Let $u:B(0,R)\subset \mathbb{R^d}\rightarrow\mathbb{R}$ be harmonic and nonnegative. Prove the following version of the Harnack inequality: $$\dfrac{R^{d-2}(R-|x|)}{(R+|x|)^{d-1}}u(0)\leq u(x)\leq \dfrac{R^{d-2}(R+|x|)}{(R-|x|)^{d-1}}u(0)$$.
I know what the usual inequality says and I have the proof of that at hand, however I do not know how to begin. Thanks in advance for any hints.
For $|y|=R$, we have $R-|x|\le |x-y|\le R+|x|$. Using the Poisson formula, we obtain (when $x\in B(0,R)$) that\begin{align} u(x) &= \frac{R^2-|x|^2}{d\omega_d R}\int_{\partial B(0,R)}\frac{u(y)}{|x-y|^d}d\sigma(y)\\ &\ge \frac{R^2-|x|^2}{(R+|x|)^d d\omega_d R}\int_{\partial B(0,R)}u(y) d\sigma(y)\\ &= \frac{R^{d-2}(R-|x|)}{(R+|x|)^{d-1}}\frac{1}{d\omega_d R^{d-1}}\int_{\partial B(0,R)}u(y) d\sigma(y)\\ &= \frac{R^{d-2}(R-|x|)}{(R+|x|)^{d-1}}u(0) \end{align} The last step uses the mean value property of the harmonic function $u$. The other estimation is similar.