Show that the Poisson kernel is harmonic as a function in x over $B_1(0)\setminus\left\{0\right\}$ Show that the Poisson-kernel
$$
P(x,\xi):=\frac{1-\lVert x\rVert^2}{\lVert x-\xi\rVert^n}\text{ for }x\in B_1(0)\subset\mathbb{R}^n, \xi\in S_1(0)
$$
is harmonic as a function in $x$ on $B_1(0)\setminus\left\{0\right\}$.
On my recent worksheet, this task is rated with very much points. So I guess it is either very difficult or requires much calculation.
Am I right that I do have to show (most likely by a rather long calculation) that for any $1\leq i\leq n$
$$
\frac{\partial^2}{\partial x_i^2}P(x,\xi)=\frac{\partial^2}{\partial x_i^2}\left(\frac{1-\sum_{i=1}^{n}x_i^2}{(\sum_{i=1}^{n}(x_i-\xi_i)^2)^{\frac{n}{2}}}\right)=0?
$$
I ask, because I do not want to start this exhausting calculation if there is maybe another way or without having the affirmation that this is constructive.
For example a continuous function that fulfills the mean value property is harmonic. Maybe this is an alternative way here?

My result for the first derivative
Consider any $1\leq i\leq n$. Then my result for $P_{x_i}$ is
$$
P_{x_i}=\frac{-2x_i\lVert x-\xi\rVert^n-(1-\lVert x\rVert^2)\frac{n}{2}\lVert x-\xi\rVert^{n-2}(2x_i-2\xi_i)}{\lVert x-\xi\rVert^{2n}}.
$$
Here I used the quotient rule. Moreover, I used the chain rule to calculate
$$
\frac{\partial}{\partial x_i}(\lVert x-\xi\rVert^n)=\frac{1}{2}n\lVert x-\xi\rVert^{n-2}(2x_i-2\xi_i).
$$
Maybe you can say me if my calculation is correct to this point.
My final result
As the second derivative I get
$$
P_{x_i x_i}=-2\lVert x-\xi\rVert^{-n}+4x_in\lVert x-\xi\rVert^{-n-2}(x_i-\xi_i)-n\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)-n(x_i-\xi_i)^2(-n-2)\lVert x-\xi\rVert^{-n-4}(1-\lVert x\rVert^2)
$$
My question is if then
$$
\Delta P=\sum_{i=1}^{n}P_{x_i x_i}=0?
$$
Maybe you can say me if this is correct. Unfortunately I do not see how I can show with that result, that $\Delta P=0$. Maybe I am blind, maybe my result is wrong. I did it again and again and I always get this second derivative. Therefore I hope that you can help me finding the mistake or my error in reasoning.
I am aware of the fact that I probably won't get any help, because it is too much calculation, but maybe someone has pity with me and my effort.
 A: Here is a vectorized approach (as in "look ma, no coordinates!"). Make sure you have the chain and product formulas written down in convenient form (they help with other calculations too):
$$\nabla(\varphi(u)) = \varphi'(u) \nabla u\tag{1}$$
$$\Delta(u) = \operatorname{div} \nabla u \tag{2}$$
$$\operatorname{div} u \mathbf F = \nabla u\cdot \mathbf F
+ u \operatorname{div} \mathbf F\tag{3}$$
$$\Delta(uv) = u\Delta v+v\Delta u+2\nabla u\cdot \nabla v \tag{4}$$
Your function is $uv$ with $u=(1-\|x\|^2)$ and $v=\|x-\xi\|^{-n}$. 
We have 
$$\nabla u = -2 x,\quad \Delta u = -2n $$
Using (1): 
$$
\begin{split}
\nabla v &= -n \|x-\xi\|^{-n-1}\nabla \|x-\xi\|  \\
 &= -n \|x-\xi\|^{-n-1}\frac{x-\xi}{\|x-\xi\|} \\ 
 &= -n \|x-\xi\|^{-n-2}(x-\xi) 
 \end{split}
 $$
Using (2) and then (3): 
$$
\begin{split}
\Delta v &= -n \operatorname{div} ( \|x-\xi\|^{-n-2}(x-\xi)) \\
 &= -n (-n-2) \|x-\xi\|^{-n-3}\frac{x-\xi}{\|x-\xi\|}\cdot (x-\xi)
  -n \|x-\xi\|^{-n-2} n \\ 
&  =2n\|x-\xi\|^{-n-2}
  \end{split}
 $$ 
Finally, combine the results using (4). For convenience, I multiply the Laplacian by $\|x-\xi\|^{ n+2}$: 
$$
\begin{split}
\|x-\xi\|^{ n+2}\Delta(uv)
 &= -2n \|x-\xi\|^{2} + (1-\|x\|^2) 2n  +4nx \cdot  (x-\xi)  \\
& =  0     
\end{split}
 $$
A: It looks like your derivatives are correct, and then you get $\Delta P = 0$ as you should, if you sum and group terms to annihilate each other.
