**Proof plan.** Use the [Poisson formula for the ball](/theorems/576) to express $u(x)$ as an [integral](/page/Integral) over $\partial B(0, R)$, bound the Poisson kernel from above and below using the triangle inequality, then evaluate the remaining integral by the [mean-value property](/theorems/31). **Step 1: Write the Poisson integral.** By [Poisson's formula for the ball](/theorems/576): \begin{align*} u(x) = \frac{R^2 - |x|^2}{n\omega_n R} \int_{\partial B(0, R)} \frac{u(\sigma)}{|x - \sigma|^n} \, d\mathcal{H}^{n-1}(\sigma). \end{align*} **Step 2: Upper bound.** For $\sigma \in \partial B(0, R)$, the triangle inequality gives $|x - \sigma| \geq R - |x|$. Since $u \geq 0$: \begin{align*} u(x) \leq \frac{R^2 - |x|^2}{n\omega_n R(R - |x|)^n} \int_{\partial B(0, R)} u(\sigma) \, d\mathcal{H}^{n-1}(\sigma). \end{align*} By the [mean-value property](/theorems/31), $\int_{\partial B(0, R)} u(\sigma) \, d\mathcal{H}^{n-1}(\sigma) = n\omega_n R^{n-1} \, u(0)$. Substituting and factoring $R^2 - |x|^2 = (R - |x|)(R + |x|)$ gives $u(x) \leq \frac{R^{n-2}(R + |x|)}{(R - |x|)^{n-1}} \, u(0)$. **Step 3: Lower bound.** Similarly, $|x - \sigma| \leq R + |x|$, so: \begin{align*} u(x) \geq \frac{R^2 - |x|^2}{n\omega_n R(R + |x|)^n} \cdot n\omega_n R^{n-1} \, u(0) = \frac{R^{n-2}(R - |x|)}{(R + |x|)^{n-1}} \, u(0). \end{align*}