tag{2}
$$
where $u_r$ is the  derivative of $u$ in the radial direction.
On the other hand, differentiating the function 
$$
I(a) = \int_{\partial B}u(aw)u((b^2/a)w) \tag{3}
$$
we get 
$$
I'(a) = \int_{\partial B}\left(u_r(aw)u((b^2/a)w)-\frac{b^2}{a^2}u(aw)u_r((b^2/a)w)\right) =0  \tag{4}
$$
according to (2). Thus, $I$ is a constant function, which implies the claim. 
A: $a<b<c \text{ and }b^2 = ac\implies a= b\cdot\dfrac{b}{c}:=b\alpha$ also $c= \dfrac{b}{\alpha}$.
Consider $f: (\dfrac{b}{c},1)\mapsto \mathbb{R}$ with
$$f(r) =\int_{|\omega|=1}u(x_0+br\xi)u(x_0+\frac{b}{r}\xi)dS(\xi)$$
Then $f$ is well defines and smooth with vanishing derivative thanks to Harmonicity of $u$ and the Green identity as follows:
\begin{align*}
f'(r) &= \int_{|\xi|=1} b\nabla u(x_0+br\xi) \cdot \xi\,\,  u(x_0+\frac{b}{r}\xi) - \frac{b}{r^2} u(x_0+br\xi)\,\,  \nabla u(x_0+\frac{b}{r}\xi)\cdot \xi d S(\xi)\\
&=  \frac{1}{r} \int_{|\xi|=1} \nabla \big[u(x_0+br\cdot)\big](\xi) \cdot \xi\,\,  u(x_0+\frac{b}{r}\xi) -  u(x_0+br\xi)\,\,  \nabla \big[u(x_0+\frac{b}{r}\cdot)\big](\xi)\cdot \xi d S(\xi)\\
&= \frac{1}{r} \int_{|\xi|<1} \Delta \big[u(x_0+br\cdot)\big](\xi) \,\,  u(x_0+\frac{b}{r}\xi) -  u(x_0+br\xi)\,\,  \Delta \big[u(x_0+\frac{b}{r}\cdot)\big](\xi) d \xi\\
&= \int_{|\xi|<1} b^2 r[\Delta u](x_0+br\xi) \,\,  u(x_0+\frac{b}{r}\xi) -  \frac{b^2}{r^3}u(x_0+br\xi)\,\,  [\Delta u](x_0+\frac{b}{r}\xi) d \xi=0
\end{align*}
Since for all $r\in (\frac{b}{c}, 1)$ we have $x_0+\frac{b}{r}\xi, x_0+ br\xi\in B(x_0, c)\subset \Omega$. So that by Harmonicity we get  $[\Delta u](x_0+br\xi)= [\Delta u](x_0+\frac{b}{r}\xi) =0$. Thus $f$ is constant on $(\frac{b}{c},1)$ and in particular we get
\begin{align*}
 \int_{|\xi|=1}u^2(x_0+b\xi)d S(\xi)= \lim_{r\to 1^-}f(r) = \lim_{r \to \frac{b}{c}^+} f(r)&= \int_{|\xi|=1} u(x_0+a\xi) u(x_0+c\xi) d S(\xi).
\end{align*}
