Define the upper half ball in $\mathbb{R}^n$ to be the set $B_r(0)^+ = B_r(0) \cap \{y_{n}>0\} .$ Denote for $y = (y',y_n)\in B_r(0)^+$, $$\tilde{y} = (y',-y_{n}).$$ Then for a harmonic function $u \in C^2(B_r(0)^+) \cap C(\overline{B_r(0)^+})$ vanishing on $\{y_{n}=0\}$, one may show using for instance, Schwarz reflection combined with the Poisson integral formula for a ball, that for each $ x\in B_r(0)^+$ one has: $$ u(x) = \frac{r^2-|x|^2}{r|S^{n-1}|}\int_{|y|=r \hspace 0.1 cm\text{and} \hspace0.1 cm y_{n} \geq 0}\ \bigg(\frac{1}{|x-y|^n}-\frac{1}{|x-\tilde{y}|^n}\bigg)u(y)d\sigma(y).$$
I'd like to use this representation formula to prove the following Harnack-type inequality for functions $u$ as above that satisfy $u \geq 0$ and $ x \in B_{\frac{r}{2}}(0)^+$, $$\frac{1}{C} u\left(0,\frac{r}{2}\right)\frac{x_{n}}{r} \leq u(x) \leq Cu\left(0,\frac{r}{2}\right)\frac{x_{n}}{r},$$ for some constant $C > 0$ depending only on $n$. Thusfar, I've tried to use the elementary triangle inequality $$\frac{1}{|x-y|^n}-\frac{1}{|x-\tilde{y}|^n} \leq \frac{2y_{n}}{|x-y|^n|x-\tilde{y}|^n} \leq 2\left(\frac{2}{r}\right)^ny_{n}$$ to bound the integrand, and then try and apply the divergence theorem to convert the integral to one on the interior of the upper half ball. However it's not clear how to proceed. It's also not clear how to get the term $u\left(0,\frac{r}{2}\right)$ to appear out of the estimates. I'd appreciate if someone could help me out here.