Here's a trick to make such calculations more manageable: give (meaningless) short names to the building blocks. For example, if we name
$$\begin{align}
N &= 1 - \lVert x\rVert^2 = \lVert \xi\rVert^2 - \lVert x\rVert^2,\\
S &= \lVert x-\xi\rVert^2,
\end{align}$$
we get $P = N\cdot S^{-n/2}$, and
$$\begin{gather}
\partial_i P = (\partial_i N)S^{-n/2} - \frac{n}{2} N(\partial_i S)S^{-(n/2+1)}\\
\partial_i^2 P = (\partial_i^2 N)S^{-n/2} -n(\partial_i N)(\partial_i S)S^{-(n/2+1)} - \frac{n}{2}N(\partial_i^2 S)S^{-(n/2+1)} + \frac{n(n+2)}{4}N(\partial_i S)^2S^{-(n/2+2)}\\
S^{n/2+1}\partial_i^2 P = (\partial_i^2 N)S - n(\partial_i N)(\partial_i S) - \frac{n}{2}N(\partial_i^2 S) + \frac{n(n+2)}{4}N(\partial_i S)^2 S^{-1}.
\end{gather}$$
Now we look at the partial derivatives of $N$ and $S$ and the needed sums,
$$\partial_i N = -2x_i;\; \partial_i^2 N = -2;\; \partial_i S = 2(x_i-\xi_i);\; \partial_i^2 S = 2;$$
which yields
$$\begin{align}
\sum_i \partial_i^2 N &= -2n\\
\sum_i (\partial_i N)(\partial_i S) &= -4\lVert x\rVert^2 +4\langle x,\xi\rangle\\
\sum_i \partial_i^2 S &= 2n\\
\sum_i (\partial_i S)^2 &= 4S
\end{align}$$
and hence
$$\begin{align}
S^{-(n/2+1)}\Delta P &= -2nS + 4n\lVert x\rVert^2 - 4n\langle x,\xi\rangle -n^2N + n(n+2)N\\
&= 2nN + 4n\lVert x\rVert^2 - 4n\langle x,\xi\rangle - 2nS\\
\frac{S^{-(n/2+1)}\Delta P}{2n} &= \lVert \xi\rVert^2 - \lVert x\rVert^2 + 2\lVert x\rVert^2 - 2\langle x,\xi\rangle - \lVert x-\xi\rVert^2\\
&= \lVert \xi-x\rVert^2 - \lVert x-\xi\rVert^2\\
&= 0.
\end{align}$$

If we start from
$$P_{x_i x_i}=\underbrace{-2\lVert x-\xi\rVert^{-n}}_A + \underbrace{4x_in\lVert x-\xi\rVert^{-n-2}(x_i-\xi_i)}_B - \underbrace{n\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)}_C - \underbrace{n(x_i-\xi_i)^2(-n-2)\lVert x-\xi\rVert^{-n-4}(1-\lVert x\rVert^2)}_D,$$
summing $A$ produces $-2n\lVert x-\xi\rVert^{-n}$, and summing $B$ leads to $4n\lVert x-\xi\rVert^{-n-2}\langle x,x-\xi\rangle$. Summing $C$ yields, including the sign, $-n^2\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)$, and since $\sum_i (x_i-\xi_i)^2 = \lVert x-\xi\rVert^2$, $D$ produces $n(n+2)\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)$. Thus
$$\begin{align}
C+D &= 2n\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)\\
A+B &= 2n\lVert x-\xi\rVert^{-n-2}\left(2\langle x,x-\xi\rangle - \langle x-\xi,x-\xi\rangle\right)\\
&= 2n\lVert x-\xi\rVert^{-n-2}\langle x+\xi,x-\xi\rangle\\
&= 2n\lVert x-\xi\rVert^{-n-2}(\lVert x\rVert^2 - \lVert\xi\rVert^2).
\end{align}$$
Since $\lVert \xi\rVert^2 = 1$, we have $A+B+C+D = 0$.
