Why does the following inequality holds for any weak solution $u\in C^1(B_1)$ of uniformly elliptic equation $D_i(a_{ij}D_ju)=0$? Now I'm studying "Elliptic Partial Differential Equations" by Q.Han and F. Lin. Throughout the section 5 of the chapter 1, $u\in C^1(B_1)$ is a weak solution of 
$$D_i(a_{ij}D_j u)=0$$
where $0<\lambda|\xi|^2\leq a_{ij}\xi_i\xi_j\leq \Lambda|\xi|^2$. In this setting, it says that for any $0<\rho<r\leq 1$,
$$\int_{B_\rho}u^2\leq c\left(\frac{\rho}{r}\right)^\mu\int_{B_r}u^2$$
where $\mu$ depends only on $n$, $\lambda$ and $\Lambda$. It is presented as a remark of the following lemma: if $u$ is such a weak solution, then we have
$$\int_{B_{R/2}}u^2\leq \theta\int_{B_R}u^2$$
where $\theta=\theta(n,\lambda,\Lambda)\in(0,1)$ and $0<R\leq1$. This lemma is straightforward by using the Poincare inequality and the Cacciopolli inequality. However, I have no idea how to get the first inequality from this lemma. (The author says it follows by iterating the result of this lemma.) Is there any one can help? 

Let $f(r)=\int_{B_r} u^2$. We can forget the whole PDE thing and just work with this nonnegative increasing function of $r$, which satisfies $$f(r/2)\le \theta f(r),\quad  0\le r\le 1\tag1$$ 
Given $0<\rho<r\le 1$, let $k$ be the largest integer such that $2^k\rho\le r$. (It's possible that $k=0$.) Applying the inequality (1) $k$ times, we find that 
$$f(2^{-k}r)\le \theta^k f(r)$$
By the monotonicity of $f$, 
$$f(\rho) \le \theta^k f(r)\tag2$$
It remains to relate $k$ to $\rho/r$.   The maximality of $k$ implies $2^{k+1}\rho>r$. Hence, $(1/2)^{k+1}<\rho/r$. Raising both sides to power $\mu = \log\theta/\log(1/2)$ yields $\theta^{k+1}<(\rho/r)^\mu$. Hence, (2) implies 
$$f(\rho) \le \theta^{-1} \left(\frac{\rho}{r}\right)^\mu f(r) $$
which has the required form.
