I have a problem with an induction step that appears to be bothersome for me to comprehend it.
Problem: Show that $f(x_1, \dots ,x_n)=(x_1^2+ \dots + x_n^2)^{1- n/2}$ for $(x_1^2 + \dots + x_n^2) \neq 0$ solves the Laplace equation $$ (\partial_{x_1}^2 + \dots + \partial_{x_n}^2)u=0$$ for $ n \geq 3$
Base case
Let $n=3$ then $f(x_1,x_2,x_3)=(x_1^2+ x_2^2+x_3^2)^{-1/2}$ since computing the partial differentials is merely an exercise in differentiation, I will post my results only: $$\partial_{x_1}^2= \frac{2x_1^2-x_2^2-x_3^2}{(x_1^2+x_2^2+x_3)^{5/2}}, \partial_{x_2}^2= \frac{2x_2^2-x_1^2-x_3^2}{(x_1^2+x_2^2+x_3)^{5/2}}, \partial_{x_3}^2= \frac{2x_3^2-x_1^2-x_2^2}{(x_1^2+x_2^2+x_3)^{5/2}} $$ which would indeed work for the base case.
Now I struggle with the induction step because my result makes less sense to me, I assume that $$f(x_1, \dots , x_n, )=(x_1^2+ \dots + x_{n}^2 ) ^{1-(n/2)}=:u $$ solves $$ (\partial_{x_1}^2 + \dots + \partial_{x_n}^2)u=0$$ and want to show that $n \leadsto n+1$. So could I try to say that $$(\partial_{x_1}^2 + \dots + \partial_{x_n}^2+\partial_{x+1}^2)u=\underbrace{(\partial_{x_1}^2 + \dots + \partial_{x_n}^2)u}_{=0}+\partial_{n+1}^2u=0 \\ \implies \partial_{n+1}^2 u =0 \implies \partial_{n+1}^2 =0, \text{ because $u \neq 0$} $$ My induction step seems very confusing to me, because it suggest to me that the last partial differential vanishes, but my induction base case suggest otherwise, i.e. that the partial differentials cancel each other out.
It is advantageous to set $r(x)^2:= x_1^2 + \dots + x_n^2$, then $f(x) = r(x)^{2-n}$, $\partial_{x_1}f = (2-n)r^{1-n} \partial_{x_1}r$. You get shorter expressions for derivatives.
Also there is no need for induction. Just compute the partial derivatives, and sum up.