For $u$ harmonic and positive in $B(0,R)$, we have $R^{n-1}\frac{R+|x|}{(R-|x|)^{n-1}}u(0) \le u(x) \le R^{n-1} \frac{R - |x|}{(R + |x|)^{n-1}} u(0)$

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This was an exercise in my functional analysis course: Let $B := B(0, R) \subset \mathbb R^n$, $u$ harmonic and positive. We have to show that $$R^{n-1} \frac{R + |x|}{(R - |x|)^{n-1}}u(0) \le u(x) \le R^{n-1} \frac{R - |x|}{(R + |x|)^{n-1}} u(0)$$ for all $x \in B$.

I do not really see how to begin. I only found that for all $x \in B$, there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset B$ so that \begin{align*} u(x) &= \frac{c_n}{\varepsilon^n} \int_{B(x, \varepsilon)} u(y)dy\\ &\le \left(\frac{R}{\varepsilon}\right)^n\frac{c_n}{R^n} \int_{B} u(y)dy\\ &= \left(\frac{R}{\varepsilon}\right)^nu(0) \end{align*} but I do not see how continue from here. Any help is welcome!