Continuous function with bounded oscillation. [Gilbarg, Trudinger ch. 8] I have the following problem (Problem 8.5 from Gilbarg, Trudinger - "Ellipic PDEs of 2nd Order"):
Let $u\in C^1(B_R(0))$ where $B_R(0)\subset\mathbb{R}^2$ is the unit disc, and suppose $u$ satisfies$$\omega(r)=osc_{\partial B_r}u = \sup_{x,y\in\partial B_r}\vert u(x)-u(y)\vert$$ where $B_r = B_r(0)$ is the disc of radius $R$ centered at the origin. Define$$D(r) = \int_{B_r}\vert \nabla u(x)\vert^2dV(x).$$
If $\omega(r)$ is non-decreasing, show that for $0<r<R$ $$\omega(r)\leq\sqrt{\pi D(R)/\log(R/r)}.$$
I feel the answer is so obvious, yet I can't see it. Any hints?
 A: The proof below appears to give $\sqrt{\pi/2}$ instead of $\sqrt{\pi}$ (could be my oversight). I divided it into parts, so you can choose how much  to read and how much to do on your own.

The basic idea is that the oscillation of $u$ on a circle is controlled by the integral of $|\nabla u|$ over that circle. Namely, 
$$2\,\omega(r)\le \int_{\partial B_r}\vert \nabla u \vert \tag1$$
where $2$ comes from the fact that we have two ways to get from maximum to minimum of $u$ on the circle. 

Now we want to relate the integral in (1) to $D(R)$. Simply integrating over $r$ and using the Cauchy-Schwarz inequality would produce unnatural-looking result. The trick is to divide by the radius before integration. 
$$\begin{split}
\int_r^R \frac{d\rho}{\rho} \int_{\partial B_\rho}\vert \nabla u \vert 
&= \int_{r<|x|<R} \frac{1}{\rho} \vert \nabla u \vert 
\\&\le \sqrt{\int_{r<|x|<R} \frac{1}{\rho^2}}\sqrt{\int_{r<|x|<R}  \vert \nabla u \vert^2 }
\\&\le \sqrt{2\pi \log(R/r)} \sqrt{D(R)}
\end{split} \tag2$$

Doing the same on the left side of (1) yields, 
$$2 \int_r^R \omega(\rho)\frac{d\rho}{\rho} \le 2\,\omega(r)\int_r^R \frac{d\rho}{\rho}  
= 2\,\omega(r) \log(R/r) \tag3$$
where the monotonicity of $\omega$ is used at the first step.

The required inequality follows by combining (1), (2) and (3).
