Interior gradient bound

By rotational symmetry we may assume that $Du(x_0)$ points in the direction of the first basis vector $e_1$. We must prove that $${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)]\nu_1\le \sup u-u(x_0)$$ where $\nu_1$ is the first component of the unit normal vector. By the mean value property $u-u(x_0)$ has zero mean on $\partial B_R$. Thus, we can add any number to $\nu_1$ without changing the integral. Let's add $1$: $${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)]\nu_1={\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)](\nu_1+1)$$ Now that the factor $\nu_1+1$ is nonnegative, we use a one-sided bound on $u-u(x_0)$: $${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)](\nu_1+1)\le (\sup u-u(x_0)){\int\!\!\!\!\!\!-}_{\partial B_R}(\nu_1+1)$$ Finally, $${\int\!\!\!\!\!\!-}_{\partial B_R}(\nu_1+1)=1$$ because $\nu_1$ has zero mean.