fa.functional analysis – How to prove the reverse Hölder inequality for Laplace equations?

Let $ uin H^1(2B) $ be a weak solution of $ Delta u=0 $ in $ 2B $, where $ B=B(0,1) $ is a ball with center $ 0 $ and radius $ 1 $. Then there exists some $ p>2 $ such that
begin{eqnarray}
left(frac{1}{|B|}int_{B}|triangledown u|^p dxright)^{1/p}leq Cleft(frac{1}{|2B|}int_{2B}|triangledown u|^2 dxright)^{1/2}.
end{eqnarray}

where $ C $ is an absolute constant.

I recently saw this problem and I want to get the solution of this problem. However, I meet with some troubles in it. Here is my try. First as $ u-frac{1}{|2B|}int_{2B}u $ is also a weak solution for the Laplace equation, then by using integration by parts on the function, I can obtain that
begin{eqnarray}
frac{1}{|B|}int_{B}|triangledown u|^2 dxleq Cleft{int_{2B}left|u-frac{1}{|2B|}int_{2B}uright|^2dxright}.
end{eqnarray}

where $ C $ is an absolute constant. Then, by using the Sobolev-Poincaré inequality I have
begin{eqnarray}
left(frac{1}{|B|}int_{B}|triangledown u|^2 dxright)^{1/2}leq Cleft(frac{1}{|2B|}int_{2B}|triangledown u|^q dxright)^{1/q}
end{eqnarray}

where $ q=frac{2d}{d+2} $. I think it is quite similar to the final result. But I cannot go further. Can you give me some hints or references?